Frank Brothers Solutions for ICSE Class 9 Physics Chapter 3 Laws Of Motion

Page No: 113

Question 1. Define the term inertia.
Answer: The property by which a body neither changes its present state of rest or of uniform motion in a straight line nor tends to change the present state is known as inertia.
In simple words: Inertia is just the "laziness" of an object. If it is sitting still, it wants to stay still; if it is moving, it wants to keep moving in the same direction unless something pushes it.

📝 Teacher's Note: Use the example of a passenger in a bus to explain this. When the bus starts or stops, the body tries to maintain its previous state of motion or rest.

🎯 Exam Tip: Always mention both "state of rest" and "state of uniform motion" in your definition to get full marks.

Question 2. Give examples of inertia of rest and inertia of motion.
Answer: A book lying on a table will remain placed at table unless it is displaced by some external force. This is an example of inertia of rest.
A ball rolling on the ground will continue to roll unless the external force, the force of friction between the ball and the ground stops it.
In simple words: A book won't move by itself (rest), and a rolling ball wouldn't stop if there was no friction (motion).

📝 Teacher's Note: Friction is the "external force" that often hides the law of inertia of motion in our daily lives. Remind students that in space, the ball would roll forever.

🎯 Exam Tip: When giving examples, clearly state which type of inertia (rest or motion) is being demonstrated.

Question 3. How is the mass of an object related to its inertia?
Answer: The greater is the MASS, the greater is the inertia of the object.
In simple words: Heavier things are harder to start moving and harder to stop because they have more inertia.

📝 Teacher's Note: Ask students if it's easier to push an empty shopping cart or a full one. This practical example perfectly illustrates that mass is a measure of inertia.

🎯 Exam Tip: Remember this key phrase: "Mass is a measure of inertia."

Question 4. Describe the two kinds of inertia with examples.
Answer: An object possess two kind of inertia, inertia of rest and inertia of motion. A book lying on a table will remain placed at table unless it is displaced by some external force. This is an example of inertia of rest.
A ball rolling on the ground will continue to roll unless the external force, the force of friction between the ball and the ground stops it.
In simple words: There are two types: one makes things stay still, and the other makes things keep moving.

📝 Teacher's Note: This is a repetition of earlier concepts but emphasizes categorization. Use a coin-on-a-card trick to demonstrate inertia of rest.

🎯 Exam Tip: If asked for "kinds of inertia," make sure to list them first before providing the examples.

Question 5. Define 1 Newton force.
Answer: 1 Newton is the force which when applied to a body of unit mass produces a unit acceleration in it. 1 newton would produce acceleration of \( 1 \text{ ms}^{-2} \) in a mass of \( 1 \text{ kg} \).
In simple words: 1 Newton is the amount of push needed to make a 1 kg weight speed up by 1 meter per second every second.

📝 Teacher's Note: Relate this to the formula \( F = m \times a \). If \( m = 1 \) and \( a = 1 \), then \( F = 1 \).

🎯 Exam Tip: Always include the units (\( \text{kg} \) and \( \text{ms}^{-2} \)) when defining a unit of force.

Question 6. State the relationship between acceleration, force, and mass.
Answer: The acceleration produced by a force in an object is directly proportional to the applied FORCE and inversely proportional to the MASS of the object.
In simple words: If you push harder (more force), you go faster (more acceleration). If the object is heavier (more mass), it speeds up slower (less acceleration).

📝 Teacher's Note: This is Newton's Second Law. Use the formula \( a = \frac{F}{m} \) to show the direct and inverse relationships visually.

🎯 Exam Tip: Use the words "directly proportional" and "inversely proportional" to show a sophisticated understanding of the relationship.

Question 7. What is the SI unit of force?
Answer: SI unit of force is Newton (N).
In simple words: Scientists use "Newtons" to measure how hard something is pushed or pulled, just like they use "meters" for distance.

📝 Teacher's Note: Remind students that "Newton" is a derived unit, equivalent to \( \text{kg} \cdot \text{m/s}^2 \).

🎯 Exam Tip: Always write the symbol 'N' in capital letters as it is named after a scientist.

Question 8. Which physical quantity is associated with \( \text{N kg}^{-1} \)?
Answer: Acceleration is the physical quantity associated with \( \text{N kg}^{-1} \).
In simple words: \( \text{N kg}^{-1} \) is just another way of writing acceleration, which tells us how fast an object is changing its speed.

📝 Teacher's Note: Show the unit derivation: \( \frac{F}{m} = \frac{\text{N}}{\text{kg}} \). Since \( F/m = a \), then \( \text{N/kg} = \text{ms}^{-2} \).

🎯 Exam Tip: Units can sometimes be tricky. If you see \( \text{N kg}^{-1} \), equate it to \( \text{ms}^{-2} \) to simplify your thinking.

Question 9. What is the relation between Newton and Dyne?
Answer: \( 1 \text{ N} = 10^5 \text{ Dyne} \).
In simple words: One Newton is much bigger than a Dyne—it takes 100,000 Dynes to make just one Newton.

📝 Teacher's Note: Explain that Dyne is the CGS (Centimeter-Gram-Second) unit while Newton is the SI unit.

🎯 Exam Tip: Memorize the power of 10 (\( 10^5 \)) for conversion questions; it is a very common objective question.

Question 10. Which would require more force to stop: a sports car or a loaded van?
Answer: As mass of loaded van is greater than sports car so it would require more force to stop.
In simple words: Because the van is heavier, it has more "momentum" and "inertia," making it harder to bring to a halt.

📝 Teacher's Note: This connects mass back to the concept of force required to change motion. It's a practical application of \( F = ma \).

🎯 Exam Tip: Identify "mass" as the deciding factor in your explanation to score full marks.

Question 11. Calculate the acceleration of a body of mass 4 kg when a force of 12 N is applied.
Answer: We know force = mass \( \times \) acceleration.
\( a = F/m = 12 \text{ N} / 4 \text{ kg} = 3 \text{ ms}^{-2} \)
so acceleration of the body would be \( 3 \text{ ms}^{-2} \).
In simple words: If you push a 4 kg box with a strength of 12 Newtons, it will speed up by 3 meters per second every second.

📝 Teacher's Note: Encourage students to always write the formula first and then substitute the values with units.

🎯 Exam Tip: Don't forget to include the unit (\( \text{ms}^{-2} \)) in your final answer.

Question 12. Distinguish between the SI and CGS units of force.
Answer: SI unit of force is Newton whereas CGS unit of force is dyne.
\( 1 \text{ newton} / 1 \text{ dyne} = 10^5 \).
In simple words: Scientists have two different "rulers" for force: the big one is Newton (SI) and the tiny one is Dyne (CGS).

📝 Teacher's Note: CGS stands for Centimeter-Gram-Second. It's an older system but still used in specific scientific contexts.

🎯 Exam Tip: If a numerical problem uses grams and centimeters, convert them to kilograms and meters first to use Newtons, or use Dynes if requested.

Question 13. What is the SI unit of momentum?
Answer: SI unit of momentum is \( \text{kgms}^{-1} \).
In simple words: Momentum is measured by how much mass is moving at what speed, so we use "kilogram-meters per second."

📝 Teacher's Note: Momentum (\( p \)) is mass (\( m \)) times velocity (\( v \)). So the unit is just the unit of mass times the unit of velocity.

🎯 Exam Tip: Be careful with the power; it is \( s^{-1} \), not \( s^{-2} \). Don't confuse it with the unit of force.

Question 14. Define momentum.
Answer: Momentum is defined as the amount of motion contained in the body. It is given by the product of the mass of the body and its velocity.
In simple words: Momentum is "moving power." It tells you how hard it is to stop something based on how heavy it is and how fast it’s going.

📝 Teacher's Note: Explain that a small bullet has high momentum because of its speed, while a slow truck has high momentum because of its mass.

🎯 Exam Tip: Provide both the qualitative definition ("amount of motion") and the mathematical definition ("product of mass and velocity").

Question 15. Which physical quantity is associated with the motion of the body?
Answer: Momentum is the physical quantity associated with the motion of the body.
In simple words: When we want to measure how much "motion" an object has, we look at its momentum.

📝 Teacher's Note: While velocity describes motion, momentum describes the "oomph" behind that motion.

🎯 Exam Tip: This is a common one-word answer question. "Momentum" is the keyword.

Question 16. Which bodies possess momentum?
Answer: Momentum is possessed by bodies in MOTION.
In simple words: Only things that are moving have momentum. If something is standing still, its momentum is zero.

📝 Teacher's Note: Use the formula \( p = mv \). If velocity (\( v \)) is zero, then momentum (\( p \)) must be zero.

🎯 Exam Tip: Capitalizing "MOTION" helps emphasize that movement is the essential requirement.

Question 17. Which has more momentum: a slow pitched soft ball or a fast pitched soft ball?
Answer: A fast pitched soft ball has more momentum.
In simple words: Since both balls have the same weight, the one going faster has more "punch" or momentum.

📝 Teacher's Note: This illustrates that for a constant mass, momentum is directly proportional to velocity.

🎯 Exam Tip: Mention that since mass is equal, higher velocity leads to higher momentum.

Question 18. What are the units of momentum in SI and CGS systems and their ratio?
Answer: SI unit of momentum is \( \text{kgms}^{-1} \) and CGS unit of momentum is \( \text{g cms}^{-1} \).
And their ratio is = \( 1000 \times 100 \text{ gms}^{-1} = 1:10^5 \).
In simple words: The SI unit is much larger than the CGS unit because a kilogram and a meter are much bigger than a gram and a centimeter.

📝 Teacher's Note: To find the ratio, convert \( \text{kg} \rightarrow \text{g} \) (\( 10^3 \)) and \( \text{m} \rightarrow \text{cm} \) (\( 10^2 \)). Multiplying these gives \( 10^5 \).

🎯 Exam Tip: Show the step-by-step conversion of units to ensure you don't miss any zeros in the ratio.

Question 19. Does a body at rest have momentum? Why?
Answer: A body at rest has zero momentum as its velocity is zero.
In simple words: No, because if you aren't moving, you don't have any "motion" to measure!

📝 Teacher's Note: Emphasize the math: \( p = m \times 0 = 0 \).

🎯 Exam Tip: Always give the reason (velocity is zero) along with the answer (zero momentum).

Question 20. State Newton’s third law of motion.
Answer: According to Newton’s third law, for every action there is always an equal and opposite reaction.
In simple words: If you push a wall, the wall pushes you back just as hard but in the opposite direction.

📝 Teacher's Note: Use the "balloon rocket" or "skating" analogy to make this law feel real to the students.

🎯 Exam Tip: Use the keywords "equal" and "opposite" to get full credit for this definition.

Question 21. What is meant by 'action'?
Answer: When a force acts on a body then this is called an action.
In simple words: "Action" is just the name we give to the first force being applied.

📝 Teacher's Note: Action and reaction are just two sides of a single interaction. We call one "action" and the response "reaction."

🎯 Exam Tip: Remember that action and reaction are simply forces.

Question 22. Can action and reaction act on the same body?
Answer: No, action and reaction never act on a same body they always act simultaneously on two different bodies.
In simple words: If I kick a ball, the action is on the ball, but the reaction is on my foot. Two different things!

📝 Teacher's Note: This is a common point of confusion. If they acted on the same body, they would cancel out, and nothing would ever move!

🎯 Exam Tip: Stress the phrase "two different bodies" in your answer.

Question 23. Which law of motion defines force?
Answer: 2nd law of motion gives the definition of force.
In simple words: The second law tells us exactly how to calculate force using mass and acceleration.

📝 Teacher's Note: Actually, the 1st law provides a qualitative definition (force as an agent of change), while the 2nd law provides the quantitative definition (\( F=ma \)).

🎯 Exam Tip: If the question asks for the mathematical definition, specify the Second Law.

Question 24. Which law explains why a swimmer pushes water backward to move forward?
Answer: Newton’s third law explains this statement.
In simple words: The swimmer pushes the water back (action), and the water pushes the swimmer forward (reaction).

📝 Teacher's Note: This is a classic example of action-reaction pairs in fluid dynamics.

🎯 Exam Tip: Identifying the specific law by name (Newton's Third Law) is essential here.

Question 25. Is force a scalar or vector quantity?
Answer: Force is a vector quantity.
In simple words: It matters which way you push! Pushing left is different from pushing right, so force has direction.

📝 Teacher's Note: Explain that vector quantities have both magnitude (strength) and direction.

🎯 Exam Tip: Always remember that direction is key for force, making it a vector.

Question 26. What does it mean if the net force on an object is zero?
Answer: This means these forces are balanced forces.
In simple words: It’s like a tug-of-war where both sides are pulling equally hard—nobody moves because the forces cancel out.

📝 Teacher's Note: Balanced forces do not change an object's state of motion; they can only change its shape (like squeezing a sponge).

🎯 Exam Tip: Use the term "balanced forces" when the resultant force is zero.

Question 27. Why do passengers fall sideways when a bus takes a sharp turn?
Answer: Passengers tend to fall sideways when the bus takes a sharp turn due to the inertia of direction.
In simple words: Your body wants to keep going straight, but the bus turns. Because your body tries to stay on its original straight path, you feel like you are being thrown to the side.

📝 Teacher's Note: This is a specific type of inertia: inertia of direction. The body resists a change in the direction of motion.

🎯 Exam Tip: Mention "inertia of direction" specifically to show you understand the different types of inertia.

Question 28. Why are passengers thrown forward when a running bus stops suddenly?
Answer: Passengers are thrown in the forward direction as the running bus stops suddenly because due to their inertia of motion, their upper body continues to be in the state of motion even though the lower body comes to rest when the bus stops.
In simple words: Your feet stop because they are touching the floor, but your top half keeps "flying" forward because it has inertia.

📝 Teacher's Note: Explain that friction between the seat and the passenger stops the lower body, while the upper body lacks that direct contact and continues forward.

🎯 Exam Tip: Contrast the "lower body" (which stops) with the "upper body" (which stays in motion) for a complete answer.

Question 29. Why do passengers fall backward when a bus starts suddenly?
Answer: Passengers tends to fall in backward direction when bus starts suddenly because due to their inertia of rest, as soon as the bus starts, their lower body comes in motion but the upper body continues to be in the state of rest.
In simple words: The bus pulls your feet forward, but your head wants to stay right where it was, so it feels like you're falling back.

📝 Teacher's Note: This is the opposite of the previous question, demonstrating inertia of rest.

🎯 Exam Tip: Use the phrase "state of rest" to describe the upper body's initial condition.

Question 30. Can internal forces change the velocity of a body?
Answer: No, internal forces cannot change the velocity of a body.
In simple words: You can't lift yourself up by your own shoelaces! You need an outside force (like the ground) to move.

📝 Teacher's Note: Internal forces occur in pairs within a system and cancel out (Third Law). Only an external unbalanced force can cause acceleration.

🎯 Exam Tip: "External force" is the requirement for changing velocity—internal forces never count.

Question 31. Explain why dust comes out of a carpet when it is beaten with a stick.
Answer: When a hanging carpet is beaten using a stick, the dust particles will start coming out of the carpet because the part of the carpet where the stick strikes, immediately comes in motion while the dust particle sticking to the carpet remains at rest. Hence a part of the carpet moves ahead along with the stick, and the dust particles fall down due to the earth’s pull.
In simple words: You hit the carpet and it moves away fast, but the dust is "lazy" (inertia of rest) and stays in place. Since the carpet isn't there to hold it anymore, it just drops to the floor.

📝 Teacher's Note: This is a great real-world demo of inertia of rest. The dust doesn't "fly out"; the carpet "moves away" from it.

🎯 Exam Tip: Mention "inertia of rest" specifically for the dust particles.

Question 32. Why do fruits and leaves fall when branches of a tree are shaken?
Answer: When we shake the branches of a tree, the fruits and leaves remain in state of rest while branches comes in motion so fruits and leaves are detached from the tree.
In simple words: The branch moves back and forth, but the heavy fruit wants to stay still. The snapping motion breaks the stem holding them together.

📝 Teacher's Note: This is again an application of inertia of rest. The branch moves, but the fruit/leaf tries to stay put.

🎯 Exam Tip: The key concept here is "inertia of rest" for the fruits/leaves.

Question 33. Which body requires more force: a 10g mass accelerating at \( 5 \text{ cm/s}^2 \) or a 20g mass accelerating at \( 2 \text{ cm/s}^2 \)?
Answer: We know force = mass \( \times \) acceleration
\( F_1 = 10 \times 5 = 50 \text{ dyne} \).
\( F_2 = 20 \times 2 = 40 \text{ dyne} \).
So first body require more force.
In simple words: Even though the second object is heavier, the first one is speeding up so much faster that it actually needs more push.

📝 Teacher's Note: This problem uses CGS units (grams and cm/\( \text{s}^2 \)). Ensure students recognize that the result is in "dynes."

🎯 Exam Tip: Always show the calculation for both cases before stating which one is greater.

Question 34. Find the force acting on a coin using the given velocity-time graph.
Answer: As we know that slope of velocity time graph gives acceleration so acceleration of the coin is \( = (v - u)/t = (0 - 24)/8 = -3 \text{ ms}^{-2} \).
And force = mass \( \times \) acceleration
Force \( = 20/1000 \times -3/100 \text{ N} \).
Force \( = -6 \times 10^{-4} \text{ N} \).
In simple words: The graph shows the coin is slowing down. By looking at the steepness of the line, we find how fast it's slowing, then multiply by its weight to find the force of friction.

📝 Teacher's Note: The negative sign indicates a retarding force (friction). Note the unit conversion: cm to m and g to kg.

🎯 Exam Tip: Be very careful with unit conversions (\( \text{cm} \rightarrow \text{m} \) and \( \text{g} \rightarrow \text{kg} \)) when working with graphs and Newton units.

Page No: 114

Question 35. Calculate the distance covered by an object starting from rest with an acceleration of \( 8 \text{ ms}^{-2} \) in 5 seconds.
Answer: Initial velocity of the object (\( u \)) \( = 0 \text{ ms}^{-1} \)
Acceleration of the object (\( a \)) \( = 8 \text{ ms}^{-2} \).
Time (\( t \)) \( = 5 \text{ s} \).
Distance covered would be \( S = ut + 1/2 \text{ at}^2 \).
\( S = 1/2 \times 8 \times 5 \times 5 = 100 \text{ m} \).
In simple words: If you start from zero and speed up very quickly for 5 seconds, you'll travel exactly 100 meters.

📝 Teacher's Note: Since \( u = 0 \), the term \( ut \) becomes zero, simplifying the calculation to \( 0.5 \text{ at}^2 \).

🎯 Exam Tip: Always state your "givens" (u, a, t) before plugging them into the equation.

Question 36. A 5-metric ton truck covers 100m in 10s starting from rest. Find the force acting on it.
Answer: Initial velocity of the truck \( = 0 \text{ ms}^{-1} \)
Distance covered by truck \( = 100 \text{ m} \)
Time taken to cover this distance \( = 10 \text{ s} \).
We know Distance covered would be \( S = ut + 1/2 \text{ at}^2 \).
\( 100 = 1/2 \times a \times 100 \)
\( a = 2 \text{ ms}^{-2} \).
Mass of truck = 5 metric tons \( = 5000 \text{ kg} \).
Force acted on truck = mass \( \times \) acceleration
Force \( = 5000 \times 2 = 10000 \text{ N} \).
In simple words: First, we use the distance and time to find out how much the truck was speeding up. Then we multiply that by the truck's huge weight to find the total force of the engine.

📝 Teacher's Note: Explain that 1 metric ton \( = 1000 \text{ kg} \). Correct units are vital for calculating Newtons.

🎯 Exam Tip: This is a two-step problem. Don't stop after finding acceleration; make sure you finish by finding the force.

Question 37. What is the use of momentum in physics?
Answer: Momentum is used for quantifying the motion of body.
In simple words: It gives us a specific number to describe how much "moving power" an object has.

📝 Teacher's Note: It combines mass and velocity into a single useful quantity for understanding collisions and impacts.

🎯 Exam Tip: Use the word "quantifying" to describe how momentum measures motion.

Question 38. Explain why a gun recoils when a bullet is fired.
Answer: When we fire a gun, a force is exerted in the forward direction as the bullets comes out; in reaction to which an equal and opposite force is act in the backward direction and hence, we feel a backward jerk on the shoulder.
In simple words: The gun pushes the bullet forward, and because of Newton's Third Law, the bullet pushes the gun back into your shoulder.

📝 Teacher's Note: This is a perfect demonstration of action-reaction pairs. The forces are equal, but the accelerations differ because the gun is much heavier than the bullet.

🎯 Exam Tip: Mention the "equal and opposite force" as the reason for the "backward jerk."

Question 39. How does a person swim according to the laws of motion?
Answer: A person applies force on water in backward direction and water according to third law of motion water apply an equal and opposite force in forward direction which helps a person to swim.
In simple words: You push the water back, so the water pushes you forward.

📝 Teacher's Note: This shows that you can't move yourself without pushing against something else.

🎯 Exam Tip: Explicitly mention Newton's Third Law when explaining biological movements like swimming or walking.

Question 40. Which law is involved in the working of a jet plane?
Answer: Newton’s third law of motion is involved in the working of a jet plane.
In simple words: The jet engine shoots hot air out the back, so the air pushes the plane forward.

📝 Teacher's Note: It's the same principle as the recoil of a gun—the exhaust gases are the "bullet" and the plane is the "gun."

🎯 Exam Tip: If asked for the principle of a rocket or jet, "Newton's Third Law" or "Conservation of Momentum" are both correct.

Question 41. Can a rocket propel itself in a vacuum?
Answer: Yes, a rocket can propel itself in a vacuum once it is given initial velocity.
In simple words: Yes! A rocket doesn't need air to "push against." It works by throwing fuel out the back, which pushes the rocket forward even in empty space.

📝 Teacher's Note: Many students think rockets need air to push against. Clarify that the reaction comes from the exhaust gases being expelled, not from the surrounding atmosphere.

🎯 Exam Tip: Remember that rockets operate on internal reaction forces, not external atmospheric support.

Question 42. If action and reaction are equal and opposite, why do objects move?
Answer: Action is equal and opposite to reaction but they act on different bodies and object moves as movement requires an unbalanced force and these are provided once inertia is overcome.
In simple words: They don't cancel out because they happen to different things. If I push a car, the car moves because it feels my push. I feel the car pushing back on me, but that doesn't stop the car from moving!

📝 Teacher's Note: This is the most crucial concept in the Third Law. Emphasize that for a body to move, we only look at the forces acting *on* that body.

🎯 Exam Tip: The phrase "they act on different bodies" is the single most important part of this answer.

Page No: 125

Question 1. Who stated the law of gravitation?
Answer: Sir Isaac Newton stated the law of gravitation.
In simple words: Isaac Newton was the scientist who figured out how gravity works for everything from falling apples to spinning planets.

📝 Teacher's Note: Use the famous (likely legendary) apple story to introduce the concept of universal gravity.

🎯 Exam Tip: Remember the full name "Sir Isaac Newton" for formal science answers.

Question 2. State the Universal Law of Gravitation.
Answer: Every object in the universe attracts every other object with a force directly proportional to the product of their masses and inversely proportional to the square of distance between them.
In simple words: Everything pulls on everything else. Heavier things pull harder, and things that are farther away pull much weaker.

📝 Teacher's Note: Introduce the formula \( F = G \frac{m_1 m_2}{r^2} \) alongside this verbal definition.

🎯 Exam Tip: Use the words "directly proportional" for mass and "inversely proportional" for the square of distance.

Question 3. Distinguish between gravity and gravitation.
Answer: Gravity is the force of attraction between the object and the earth whereas gravitation refers to the force of attraction that exists between any two bodies that possess mass.
In simple words: Gravitation is the general pull between any two things (like two magnets). "Gravity" is the special name we use when one of those things is the Earth.

📝 Teacher's Note: Gravitation is the universal phenomenon; gravity is the local Earth-based case of it.

🎯 Exam Tip: Gravity always involves the Earth as one of the two objects.

Question 4. What is meant by acceleration due to gravity?
Answer: Acceleration due to gravity is the acceleration experienced by a body during free fall.
In simple words: It’s how much a falling object speeds up every second just because the Earth is pulling it down.

📝 Teacher's Note: On Earth, this value (\( g \)) is approximately \( 9.8 \text{ ms}^{-2} \).

🎯 Exam Tip: Mention "free fall" to correctly context the acceleration.

Question 5. What is the formula for 'g'?
Answer: \( g = GM/R^2 \).
In simple words: This formula tells us how strong gravity is on a planet based on the planet's mass and how big it is.

📝 Teacher's Note: Here \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

🎯 Exam Tip: Ensure students know that \( g \) depends on the planet's mass and radius, but NOT on the mass of the falling object.

Question 6. Describe the properties of the universal constant of gravitation 'G'.
Answer: We know that law of gravitation is \( F = G ( m_1 \times m_2)/R^2 \).
Here \( G \) is universal constant and is called constant of gravitation. It does not depend upon the value of \( m_1 \), \( m_2 \) or \( R \).
Its value is same between any two objects in the universe.
In simple words: 'G' is a special number that never changes, no matter where you are in space or what objects you are looking at.

📝 Teacher's Note: Distinguish between \( g \) (which changes on different planets) and \( G \) (which is universal and constant everywhere).

🎯 Exam Tip: Emphasize that \( G \) is a "constant"—it does not change with location or mass.

Question 7. What is the SI unit of the constant of gravitation?
Answer: SI unit of constant of gravitation is \( \text{Nm}^2\text{kg}^{-2} \).
In simple words: This complex unit is just what scientists use to make the math balance out when calculating gravitational force.

📝 Teacher's Note: Derive the unit from \( G = \frac{Fr^2}{m_1 m_2} \rightarrow \frac{\text{N} \cdot \text{m}^2}{\text{kg} \cdot \text{kg}} \).

🎯 Exam Tip: Memorize this unit carefully; the powers (2 and -2) are frequently tested.

Question 8. How does the gravitational force change with distance?
Answer: We know that law of gravitation is \( F = G ( m_1 \times m_2)/R^2 \).

• If distance between them is halved then put \( R = R/2 \).
  \( F = 4 \times G( m_1 \times m_2)/ R^2 \).
  \( F_1 = 4 \text{ F} \).
• If distance between them is doubled then put \( R = 2\text{R} \).
  \( F = G( m_1 \times m_2)/ 4\text{R}^2 \).
  \( F_1 = F/4 \).
• If distance between them is made four times then put \( R = 4\text{R} \).
  \( F = G( m_1 \times m_2)/16 \text{ R}^2 \).
  \( F_1 = F/16 \).
• If distance between them is infinite then put \( R = \text{infinite} \).
  \( F = G( m_1 \times m_2)/ R^2 \).
  \( F_1 = 0 \).
• If distance between them is almost zero then put \( R = 0 \).
  \( F = G( m_1 \times m_2)/ 0 \).
  \( F_1 = \text{infinite} \).

In simple words: Gravity gets much weaker as you move away. If you double the distance, gravity is 4 times weaker. If you go half as far, gravity is 4 times stronger!

📝 Teacher's Note: This is the "Inverse Square Law." Use a grid or a light bulb analogy to explain how intensity drops off with distance.

🎯 Exam Tip: Be prepared for numericals asking "What happens to force if distance is doubled/tripled?" Squaring the change in distance and putting it in the denominator is the shortcut.

Question 9. In what direction do objects attract each other?
Answer: All objects in the universe attract each other along the line joining their CENTRES.
In simple words: Gravity always pulls in a straight line toward the very center of an object.

📝 Teacher's Note: This is why gravity is called a "central force." It always acts through the center of mass.

🎯 Exam Tip: "Line joining their centers" is the technical phrase examiners look for.

Question 10. What is the force of attraction between two material objects called?
Answer: The force of attraction between any two material objects is called FORCE OF GRAVITATION.
In simple words: Any two things with weight pull on each other, and we call that pull gravitation.

📝 Teacher's Note: Emphasize that *every* object with mass exerts this force, even people, though the force is too small to feel.

🎯 Exam Tip: Use the term "gravitation" generally and "gravity" specifically for Earth.

Question 11. What is the gravitational force of the Earth called?
Answer: The gravitational force of the earth is called earth’s GRAVITY.
In simple words: Earth’s pull has its own special name: Gravity.

📝 Teacher's Note: This reinforces the distinction made in earlier solutions.

🎯 Exam Tip: Gravity is a "particular case" of gravitation.

Question 12. Is gravity a unique force or a type of gravitation?
Answer: Gravity is a particular case of GRAVITATIONAL FORCE OF EARTH.
In simple words: Gravity is just gravitation, but it’s what we call it when the Earth is doing the pulling.

📝 Teacher's Note: Scientifically, they are the same force, but terminology differs based on the objects involved.

🎯 Exam Tip: Don't get confused—gravity *is* a gravitational force.

Question 13. What is the general value of 'G'?
Answer: The value of \( G \) is extremely SMALL.
In simple words: The number for \( G \) is so tiny (\( 0.0000000000667 \)) that you only feel gravity when at least one of the objects is as big as a planet.

📝 Teacher's Note: Since \( G \) is \( 6.67 \times 10^{-11} \), the product \( G \times m_1 \times m_2 \) stays small unless the masses are huge.

🎯 Exam Tip: If asked why we don't feel gravity between two people, the answer is "the value of G is extremely small."

Question 14. Does the law of gravitation apply to the sun and moon?
Answer: Yes the law of gravitation is also applicable in case of the sun and moon.
In simple words: Yes! The same rule that makes an apple fall makes the moon orbit the Earth and the Earth orbit the sun.

📝 Teacher's Note: This is why the law is called "Universal." It applies to everything in space just as it applies on Earth.

🎯 Exam Tip: The word "Universal" means there are no exceptions in the entire universe.

Question 15. Calculate the force of gravity acting on a 100 kg person.
Answer: We know that law of gravitation is \( F = G ( m_1 \times m_2)/R^2 \).
Mass of earth \( = 6 \times 10^{24} \text{ kg} \).
Mass of the person \( = 100 \text{ kg} \).
\( G = 6.7 \times 10^{-11} \text{ Nm}^2\text{kg}^{-2} \).
Radius of earth \( = 6.4 \times 10^6 \text{ m} \).
\( F = (6.7 \times 10^{-11} \times 100 \times 6 \times 10^{24} )/ (6.4 \times 6.4 \times 10^{12}) = 981.4 \text{ N} \)
Force of gravity due to earth acting on a 100 kg person is \( 981.4 \text{ N} \).
In simple words: By using the weight of the Earth and its size, we can calculate that a 100 kg person is pulled down with a force of about 981 Newtons.

📝 Teacher's Note: Note that the radius used in the OCR (\( 10^{14} \)) is a typo; it should be \( 6.4 \times 10^6 \text{ meters} \) (or \( 6400 \text{ km} \)). The final result of \( \sim 980 \text{ N} \) is correct for a \( 100 \text{ kg} \) mass.

🎯 Exam Tip: \( 980 \text{ N} \) is the standard weight for \( 100 \text{ kg} \) on Earth (\( 100 \times 9.8 \)). Use this as a quick check for your calculation.

Question 16. Why do objects fall towards the Earth?
Answer: Objects fall towards the earth due to force of gravitation.
In simple words: The Earth is huge and pulls everything toward its center.

📝 Teacher's Note: Gravity is an attractive force that acts between any object and the Earth.

🎯 Exam Tip: This is a fundamental concept; identify the specific force (gravitation/gravity).

Question 17. Why do we not feel the gravitational force between two persons?
Answer: Because the masses of persons are not large enough to overcome the value of small constant of gravitation so the force of gravitation is very small and negligible to feel.
In simple words: Gravity is a very weak force. Unless you are as big as a planet, the pull between you and another person is so tiny that it's impossible to notice.

📝 Teacher's Note: Remind students that the value of \( G \) is \( 6.67 \times 10^{-11} \), which is why we only feel gravity when one of the masses (like Earth) is huge.

🎯 Exam Tip: Use the term "negligible" to describe forces that are too small to have a measurable effect.

Question 18. A ball is thrown vertically upwards with a speed of \( 4.9 \text{ ms}^{-1} \). Calculate its maximum height and the time taken to reach the ground.
Answer: Initial speed of ball is \( = 4.9 \text{ ms}^{-1} \).
Acceleration due to gravity \( = -9.8 \text{ ms}^{-2} \).

• We know \( v^2 - u^2 = 2as \)
  At highest point final velocity is zero so
  \( 0 - 4.9 \times 4.9 = 2 \times (-9.8) S \)
  \( \implies S = 1.125 \text{ m} \)
• We know \( v = u + at \)
  \( 0 = 4.9 - 9.8 t \)
  \( \implies T = 0.5 \text{ sec.} \)
• for highest point initial velocity is zero
  Acceleration due to gravity is \( = 9.8 \text{ ms}^{-2} \).
  Final velocity at ground is \( v \)
  \( V^2 - 0 = 2 \times 9.8 \times 1.125 \)
  \( \implies V = 4.9 \text{ ms}^{-1} \).

Time taken to reach ground from highest point
\( V = u + at \)
\( 4.9 = 0 + 9.8 t \)
\( \implies T = 4.9/9.8 = 0.5 \text{ sec.} \)
So time of ascent is equal to time descent.
In simple words: When you throw a ball up, it slows down until it stops at the top, then speeds back up as it falls. It takes exactly the same amount of time to go up as it does to fall back down.

📝 Teacher's Note: This problem is a classic example of motion under gravity. Point out that acceleration is negative going up and positive coming down.

🎯 Exam Tip: Remember that final velocity at the highest point is always zero. This is a key "hidden" value in physics problems.

Question 19. Write the formula for acceleration due to gravity.
Answer: \( g = GM/R^2 \).
In simple words: This formula tells us how strong gravity is on a planet based on the planet's mass and how big it is.

📝 Teacher's Note: Explain that \( G \) is a universal constant, but \( g \) changes depending on which planet you are on.

🎯 Exam Tip: Capital \( M \) is for the planet's mass and \( R \) is for the planet's radius. Don't mix them up with the object's mass.

Question 20. What is the value of \( g \) at the surface of the Earth?
Answer: Value of the \( g \) at the surface of the earth is \( 9.8 \text{ ms}^{-2} \).
In simple words: Anything you drop on Earth will speed up by about 10 meters per second every second as it falls.

📝 Teacher's Note: While \( 9.8 \) is the standard, mention that it varies slightly depending on where you are on Earth (poles vs equator).

🎯 Exam Tip: Always include the units (\( \text{ms}^{-2} \)) for acceleration to get full credit.

Question 21. How do mass and weight change on different planets?
Answer: Mass of the body is constant at all positions so mass will not change. But weight will change as gravity on the surface of earth is almost 6 times than on the surface of the moon, so its weight will increase almost 6 times on the surface of earth.
In simple words: Your mass is the "stuff" you are made of, which never changes. Your weight is how hard a planet pulls on you, so you'd feel lighter on the moon but stay the same size.

📝 Teacher's Note: Use the "suitcase" analogy: the number of clothes inside is mass, but how heavy the bag feels to lift is weight.

🎯 Exam Tip: Examiners love to ask if mass changes on the moon. The answer is always "No"! Only weight changes.

Question 22. Where would you weigh more, Earth or Moon?
Answer: We will weigh more on the surface of the earth.
In simple words: Earth is much heavier than the moon, so it pulls you down with more force, making you weigh more here.

📝 Teacher's Note: Gravity depends on mass. Since Earth has much more mass than the moon, its gravity is stronger.

🎯 Exam Tip: Weight is directly proportional to the acceleration due to gravity (\( g \)).

Question 23. Name the instrument used to measure mass.
Answer: Beam balance is used to measure the mass of a body.
In simple words: A beam balance compares two weights, so it doesn't matter how strong gravity is; it just measures the "stuff" inside.

📝 Teacher's Note: Contrast this with a spring scale. A beam balance works even on the moon because gravity affects both sides equally.

🎯 Exam Tip: Associate "Beam Balance" with "Mass" and "Spring Scale" with "Weight".

Question 24. Name the instrument used to measure weight.
Answer: Spring scale is used to measure the weight of a body.
In simple words: A spring scale measures how much gravity stretches a spring. If gravity is weak, the spring won't stretch as much.

📝 Teacher's Note: This is a common point of confusion. Weight is a force, so we use a scale that measures force (tension in a spring).

🎯 Exam Tip: Remember that weight is measured in Newtons, which is what a spring scale typically reads.

Question 25. Where is weight greater: at the poles or the equator?
Answer: The weight is greater at the poles than the equator.
In simple words: Earth isn't a perfect ball; it's a bit flat at the top and bottom. At the poles, you are closer to the center of the Earth, so gravity pulls on you harder.

📝 Teacher's Note: Earth is an oblate spheroid. The distance \( R \) is smaller at the poles, and since \( g = GM/R^2 \), a smaller \( R \) means a larger \( g \).

🎯 Exam Tip: If the question asks "why," explain that the Earth's radius is shorter at the poles.

Question 26. What is the relationship between Newton and kgwt?
Answer: Newton \( 1 \text{N} = 9.8 \text{ kgwt} \).
In simple words: This is a way to convert between scientific units of force (Newtons) and everyday units of weight (kilogram-weight).

📝 Teacher's Note: Usually, we say \( 1 \text{ kgwt} = 9.8 \text{ N} \). Ensure students understand that kgwt is a unit of force, not just mass.

🎯 Exam Tip: Check the conversion factor carefully; multiplying by \( 9.8 \) is usually the key.

Question 27. Why do we weigh more on the Earth's surface compared to high altitudes?
Answer: We will weigh more on earth surface as value of \( g \) is greater on earth surface.
In simple words: The closer you are to the center of the Earth, the stronger the pull. When you go up high, you are farther away, so gravity gets weaker.

📝 Teacher's Note: This demonstrates the inverse relationship between gravitational force and distance (\( R \)).

🎯 Exam Tip: Higher altitude means larger distance from Earth's center, which leads to lower gravity.

Question 28. Does gravitational force depend on the medium between objects?
Answer: No, the force of gravitation between two objects does not depend on the medium between them.
In simple words: Gravity doesn't care if there is water, air, or a wall between two objects; the pull stays exactly the same.

📝 Teacher's Note: This is a unique property of gravity compared to electric or magnetic forces, which *do* depend on the medium.

🎯 Exam Tip: Use "medium independence" as a keyword to describe this property of gravitation.

Question 29. If both masses are doubled and the distance between them is also doubled, what happens to the gravitational force?
Answer: we know that law of gravitation.
\( F = G ( m_1 \times m_2)/R^2 \).
Now \( m_1 = 2 m_1 \)
\( m_2 = 2 m_2 \)
\( R = 2 R \)
\( \implies F_1 = G ( 2m_1 \times 2m_2)/(2R)^2 \).
\( \implies F_1 = F \)
So force between them remains same.
In simple words: Doubling the masses makes the pull stronger, but doubling the distance makes it weaker. In this case, the two changes perfectly cancel each other out!

📝 Teacher's Note: This is a great exercise in proportional reasoning. The \( 2 \times 2 \) in the numerator is cancelled by the \( 2^2 \) in the denominator.

🎯 Exam Tip: Don't forget to square the distance when it changes. Many students forget that \( (2R)^2 \) becomes \( 4R^2 \).

Exercise Page No: 126

Question 30. Do all freely falling bodies have the same force acting on them?
Answer: Yes, in absence of gravity all freely falling body have same force acting on them.
In simple words: This means that without other things like air getting in the way, everything falls at the exact same rate.

📝 Teacher's Note: Actually, the force is different (Force = mass \( \times \) g), but the acceleration is the same for all objects in free fall. The OCR text here is a bit misleading—clarify that "acceleration" is what is identical.

🎯 Exam Tip: Remember Galileo's experiment: a feather and a hammer fall together in a vacuum.

Question 31. Explain how \( g \) depends on mass and distance.
Answer: \( g = GM/R^2 \)
it means acceleration due to gravity is directly proportional to the mass of body and inversely proportional to the square of distance between earth and object.
In simple words: More mass means a stronger pull, but more distance means a much weaker pull.

📝 Teacher's Note: Be careful—here \( M \) is the mass of the Earth, not the object. The object's own mass does not affect how fast it falls.

🎯 Exam Tip: "Inversely proportional to the square" means if you triple the distance, the gravity becomes 9 times weaker.

Question 35. Does a freely falling body have constant acceleration?
Answer: Yes a body falling freely near the earth surface has a constant acceleration.
In simple words: As long as you are close to the ground, gravity pulls on you with a steady, unchanging strength.

📝 Teacher's Note: Near the surface, we assume \( g = 9.8 \text{ ms}^{-2} \) is constant. This allows us to use standard equations of motion.

🎯 Exam Tip: In "free fall" problems, you can always assume acceleration is constant and equal to \( g \).

Question 36. Where would \( g \) be maximum: at the poles or the equator?
Answer: As we know
\( g = 1/R^2 \)
so value of \( g \) is more at poles than equator so value of \( g \) is maximum near a camp site in Antarctica as this lie on the pole.
In simple words: Because the Earth bulges at the middle, the poles are closer to the center. Being closer means you feel more gravity.

📝 Teacher's Note: Use a drawing of Earth as a "squashed ball" to show that the polar radius is shorter than the equatorial radius.

🎯 Exam Tip: Mentioning Antarctica is a clever way to answer questions about the poles in real-world scenarios.

Question 37. At what height \( H \) is the value of \( g \) half that of the Earth's surface?
Answer: As we know
\( g = GM/R^2 \)
So let at height \( H \) the value of \( g \) is half that of the earth surface
then \( g \) at \( R + H \) would be equal to
\( g' = GM/(R+H)^2 \).
Now \( g'/g = 1/2 \)
\( 1/2 = R^2 / (R + H)^2 \)
\( (R + H)^2 = 2 R^2 \)
\( R + H = \sqrt{2} R \)
\( \implies H = (\sqrt{2} - 1) R \).
In simple words: To feel only half the gravity of Earth, you would have to fly up to a height that is about 41% of the Earth's radius (around 2,600 kilometers high).

📝 Teacher's Note: This involves taking the square root of both sides. It's a standard derivation to show how gravity drops off.

🎯 Exam Tip: The final answer \( H = (\sqrt{2}-1)R \) is the most simplified form and scores full marks.

Exercise Page No: 128

Question 1. Define force.
Answer: Force is that external agency which tends to change the state of rest or the state of motion of a body.
In simple words: A force is a push or a pull. It's what makes things start moving, stop moving, or change direction.

📝 Teacher's Note: Remind students that a force doesn't *always* cause movement; it "tends" to change it (like pushing against a heavy wall).

🎯 Exam Tip: Mention both "state of rest" and "state of motion" in your definition.

Question 2. Define 1 Newton force.
Answer: 1 Newton is the force which when applied to a body of unit mass produces a unit acceleration in it.
In simple words: 1 Newton is the amount of push needed to make a 1 kg block speed up by 1 meter per second every second.

📝 Teacher's Note: This is based on the formula \( F = m \times a \). If mass is 1 and acceleration is 1, force is 1.

🎯 Exam Tip: "Unit mass" means 1 kg and "unit acceleration" means \( 1 \text{ ms}^{-2} \).

Question 3. What are the SI and CGS units of force? Give their relationship.
Answer: Newton is the SI unit of force whereas dyne is the CGS unit of force.
\( 1 \text{ N} = 10^5 \text{ dyne} \).
In simple words: Scientists use Newtons for big forces and Dynes for tiny ones. 1 Newton is equal to 100,000 Dynes.

📝 Teacher's Note: CGS stands for Centimeter-Gram-Second. It's an older system but still common in some science labs.

🎯 Exam Tip: Memorize the power of 10 (\( 10^5 \)) as it often appears in objective questions.

Question 4. Is force a vector quantity?
Answer: No, force is a vector quantity.
In simple words: (Correction: Force IS a vector). This means direction matters—pushing something left is very different from pushing it right!

📝 Teacher's Note: The OCR says "No," but force is definitely a vector. Explain that a vector has both a size (magnitude) and a direction.

🎯 Exam Tip: Always specify the direction when writing down a force value.

Question 5. What can a force do to an object?
Answer: A force can produce MOTION in an object at rest. It can ACCELERATE an object and can change its DIRECTION of motion.
In simple words: Force can start things, speed them up, or turn them around.

📝 Teacher's Note: Ask students for examples: hitting a ball (starting motion), pressing gas (accelerating), turning a steering wheel (direction).

🎯 Exam Tip: List at least three distinct effects of force to get full marks on this question.

Question 6. List four common effects of force.
Answer:

• force changes the shape of skin.
• force produces stretching in the rubber.
• force provides retardation to the car and finally stops the car.
• force decreases the momentum of ball and finally stops the ball.

In simple words: Force can squish things, stretch things, slow things down, or bring them to a complete stop.

📝 Teacher's Note: These examples cover changes in shape (rubber) and changes in motion (car/ball).

🎯 Exam Tip: "Retardation" is just another word for slowing down or negative acceleration.

Question 7. Does every force produce motion?
Answer: No, every force does not produce motion in every type of body.
In simple words: If you push a giant brick wall, you are using force, but the wall isn't going anywhere!

📝 Teacher's Note: This happens when the applied force is balanced by a reaction force (like static friction or the wall's structure).

🎯 Exam Tip: Movement only happens if there is an *unbalanced* force.

Question 8. What does inertia depend on?
Answer: The amount of inertia of a body depends on its MASS.
In simple words: Heavier things are harder to move or stop because they have more inertia.

📝 Teacher's Note: Mass is the quantitative measure of inertia. The more matter an object has, the more it resists changes in its motion.

🎯 Exam Tip: Remember: Mass = Inertia. They are directly linked.

Question 9. How can you change the direction of a moving object?
Answer: You can change the direction in which an object is moving by APPLYING FORCE ON IT.
In simple words: If a ball is rolling toward you, you have to push or kick it to make it go a different way.

📝 Teacher's Note: This relates back to the First Law—an object will go in a straight line forever unless a force makes it turn.

🎯 Exam Tip: Mention that an external unbalanced force is required to change direction.

Question 10. Why does a person in a moving car have inertia?
Answer: A man riding on a car has INERTIA of motion.
In simple words: Because your body is moving at the same speed as the car, it wants to keep moving even if the car stops suddenly.

📝 Teacher's Note: This is why seatbelts are so important—they provide the external force to stop your body's inertia.

🎯 Exam Tip: Specifically name it "Inertia of Motion" for moving objects.

Question 11. Define inertia of rest.
Answer: When a body is at rest, it will continue to remain at rest unless some external force is applied to change its state of rest. This property of body is called inertia of rest.
In simple words: It’s the "laziness" of objects. If something is sitting still, it won't move until something else pushes it.

📝 Teacher's Note: Use the example of a coin sitting on a card; if you flick the card, the coin stays in place due to inertia of rest.

🎯 Exam Tip: "External force" is a required phrase in any definition involving inertia.

Question 12. Give examples of action and reaction pairs.
Answer:

• Weight of the book is action and normal force applied by table on book is reaction.
• Force applied by man on ground is action and force of friction is the reaction.
• Force applied by hammer on nail is action and normal force applied by nail on hammer is reaction to this force.
• Firing of bullet is the action and recoiling of gun is the reaction.
• Force applied by us on wall is action and opposite force applied by wall on us or we can say that resistance of wall to our force is reaction.

In simple words: For every push, there is a push back. If you lean on a wall, the wall pushes back on you exactly as hard.

📝 Teacher's Note: These are applications of Newton's Third Law. Emphasize that action and reaction act on *different* bodies.

🎯 Exam Tip: When identifying pairs, always state which force is the action and which is the reaction.

Question 13. Compare inertia of rest and inertia of motion with examples.
Answer: A book lying on a table will remain placed on table unless it is displaced by some external force. This is an example of inertia of rest.
A ball rolling on the ground will continue to roll unless the external force, the force of friction between the ball and the ground stops it. This is an example of inertia of motion.
In simple words: Inertia of rest makes things stay still; inertia of motion makes moving things keep going.

📝 Teacher's Note: These both stem from Newton's First Law. Friction is often the "hidden" force that stops inertia of motion on Earth.

🎯 Exam Tip: Be sure to distinguish clearly between "rest" and "motion" in your examples.

Question 14. What causes motion in a body?
Answer: Unbalance external force causes motion in the body.
In simple words: If two people push a box equally from opposite sides, nothing moves. It only moves if one person pushes harder than the other.

📝 Teacher's Note: Balanced forces result in zero acceleration. Only an *unbalanced* net force creates motion.

🎯 Exam Tip: Use the word "unbalanced" to show the force is not cancelled out.

Question 15. Define linear momentum.
Answer: Linear Momentum is defined as the amount of motion contained in a body. It is given by the product of the mass of the body and its velocity.
In simple words: Think of it as "oomph." A big truck moving slowly and a small car moving fast can have the same momentum.

📝 Teacher's Note: Momentum (\( p \)) is a vector quantity. Its formula is \( p = m \times v \).

🎯 Exam Tip: Momentum combines both mass and velocity; don't mention one without the other.

Question 16. What is the SI unit of momentum?
Answer: SI unit of momentum is \( \text{kgms}^{-1} \).
In simple words: It’s a combined unit: kilograms for how heavy the object is, and meters per second for how fast it’s going.

📝 Teacher's Note: This unit is derived directly from the formula \( p = mv \), where \( m \) is in \( \text{kg} \) and \( v \) is in \( \text{m/s} \).

🎯 Exam Tip: Pay attention to the negative exponent: \( \text{s}^{-1} \) is correct for units of speed/velocity.

Question 17. How does Newton's first law define force?
Answer: According to Newton’s first law force is that external agency which tends to change the state of rest or the state of motion of a body.
In simple words: The first law says that if something is changing its motion, a force must be the reason why.

📝 Teacher's Note: This is the qualitative definition of force. The second law provides the quantitative (math) definition.

🎯 Exam Tip: Link the first law directly to the concept of "changing state".

Question 18. State Newton's first law of motion.
Answer: According to Newton’s first law, everybody continues in its state of rest or in uniform motion in a straight line unless compelled by some external force to act otherwise.
In simple words: Things keep doing what they are already doing unless they are forced to change.

📝 Teacher's Note: This is also known as the Law of Inertia. It explains what happens when the net force on an object is zero.

🎯 Exam Tip: Don't forget the phrase "in a straight line"—direction is part of the law!

Exercise Page No: 129

Question 19. Which has maximum inertia: a car, a bicycle, or a truck?
Answer: Out of all these, as mass of truck is greatest and mass is measure of inertia so a truck has maximum inertia.
In simple words: The truck is the heaviest, so it's the hardest to start moving or bring to a stop.

📝 Teacher's Note: Remind students that inertia is directly proportional to mass. Larger mass = larger inertia.

🎯 Exam Tip: Identify mass as the deciding factor in your explanation.

Question 20. Why is it advantageous to run before taking a long jump?
Answer: It is advantageous to run before taking a long jump because after running we get motion of inertia which helps in long jumping.
In simple words: When you run first, your body builds up "inertia of motion." This means your body wants to keep moving forward even after you jump, helping you land further away.

📝 Teacher's Note: This is a classic example of inertia of motion. Use the analogy of a car that takes time to stop even after the brakes are applied to explain how the body carries its state of motion into the jump.

🎯 Exam Tip: The key phrase to use is "inertia of motion." Examiners look specifically for this technical term in your explanation.

Question 21. Why does a ball moving on a table top stop eventually?
Answer: Ball moving on a table top stops eventually due to force of friction between the ball surface and table surface.
In simple words: Even though the table looks smooth, it has tiny bumps that rub against the ball. This rubbing is called friction, and it pushes against the ball until it stops.

📝 Teacher's Note: Explain that according to Newton's First Law, the ball would roll forever if no external force acted on it. Friction is that external force.

🎯 Exam Tip: Always mention that friction acts in the opposite direction to the motion of the object.

Question 22. Define force in terms of momentum.
Answer: Force is equal to the rate of change of linear momentum.
In simple words: Force is just a measure of how quickly you can change how much "moving power" (momentum) an object has.

📝 Teacher's Note: Introduce the formula \( F = \frac{\Delta p}{\Delta t} \) to show that force is the derivative of momentum with respect to time.

🎯 Exam Tip: If the question asks for the definition according to the second law, this "rate of change of momentum" statement is the most accurate answer.

Question 23. State Newton's second law of motion and explain if the first law is contained within it.
Answer: According to newton second law of motion, when a force acts on a body, the rate of change in momentum of a body is equal to the product of mass of the body and acceleration produced in it.
Yes, Newton’s first law is contained in the second law as if force is zero then acceleration would be zero which means body would remain in its state of rest or in state of constant motion.
In simple words: The second law tells us that force equals mass times acceleration (\( F = ma \)). If the force (\( F \)) is zero, then the acceleration (\( a \)) must be zero, which is exactly what the first law says!

📝 Teacher's Note: This is a higher-level conceptual question. Show that if \( F = 0 \), then \( ma = 0 \), which implies \( a = 0 \). Since acceleration is change in velocity, zero acceleration means no change in velocity.

🎯 Exam Tip: To prove the first law is in the second, start with the formula \( F = ma \) and set \( F = 0 \). This logical step-by-step approach earns full marks.

Question 24. Define 1 Newton and its relation to dyne.
Answer: 1 Newton is the force which when applied to a body of unit mass produces a unit acceleration in it.
1 newton / 1dyne = \( 10^5 \).
In simple words: 1 Newton is the amount of push needed to make a 1 kg object speed up by 1 meter per second every second. It is much stronger than a dyne—it takes 100,000 dynes to equal just 1 Newton.

📝 Teacher's Note: "Unit mass" means \( 1 \text{ kg} \) and "unit acceleration" means \( 1 \text{ ms}^{-2} \). Use the SI vs. CGS comparison to explain why the units have different names.

🎯 Exam Tip: Memorize the conversion factor \( 10^5 \) as it is frequently asked in objective and numerical questions.

Question 25. Show the mathematical derivation of the relation between Newton and Dyne.
Answer: 1 newton = \( 1 \text{ kg} \times 1 \text{ ms}^{-1} = 1000 \text{ g} \times 100 \text{ cms}^{-1} = 10^5 \text{ cms}^{-1} \).
1 dyne \( 1 \text{ g} \times 1 \text{ cms}^{-1} = 1 \text{ cms}^{-1} \).
So 1 newton = \( 10^5 \text{ dyne} \).
In simple words: Since a kilogram is 1,000 grams and a meter is 100 centimeters, multiplying them (\( 1000 \times 100 \)) gives us 100,000. That's why one Newton is 100,000 Dynes.

📝 Teacher's Note: The units in the original text (ms-1) are typically used for velocity, but in the context of force definitions, they represent acceleration (\( \text{ms}^{-2} \)). Clarify this distinction if students get confused by the notation.

🎯 Exam Tip: When showing this derivation, clearly label the conversion of \( \text{kg} \rightarrow \text{g} \) and \( \text{m} \rightarrow \text{cm} \).

Question 26. Will a body move if two equal and opposite forces act on it?
Answer: No, the body will not move as the two forces are equal and opposite and they constitute balanced forces.
In simple words: Imagine a tug-of-war where both sides pull exactly as hard. The rope stays in the middle because the forces cancel each other out.

📝 Teacher's Note: This situation results in a net force of zero. Remind students that an "unbalanced" force is required to change the state of motion.

🎯 Exam Tip: Use the term "balanced forces" to describe why there is no movement.

Question 27. What is the effect of balanced forces on the motion of a body?
Answer: As these forces are balanced so they will not affect the motion and motion of the body will remain unaffected.
In simple words: Balanced forces don't change anything. If the object was sitting still, it stays still. If it was moving, it keeps moving at the same speed.

📝 Teacher's Note: Use the example of a book on a table. The downward pull of gravity and the upward push of the table are balanced, so the book stays put.

🎯 Exam Tip: Clearly state that balanced forces result in "zero acceleration."

Question 28. Explain the principle of a rocket's motion.
Answer: According to Newton’s third law, for every action there is always an equal and opposite reaction. Rocket works on the same principle. The exhaust gases produced as the result of the combustion of the fuel are forced out at one end of the rocket. As a reaction , the main rocket moves in the opposite direction.
In simple words: A rocket works like a balloon you let go of. The air (gas) shoots out the bottom (action), and this push makes the rocket fly up (reaction).

📝 Teacher's Note: Emphasize that the rocket does not "push against the air" but works due to the internal reaction between the expelled fuel and the rocket body.

🎯 Exam Tip: Always mention "Newton's Third Law" and the "action-reaction pair" (gases vs. rocket) to score full marks.

Question 29. If a boy pushes a wall with a force of 30 N, what is the force exerted by the wall on the boy?
Answer: According to Newton’s third law, every action has equal and opposite reaction so force exerted by the wall on the boy is 30 N.
In simple words: If you push a wall with 30 N, the wall pushes you back with exactly 30 N. It's like a mirror for forces!

📝 Teacher's Note: This is a direct application of the third law. It helps students understand that forces always exist in pairs.

🎯 Exam Tip: The magnitude of the force remains the same, but the direction is opposite.

Question 30. Who stated the law of inertia?
Answer: Newton stated the law of inertia.
In simple words: Isaac Newton was the scientist who explained why things keep doing what they are already doing (staying still or moving).

📝 Teacher's Note: While Galileo first conceived the idea, Newton formalized it as his First Law of Motion.

🎯 Exam Tip: Link "Law of Inertia" directly to "Newton's First Law."

Question 31. State the universal law of gravitation.
Answer: Every object in the universe attracts every other object with a force directly proportional to the product of their masses and inversely proportional to the square of distance between them.
Law of gravitation is called universal because it applies to all bodies of universe.
In simple words: Everything in space pulls on everything else. Heavier things pull harder, and things that are closer together pull harder too. It works for an apple on Earth just like it works for planets in space.

📝 Teacher's Note: Write the formula \( F = \frac{G m_1 m_2}{r^2} \) on the board while explaining this. Explain why "G" is called the Universal Gravitational Constant.

🎯 Exam Tip: Use the words "directly proportional" for mass and "inversely proportional" for distance to demonstrate a clear scientific understanding.

Question 32. Distinguish between gravity and gravitation.
Answer: Gravity is the force of attraction between the object and the earth whereas gravitation refers to the force of attraction that exists between any two bodies that possess mass.
In simple words: Gravitation is the general name for the pull between any two things (like two magnets). Gravity is the special name we use when the pull is specifically from the Earth.

📝 Teacher's Note: Gravity is just a local case of universal gravitation. All gravity is gravitation, but not all gravitation is gravity.

🎯 Exam Tip: Clearly state that gravity always involves the Earth as one of the two objects.

Question 33. Why would a person weigh more at Delhi than at Shimla?
Answer: Person will weigh more at Delhi as we know that gravity decreases with increase in height. Now as Shimla is at a height from Delhi so weight is less in Shimla and more in Delhi.
In simple words: Gravity gets a little bit weaker the higher up you go from the Earth's center. Since Shimla is high up in the mountains, the Earth's pull is slightly weaker there than in flat Delhi.

📝 Teacher's Note: Weight is \( m \times g \). Since \( g \) (acceleration due to gravity) decreases with altitude, weight also decreases as you go higher.

🎯 Exam Tip: The key relationship is that gravity is inversely proportional to altitude (height from Earth's center).

Question 34. Name the instrument used to measure the weight of a body.
Answer: Spring scale is used to measure the weight of a body.
In simple words: A spring scale measures how much gravity stretches a spring. The more weight, the more the spring stretches!

📝 Teacher's Note: Distinguish between a beam balance (used for mass) and a spring scale (used for weight/force).

🎯 Exam Tip: Remember that a spring scale measures the *force* of gravity acting on an object.

Question 35. Explain the importance of gravity in the universe.
Answer: Gravity is another kind of FORCE. It exerts all through the UNIVERSE. The sun’s gravity keeps the PLANETS in their orbits. Gravity can only be felt with very large MASS.
In simple words: Gravity is like the glue of the universe. It keeps the Earth spinning around the Sun and holds the planets in their place. We only really notice it with huge things like stars and planets.

📝 Teacher's Note: Explain that without gravity, planets would fly off into space in straight lines. Gravity provides the centripetal force for orbital motion.

🎯 Exam Tip: Use the word "orbit" to explain how gravity controls the motion of planets.

Question 36. Provide three pieces of evidence for the existence of gravitational force.
Answer:
(i) Objects fall on the earth due to gravitational force between the earth and object.
(ii) Atmosphere doesnot escape because molecules of atmosphere are attracted by earth due to gravitational force of earth.
(iii) A moon rocket needs to reach a certain velocity because during its motion earth attracts the rocket towards it by its gravitational force.
In simple words: We know gravity exists because: things drop when you let them go, we have air to breathe because Earth holds it down, and rockets have to work hard to "break away" from Earth's pull.

📝 Teacher's Note: Use the atmosphere example to explain why the moon has no air (too little gravity to hold it). This makes the concept more concrete.

🎯 Exam Tip: Listing multiple varied examples (falling objects, atmosphere, rockets) shows a thorough understanding of the concept.

Question 37. Distinguish between 'g' and 'G'.
Answer: ‘g’ is acceleration due to earth’s gravity and ‘G’ is universal gravitational constant.
In simple words: 'g' is how fast things fall on Earth (\( 9.8 \text{ ms}^{-2} \)), but 'G' is a special number that never changes anywhere in the whole universe.

📝 Teacher's Note: Explain that \( g \) varies by location (like on the Moon or atop Everest), whereas \( G \) is a fundamental constant (\( 6.67 \times 10^{-11} \text{ Nm}^2\text{kg}^{-2} \)).

🎯 Exam Tip: Never use the two symbols interchangeably. 'g' is a variable acceleration, while 'G' is a universal constant.

Question 38. What is meant by free fall?
Answer: Free fall means motion of a body under the gravity of earth only.
In simple words: Free fall is when you drop something and only gravity is pulling it down. There is no engine or air pushing it another way.

📝 Teacher's Note: In ideal free fall calculations, we ignore air resistance. All objects in free fall accelerate at the same rate, regardless of their mass.

🎯 Exam Tip: The keyword is "only gravity." If other forces like air resistance are mentioned, it is not pure free fall.

Question 39. Is there a gravitational force between us and a book? Why don't we feel it?
Answer: Yes, we have a gravitational force of attraction between us and a book. But our mass is very small so the force between us and book is very small almost negligible.
In simple words: Gravity works for everything with weight, so you and your book are pulling on each other! But you are both so small compared to a planet that the pull is too weak to notice.

📝 Teacher's Note: Revisit the formula \( F = \frac{G m_1 m_2}{r^2} \). Since \( G \) is very tiny (\( 10^{-11} \)), the force only becomes noticeable when at least one mass is very large (like Earth).

🎯 Exam Tip: Use the word "negligible" to describe forces that are too small to cause visible motion.

Question 40. Does Earth's gravity affect the Moon?
Answer: Yes, the force of gravitation of earth affects the motion of moon, because moon is revolving around earth and centripetal force for this revolution is provided by earth’s gravitation.
In simple words: Yes, Earth's gravity pulls on the Moon like an invisible string. This pull (centripetal force) keeps the Moon spinning in a circle around us instead of flying off into space.

📝 Teacher's Note: This is a great way to link gravity with circular motion. Explain that gravity acts as the centripetal force in orbits.

🎯 Exam Tip: Mention "centripetal force" to explain *how* gravity keeps the moon in orbit.

Question 41. Distinguish between inertial mass and gravitational mass.
Answer: Inertial mass is measure of inertia of the object. According to second law of motion \( F = m \times a \)
\( m = F/a \) and this mass is called as inertial mass.
Newton law of gravitation gives another definition of mass.
\( F = \frac{G m_1 m_2}{R^2} \)
Thus \( m_2 \) is the mass of the body by which another body of mass \( m_1 \) attracts it towards it by law of gravitation. This mass is called gravitational mass.
In simple words: Inertial mass is how hard it is to push something. Gravitational mass is how hard something pulls on other things with gravity. Even though they come from different rules, they are actually the same amount for any object!

📝 Teacher's Note: This is a sophisticated physics concept. Inertial mass relates to resistance to motion (\( F=ma \)), while gravitational mass relates to gravitational pull (\( F=mg \)).

🎯 Exam Tip: If asked for a definition, relate inertial mass to the Second Law and gravitational mass to the Law of Gravitation.

Question 42. Summarize the key concepts of gravitation and gravity.
Answer: Newton law of gravitation is that Every object in the universe attracts every other object with a force directly proportional to the product of their masses and inversely proportional to the square of distance between them.

• Gravity is the force of attraction between the object and the earth whereas gravitation refers to the force of attraction that exists between any two bodies that possess mass.
• ‘\( g \)’ is acceleration due to earth’s gravity and ‘\( G \)’ is universal gravitational constant.

In simple words: Gravitation is the general pull between any two objects. Gravity is the Earth's specific pull. \( G \) is a constant number for the whole universe, but \( g \) is how fast things fall on a specific planet.

📝 Teacher's Note: This summary helps students keep the four main terms (\( G \), \( g \), gravity, gravitation) distinct in their minds.

🎯 Exam Tip: Never confuse \( g \) (small letter, local acceleration) with \( G \) (capital letter, universal constant).

Question 43. Does an apple attract the Earth? If so, why does the Earth not move?
Answer: Yes, it is true that apple attracts the earth towards it with same force but the mass of earth is so huge that acceleration produced in it due to this force is very much small and negligible to notice.
In simple words: The pull is equal on both sides! But because the Earth is so giant and heavy, that tiny pull doesn't make it budge, while the same pull makes the tiny apple fall fast.

📝 Teacher's Note: This is a perfect example of Newton's Third Law combined with the Second Law (\( a = F/m \)). Equal force (\( F \)) but huge mass (\( m \)) means tiny acceleration (\( a \)).

🎯 Exam Tip: Mention that the force is "the same" but the "mass is huge".

Question 44. Explain how \( g \) decreases with higher altitudes.
Answer: we know that law of gravitation is
\( F = G ( m_1 \times m_2)/R^2 \)
Here the \( F \) is force due to attraction and this force is equal to weight of the body \( m_2g \).
So \( m_2g = G ( m_1 \times m_2)/R^2 \)
\( \implies g = (G \times m_2)/R^2 \).
Here \( R \) is the distance between earth centre and the object centre. Now if we go on higher altitude say ‘\( H \)’ then this \( R \) would increase to \( (R + H) \)
And value of gravity at height \( H \) becomes
\( g’ = (G \times m_2)/( R +H)^2 \).
As denominator increases so \( g’ \) would be less than \( g \) and hence we can say that gravity decreases on higher altitudes.
In simple words: Gravity gets weaker the farther you are from Earth's center. When you go high up, you are increasing the distance, which makes the pull drop.

📝 Teacher's Note: Use the math to show that because \( R \) is on the bottom of the fraction (denominator), making it bigger makes the whole answer smaller.

🎯 Exam Tip: Focus on the increase in distance from the center of mass as the reason for the decrease in \( g \).

Question 45. Find the force exerted by a 20N block on its support and the tension in the string.
Answer:

• The force exerted by the block on is the weight of box and that is equal to \( 20 \text{N} \).
• The force exerted by string on block is equal to the tension in the string and this is also equal to the \( 20 \text{N} \).

In simple words: If a box is hanging still, the string has to pull up exactly as hard as the box's weight pulls down. So, both forces are the same.

📝 Teacher's Note: This is an example of balanced forces in equilibrium. If the box isn't moving, the net force is zero.

🎯 Exam Tip: In a static system, weight and tension are usually equal and opposite.

Question 46. Two bodies of mass are tied together and a force of 4N is applied. If they individually accelerate at \( 8 \text{ ms}^{-2} \) and \( 20 \text{ ms}^{-2} \) under 4N, what is their acceleration together?
Answer: we know \( F = m \times a \)
\( \implies m = F/a \)
so we can calculate mass of each body
Mass of body 1 \( m_1 = 4/8 = 0.5 \text{ kg} \).
Mass of body 2 \( m_2 = 4/20 = 0.2 \text{ kg} \).
Total mass when two masses are tied together \( M = 0.5 + 0.2 = 0.7 \text{ kg} \).
Now as force is acting on total mass so acceleration produced is
\( a = 4/0.7 = 5.71 \text{ ms}^{-2} \).
In simple words: First, we figure out how heavy each object is by seeing how fast they move alone. Then we add their weights together and see how fast the whole heavy group moves when pushed with the same force.

📝 Teacher's Note: This is a multi-step problem. Remind students to find the masses first, as mass is an additive property, but acceleration is not.

🎯 Exam Tip: Always calculate individual masses before trying to find the combined acceleration. You cannot just average the accelerations!

Page No: 130

Question 47. From the given speed-time graph, calculate the acceleration and the force acting on a body of mass 100 g.
Answer: As we know slope of speed time graph gives acceleration so we can find acceleration of body
\( a = (v - u)/t = (10 - 0)/(4 - 0) = 5/2 = 2.5 \text{ ms}^{-2} \).
Mass of the body is \( 100 \text{g} = 0.1 \text{ kg} \).
force = mass \( \times \) acceleration.
Force \( = 0.1 \times 2.5 = 0.25 \text{ N} \).
In simple words: Acceleration is how much the speed increases every second. From the graph, it increases by 2.5 each second. Multiplying this by the weight of the object (in kg) gives us the force.

📝 Teacher's Note: Remind students that the slope of a speed-time graph represents acceleration. Always convert mass to kilograms (SI unit) before calculating force in Newtons.

🎯 Exam Tip: To get full marks, show the step of converting grams to kilograms. Use the coordinates from the graph to show how you calculated the slope.

Question 48. The initial speed of a body is \( 5 \text{ ms}^{-1} \). It acquires a speed of \( 8 \text{ ms}^{-1} \) in 2s. If the force applied is 0.9 N, find the mass of the body in grams.
Answer: Initial speed of body \( = 5 \text{ ms}^{-1} \)
Final speed of body \( = 8 \text{ ms}^{-1} \)
Time taken to acquire this speed \( = 2 \text{s} \).
Acceleration of body \( = (v - u)/t \)
\( a = (8 - 5)/2 = 1.5 \text{ ms}^{-2} \).
Force applied on body \( = 0.9 \text{ N} \).
we know \( F = m \times a \).
\( m = f/a = 0.9/1.5 = 0.6 \text{ kg} \)
mass of the body is \( 600 \text{ gm} \).
In simple words: First, we find out how fast the object is speeding up (acceleration). Then, using the push (force), we can figure out how heavy it is. Since the question asks for grams, we convert our answer from kg at the end.

📝 Teacher's Note: This problem connects kinematics (\( v, u, t \)) with dynamics (\( F, m \)). Ensure students can transition between \( \text{kg} \) and \( \text{g} \) correctly by multiplying by 1000.

🎯 Exam Tip: Read the question carefully to see which unit the final answer should be in. If it says "grams", failing to convert from \( 0.6 \text{ kg} \) to \( 600 \text{ g} \) will lose you marks.

Question 49. Based on the following table, describe the motion of the object.

Answer: So object moves with increasing acceleration.
In simple words: By looking at the acceleration column, we can see the numbers are getting bigger (1, 3, 5). This means the object is speeding up faster and faster every second.

📝 Teacher's Note: This is an example of non-uniform acceleration. It is important for students to recognize that if the change in velocity per unit time is not constant, the acceleration itself is changing.

🎯 Exam Tip: To prove acceleration is increasing, show the calculation for at least two different time intervals from the table.

Question 50. Define impulsive force.
Answer: The force that acts on a body for a very short time but produces a large change in its momentum, is known as impulsive force.
In simple words: Imagine a quick "hit," like a bat hitting a ball. It's a very strong push that happens in a split second and makes the ball fly away fast.

📝 Teacher's Note: Use everyday examples like kicking a football or hitting a nail with a hammer to illustrate that "short time" doesn't mean "weak force."

🎯 Exam Tip: The two keywords examiners look for are "very short time" and "large change in momentum." Make sure to include both.

Question 51. A body of mass 20 kg is initially at rest. A force of 100 N acts on it. How long will it take to reach a final velocity of \( 100 \text{ ms}^{-1} \)?
Answer: initial velocity of body \( = 0 \text{ ms}^{-1} \).
Final velocity of body \( = 100 \text{ ms}^{-1} \).
Mass of body \( = 20 \text{ kg} \).
Force applied \( = 100 \text{N} \).
We know that
\( F \times t = m (v – u) \)
\( 100 t = 20 (100 - 0) \)

\( \implies T = 2000/100 = 20 \text{ s} \).
In simple words: We use the rule that the push (Force) times Time equals the total change in motion (Momentum). By plugging in our numbers, we find it takes 20 seconds for the object to reach that high speed.

📝 Teacher's Note: This formula is derived from Newton's Second Law (\( F = ma \)). It is often called the Impulse-Momentum Theorem.

🎯 Exam Tip: Always state your "givens" (initial velocity, mass, etc.) at the start of your answer. This makes the math easier to follow and helps earn partial credit.

Question 52. What is the SI unit of retardation?
Answer: SI unit of retardation is \( \text{ms}^{-2} \).
In simple words: Retardation is just "slowing down." Since it is still a type of acceleration, it uses the same units.

📝 Teacher's Note: Retardation is negative acceleration. Emphasize that the units remain identical whether an object is speeding up or slowing down.

🎯 Exam Tip: Do not use a negative sign in front of the units. The word "retardation" itself implies that the value of the acceleration is negative.

Question 53. How is the force applied to a body related to its mass and acceleration?
Answer: Force applied is equal to the product of mass and acceleration produced in the body.
\( F = \text{mass} \times \text{acceleration} \).
In simple words: To find out how hard you are pushing something, you just multiply how heavy it is by how fast it is speeding up.

📝 Teacher's Note: This is the fundamental equation of classical mechanics (\( F = ma \)). It shows that for a constant mass, force and acceleration are directly proportional.

🎯 Exam Tip: Be ready to rearrange this formula to find mass (\( m = F/a \)) or acceleration (\( a = F/m \)).

Question 54. A force of 0.02 N acts on a body of mass 400 g. Calculate the acceleration produced.
Answer: According to Newton’s second law of motion, when a force acts on a body, the rate of change in momentum of a body equals the product of mass of the body and acceleration produced in it due to that force, provided the mass remains constant.
Mass of body \( = 400 \text{ g} = 0.4 \text{kg} \)
Force applied on body \( = 0.02 \text{ N} \)
Acceleration \( = \text{force/mass} = 0.02/0.4 = 0.05 \text{ ms}^{-2} \).
In simple words: First, convert the weight of the object to kilograms. Then, divide the push (Force) by the weight to see how much it speeds up.

📝 Teacher's Note: Converting grams to kilograms (\( 400 \text{ g} = 0.4 \text{ kg} \)) is the most common mistake students make. Remind them that Force in Newtons requires Mass in Kilograms.

🎯 Exam Tip: Always write the formula you are using before doing the calculation. Even if you make a calculation error, you will get marks for the method.

Question 55. Define linear momentum. If a force produces an acceleration of \( 10 \text{ ms}^{-2} \) on a 1 kg body, find the acceleration produced by the same force on a 4 kg body.
Answer: Linear Momentum is defined as the physical quantity which is associated with bodies in linear motion. It is given by the product of the mass of the body and its velocity.
Mass of body \( = 1 \text{ kg} \)
Acceleration produced \( = 10 \text{ ms}^{-2} \).
Force applied would be \( = 1 \times 10 \text{ N} = 10 \text{ N} \).
Mass of second body \( = 4 \text{ kg} \).
As same force has to be applied on second body so force \( = 10 \text{N} \).
Acceleration produced is \( = F/M = 10/4 = 2.5 \text{ ms}^{-2} \).
In simple words: Momentum is just a measure of how much an object is moving. For the second part: if you push a heavy thing (4 kg) with the same strength you used for a light thing (1 kg), it will speed up 4 times slower.

📝 Teacher's Note: This problem illustrates the inverse relationship between mass and acceleration for a constant force. As mass increases, acceleration decreases proportionately.

🎯 Exam Tip: When a question says "same force," calculate the force from the first set of data first, then use that number for the second part.

Question 56. Two bodies P and Q have masses m and 2m respectively, and velocities 2v and v respectively. Find the ratio of: (i) inertia, (ii) momentum, (iii) force required to stop them in same time.
Answer: Mass of P is \( m_1 = m \).
Velocity of P is \( v_1 = 2 v \).
Mass of Q is \( m_2 = 2 m \).
Velocity of Q is \( v_2 = v \).

(i) inertia of P/inertia of Q \( = m_1/m_2 = 1/2 \).
So ratio of inertia of two bodies is 1:2.

(ii) Momentum of P/momentum of Q \( = m_1v_1 / m_2v_2 = 1 \).
So ratio of momentum of two bodies is 1:1.

(iii) As force required to stop them is equal to change in their momentum from moving to rest.
So ratio would be same as the ratio of their momentum i.e 1: 1.
In simple words: (i) Inertia is just mass, so the heavier one has twice the inertia. (ii) Since P is twice as fast and Q is twice as heavy, their total moving power (momentum) balances out to be exactly the same. (iii) Because they have the same momentum, you need the same amount of force to stop them.

📝 Teacher's Note: Use this example to show that inertia only depends on mass, while momentum depends on both mass and velocity. This is a very common conceptual comparison in physics.

🎯 Exam Tip: For ratios, always simplify to the lowest terms (e.g., use 1:2 instead of 2:4). Clearly label each sub-part of your answer.

Question 57. Prove that force is equal to the product of mass and acceleration.
Answer: According to newton second law
\( F = m \times a \)
\( a = (v – u)/t \).

\( \implies F = m(v -u)/t \)

\( \implies F = (mv – mu)/t \)
As \( F = m \times a \)
\( ma = (mv – mu)/t \)
so rate of change of momentum = mass \( \times \) acceleration.
This relation holds good when mass remains constant during motion.
In simple words: We start with the idea that force changes an object's velocity over time. By looking at how the momentum (mass times speed) changes every second, we see that it works out to just mass times acceleration.

📝 Teacher's Note: This derivation shows how Newton's Second Law (\( F = ma \)) is a specific case of the "Rate of Change of Momentum" law when mass is constant.

🎯 Exam Tip: Mention the condition "when mass remains constant" to show a complete understanding. If mass were changing (like in a rocket), the formula would be different.

Question 58. State the Law of Conservation of Momentum and prove it using a diagram.
Answer: Conservation of momentum in case of a collision between two bodies means the total momentum before and after collision remains unchanged or conserved, provided no net force acts on the system.
Consider two bodies A and B having masses \( m_1 \) and \( m_2 \) and initial velocities \( u_1 \) and \( u_2 \) respectively. The bodies collide head on with each other and their collision lasts for t seconds. Suppose the velocities of A and B after collision are \( v_1 \) and \( v_2 \) respectively. Assume that no external forces are acting on the bodies.
Rate of change of momentum of ball \( A = m_1(v_1 - u_1)/t \).
Rate of change of momentum of ball \( B = m_2(v_2 - u_2)/t \).
If \( F_{AB} \) is the force exerted by A on B and \( F_{BA} \) is the force exerted by B on A, we can write
\( F_{AB} = m_1(v_1 - u_1)/t \).
\( F_{BA} = m_2(v_2 - u_2)/t \).

\( \implies F_{AB} = - F_{BA} \)

\( \implies m_1(v_1 - u_1)/t = - m_2(v_2 - u_2)/t \).

\( \implies m_1(v_1 - u_1) = - m_2(v_2 - u_2) \).

\( \implies m_1 v_1 + m_2 v_2 = m_1 u_1 + m_2 u_2 \).
So total momentum after collision = total momentum before collision.
This proves conservation of momentum during collision.
In simple words: When two objects crash, the total "moving energy" (momentum) they had before they hit is exactly the same as the total they have after they hit. One might slow down, but the other will speed up by the same amount to keep the total the same.

📝 Teacher's Note: This proof uses Newton's Third Law (\( F_{AB} = -F_{BA} \)). Emphasize that for the conservation to hold, the system must be isolated from external forces like friction.

🎯 Exam Tip: This is a high-weightage question. Memorizing the final equation \( m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 \) and being able to explain that the negative sign comes from the opposite direction of forces is key.

Question 59. Explain Newton's third law using a balloon as an example.
Answer: According to newton third law, for every action there is always an equal and opposite reaction.
To demonstrate newton third law blow a balloon and hold its neck tightly facing downwards. When we release the balloon, the balloon will moves up instead of falling to the ground. As air is releasing from bottom of balloon and this air apply equal and opposite force to the balloon and this force helps balloon to move upwards.
In simple words: When you let go of a balloon, the air shooting out the bottom pushes the balloon up. The air going down is the "action," and the balloon going up is the "reaction."

📝 Teacher's Note: This is a great interactive demonstration for the classroom. It shows that motion is produced by the reaction force acting *on* the object (the balloon).

🎯 Exam Tip: Clearly identify the action (air being expelled) and the reaction (force on the balloon moving it forward/upward) to demonstrate a full understanding of the law.

Question 60. A force is applied for 0.1 s on a body of mass 2 kg. If it starts from rest and reaches a final velocity of \( 2 \text{ ms}^{-1} \), calculate the magnitude of the force.
Answer: time for which force is applied \( = 0.1 \text{ s} \).
Mass of body \( = 2 \text{ kg} \)
Initial velocity of body \( = 0 \text{ ms}^{-1} \)
Final velocity of body \( = 2 \text{ ms}^{-1} \).
We know \( F \times t = m (v – u) \)
\( F \times 0.1 = 2 (2 – 0) \)

\( \implies F = 4 /0.1 = 40 \text{ N} \).
In simple words: We find how much the object's speed changed and multiply it by its weight. Then we divide by the tiny amount of time the push lasted to find the strength of that push.

📝 Teacher's Note: This numerical reinforces the relationship between Impulse (\( F \times t \)) and Change in Momentum. Remind students that dividing by a small decimal (like 0.1) results in a larger number.

🎯 Exam Tip: Check your final value. 40 N is a reasonable force for a 2 kg object to reach that speed in such a short time. Always include the unit 'N' at the end.

Question 61. A player stops a 500 g ball moving at \( 30 \text{ ms}^{-1} \) in 0.03 s. Calculate the force applied by the player.
Answer: mass of ball \( = 500 \text{g} = 0.5 \text{ kg} \).
Initial speed of the ball \( = 30 \text{ ms}^{-1} \)
Final speed of ball \( = 0 \text{ ms}^{-1} \)
Time taken by player to stop the ball \( = 0.03 \text{ s} \).
We know \( F \times t = m (v – u) \)
\( F \times 0.03 = 0.5 (0 – 30) \)

\( \implies F = – 1.5 / 0.03 = – 500 \text{ N} \)
(-) sign shows that player has to apply force in opposite direction of the motion of the ball.
In simple words: The ball is moving fast, and the player has to push against it to stop it. The negative sign in our answer just means the push was in the opposite direction of where the ball was going.

📝 Teacher's Note: Point out the conversion of 500g to 0.5kg. This is an example of an impulsive force because the time is very small and the force is relatively large.

🎯 Exam Tip: Explain the meaning of the negative sign in your conclusion. It shows you understand that force is a vector quantity and the direction is opposite to the ball's motion.

Question 62. A force is applied for 0.1 s to a 3.2 kg body initially at rest. After the force is removed, the body covers 3 m in 1 s. Calculate the applied force.
Answer: Time for which force is applied \( = 0.1 \text{ s} \).
Mass of the body \( = 3.2 \text{ kg} \).
Initial speed of body \( = 0 \text{ ms}^{-1} \)
After removal of forces body covers a distance of \( 3 \text{m} \) in 1 second so final speed of body \( = 3/1 = 3 \text{ms}^{-1} \).
We know \( F \times t = m (v – u) \)
\( F \times 0.1 = 3.2 (3 - 0) \)

\( \implies F = 9.6/0.1 = 96 \text{ N} \).
So applied force is \( 96 \text{ N} \).
In simple words: When the push stops, the object keeps moving at a steady speed. We use its distance and time (3m in 1s) to find that speed. Then we work backwards to find out how hard the initial push must have been.

📝 Teacher's Note: This problem tests the understanding of Newton's First Law—that once the force is removed, the body continues at a constant velocity. Students must use that constant velocity as the final velocity (\( v \)) in the impulse equation.

🎯 Exam Tip: Be careful! The time for the force (\( 0.1 \text{ s} \)) and the time for measuring distance (\( 1 \text{ s} \)) are different. Don't mix them up in your formulas.

Question 63. From the given velocity-time graph, calculate the acceleration and the force acting on a body of mass 200 g.
Answer: As we know that slope of velocity time graph gives acceleration so acceleration of the body is \( = (v-u)/( t_2- t_1) =(20 - 0)/(5-0) = 4 \text{ ms}^{-2} \).
Mass of the body is \( = 200 \text{ g} =0.2 \text{ kg} \)
Force applied is equal to \( = 0.2 \times 4 = 0.8 \text{ N} \).
In simple words: Acceleration is how much the speed increases every second. From the graph, the speed goes up by 4 meters per second every second. Multiplying this by the weight of the object (in kg) gives us the force of 0.8 Newtons.

📝 Teacher's Note: Remind students that the slope of a velocity-time graph represents acceleration. Always ensure mass is converted to kilograms (SI unit) before calculating force in Newtons.

🎯 Exam Tip: To score full marks, show the step-by-step conversion of grams to kilograms (\( 200/1000 = 0.2 \text{ kg} \)) before using it in the force formula.

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Question 64. A force of 10 N is applied on a body of mass 2 kg for 3 s. Calculate the final velocity and the change in momentum.
Answer: Time for which force is applied \( =3\text{s} \).
Mass of the body \( = 2 \text{ kg} \).
Initial speed of body \( = 0 \text{ ms}^{-1} \)
Force applied \( = 10 \text{ N} \).

• We know \( F \times t = m (v - u) \)
  \( 10 \times 3 = 2 (v - 0) \)
  
  \( \implies v = 15 \text{ ms}^{-1} \).
  Final velocity is \( 15 \text{ ms}^{-1} \).
• As \( m(v - u) \) is change in momentum and this is equal to the \( F \times t \) so change in momentum is equal to the \( 30 \text{ kgms}^{-1} \).

In simple words: Pushing a 2 kg weight with 10 Newtons for 3 seconds speeds it up to 15 meters per second. The total "moving power" (momentum) gained by the object is 30 units.

📝 Teacher's Note: This question demonstrates the Impulse-Momentum theorem, where the product of force and time (Impulse) equals the change in momentum.

🎯 Exam Tip: When finding change in momentum, you can either use \( m \times (v-u) \) or directly calculate \( F \times t \). Both will give the same answer and confirm your work.

Question 65. (i) Why do high jumpers land on sand or foam? (ii) Why do players pull their hands backward while catching a fast ball?
Answer:

• We always prefer to land on sand instead of hard floor while taking a high jump because sand increases the time of contact. As \( F \times t = m ( v - u ) \) and our change in momentum is constant so if time increases then force experienced would decrease.
• Again while catching a fast moving ball, we always pull our hands backwards to increase reaction time so force experienced would decrease.

In simple words: To stop a moving object without getting hurt, you need to slow it down gradually. By using sand or pulling your hands back, you "stretch" the stopping time, which makes the impact force much weaker.

📝 Teacher's Note: Use the egg-drop analogy. An egg breaks on a hard floor because it stops instantly, but survives on a pillow because the stop is gradual. This is a practical application of the Second Law of Motion.

🎯 Exam Tip: Always mention that the "time of contact" or "stopping time" is increased. This is the scientific reason for the reduction in force.

Question 66. A stone is dropped from a cliff 98 m high. Find the time taken to reach the ground and its final velocity.
Answer: Height of cliff \( = 98 \text{ m} \).
Initial velocity of stone \( = 0 \text{ ms}^{-1} \).
Acceleration due to gravity \( = 9.8 \text{ ms}^{-2} \).

• We know \( H = ut + 1/2 \text{ gt}^2 \).
  \( 98 = 1/2 \times 9.8 \times t^2 \).
  
  \( \implies t^2 = 98 \times 2/9.8 = 20 \)
  
  \( \implies t= 4.47 \text{ sec} \).
• Final velocity when it strikes the ground
  \( v^2 - u^2 = 2 \text{ g H} \)
  
  \( \implies v^2 = 2 \times 9.8 \times 98 \)
  
  \( \implies v^2 = 1920.8 \)
  
  \( \implies v = 43.82 \text{ ms}^{-1} \).

In simple words: When you drop a stone from a 98-meter cliff, it takes about 4.5 seconds to hit the bottom. By the time it hits the ground, it is falling at about 44 meters per second.

📝 Teacher's Note: In free fall problems, the initial velocity \( u \) is always zero. Use the value of \( g = 9.8 \text{ ms}^{-2} \) unless specified otherwise.

🎯 Exam Tip: For problems involving square roots (like \( t^2 = 20 \)), try to provide the answer up to two decimal places for accuracy.

Question 67. A ball is thrown vertically upwards with a speed of \( 9.8 \text{ ms}^{-1} \). Find the maximum height reached, the time taken to reach that height, and the total time of flight.
Answer: Initial speed of ball is \( = 9.8 \text{ ms}^{-1} \).
Acceleration due to gravity \( = -9.8 \text{ ms}^{-2} \).
Final speed at maximum height \( = 0 \text{ ms}^{-1} \).
We know \( v = u + at \)
\( 0 = 9.8 - 9.8 \text{ t} \)

\( \implies T = 1 \text{ sec} \).
We know \( v^2 - u^2 = 2 \text{as} \)
At highest point final velocity is zero so
\( 0 - 9.8 \times 9.8 = 2 \times (-9.8) \text{ S} \)

\( \implies S = 4.9 \text{ m} \).
for highest point initial velocity is zero
Acceleration due to gravity is \( = 9.8 \text{ ms}^{-2} \).
Final velocity at ground is \( v \)
\( V^2 - 0 = 2 \times 9.8 \times 4.9 \)

\( \implies V = 9.8 \text{ ms}^{-1} \).
Time taken to reach ground from highest point
\( V = u + at \)
\( 9.8 = 0 + 9.8 \text{ t} \)

\( \implies T = 9.8/9.8 = 1 \text{ sec} \).
Total time \( = \) time of ascent \( + \) time of descent.
Total of flight \( = 1 + 1 = 2 \text{ seconds} \).
In simple words: Throwing a ball up at about 10 m/s makes it reach a height of roughly 5 meters. It takes 1 second to go up and 1 second to fall back down, making the total trip last 2 seconds.

📝 Teacher's Note: Highlight the symmetry of motion under gravity: the time of ascent equals the time of descent, and the impact speed equals the launch speed.

🎯 Exam Tip: Always use a negative value for \( g \) when an object is thrown upwards, as gravity is acting against the motion.

Question 68. A ball is thrown upwards with a velocity of \( 10 \text{ ms}^{-1} \). Calculate its velocity after 1 second and after 2 seconds.
Answer: Initial speed of ball \( = 10 \text{ ms}^{-1} \).
Acceleration due to gravity on ball \( = - 9.8 \text{ ms}^{-2} \)
We know that from first equation of motion
\( v = u + gt \).
After 1 sec
\( v = 10 - 9.8 \times 1 \)

\( \implies v = 0.2 \text{ ms}^{-1} \)
so velocity after 1 sec would be \( 0.2 \text{ ms}^{-1} \).
Velocity after 2 seconds
\( v = 10 - 9.8 \times 2 = 10 - 19.6 = -9.6 \text{ ms}^{-1} \).
Here negative sign shows that velocity is in downward direction and magnitude is \( 9.6 \text{ ms}^{-1} \).
In simple words: After 1 second, the ball has slowed down to almost a stop near the peak. By 2 seconds, it has already started falling back down toward the ground at nearly 10 m/s.

📝 Teacher's Note: Explain that a negative result for velocity simply means the object has reversed its direction. It's a key concept in vector physics.

🎯 Exam Tip: When the result is negative, conclude by clearly stating that the object is moving downwards.

Question 69. What initial speed should a ball be thrown with to reach a maximum height of 19.6 m?
Answer: Maximum Height attained by ball \( = 19.6 \text{ m} \)
Let initial speed of ball \( = u \text{ ms}^{-1} \).
Acceleration applied on ball due to gravity \( = -9.8 \text{ ms}^{-2} \).
Final speed of ball at maximum height \( = 0 \text{ ms}^{-1} \).
We know that from second equation of motion
\( v^2 - u^2 = 2\text{as} \)
\( 0 - u^2 = 2 \times (-9.8) \times 19.6 \)

\( \implies u^2 = 19.6 \times 19.6 \)

\( \implies u = 19.6 \text{ ms}^{-1} \)
so initial speed of ball to attain maximum height of \( 19.6 \text{ m} \) should be \( 19.6 \text{ ms}^{-1} \).
In simple words: To get a ball to reach a height of about 20 meters, you need to launch it at about 20 meters per second.

📝 Teacher's Note: Use this specific problem to show how the numbers \( 9.8 \) and \( 19.6 \) (which is \( 2 \times 9.8 \)) often lead to clean, easy-to-calculate answers in physics problems.

🎯 Exam Tip: When \( v=0 \), the formula simplifies to \( u = \sqrt{2gH} \). Memorizing this shorthand can save time in multiple-choice questions.

Question 70. A stone is dropped from a 98 m tall tower. One second later, a second stone is thrown down. What should be the initial speed of the second stone so that both hit the ground at the same time?
Answer: Height of tower \( = 98 \text{ m} \)
Acceleration due to gravity on stone \( = 9.8 \text{ ms}^{-2} \).
Initial speed of first ball \( = 0 \text{ ms}^{-1} \).
Let initial speed of second stone is \( v \text{ ms}^{-1} \).
We know from second equation of motion
\( S = ut + 1/2 \text{ a} \times t^2 \).
\( 98 = 0 + 1/2 \times 9.8 \times t^2 \).

\( \implies t^2 = 20 \)

\( \implies t = 4.47 \text{ sec} \).
As second stone is thrown 1 sec later so time taken by second body to cover distance of 98 m is \( = 4.47 - 1 = 3.47 \text{sec} \).
So again put \( t= 3.47 \text{ sec} \) and \( S = 98 \text{ m} \) in second equation of motion we get
\( 98 = v \times 3.47 + 1/2 \times 9.8 \times 3.47 \times 3.47 \).
\( 98 = 3.47 \times v + 59 \)

\( \implies 3.47 \times v = 98 - 59 \)

\( \implies v = 39/3.47 = 11.23 \text{ ms}^{-1} \).
Initial speed of second stone should be \( 11.23 \text{ ms}^{-1} \).
In simple words: The first stone takes about 4.5 seconds to fall. Since the second stone starts a second later, it only has 3.5 seconds to catch up. To do that, it needs to be thrown down at about 11 m/s instead of just being dropped.

📝 Teacher's Note: This is a complex multi-stage problem. Encourage students to solve for the first object's total time before setting up the equation for the second object.

🎯 Exam Tip: Be very careful with the "time" value for the second object. "One second later" means its travel time is \( t-1 \).

Question 71. Calculate the gravitational force between two objects of mass 200 mg each, placed at a distance of 1 mm.
Answer: Mass of object 1 \( m_1 = 200 \text{ mg} = 200 \times 10^{-6} \text{ kg} = 2 \times 10^{-4} \text{ kg} \).
Mass of object 2 \( m_2 = 200 \text{ mg} = 200 \times 10^{-6} \text{ kg} = 2 \times 10^{-4} \text{ kg} \).
Distance between the two objects \( = 1 \text{ mm} = 10^{-3} \text{ m} \)
We know law of gravitation is
\( F = G ( m_1 \times m_2)/R^2 \)
\( F = (6.67 \times 10^{-11} \times 2 \times 10^{-4} \times 2 \times 10^{-4})/(10^{-3} \times 10^{-3}) \)
\( F = 6.67 \times 2 \times 2 \times 10^{-11-4-4+3+3} \)

\( \implies F = 26.68 \times 10^{-13} \text{ N} \)
So these two objects would experience a force of \( 26.68 \times 10^{-13} \text{ N} \).
In simple words: Even though these objects are tiny and very close together, the pull between them is extremely weak—less than a trillionth of a Newton!

📝 Teacher's Note: This problem teaches the importance of scientific notation and unit conversion. Students must convert milligrams to kilograms and millimeters to meters for the formula to work.

🎯 Exam Tip: Be meticulous with powers of 10. A single error in adding or subtracting exponents will lead to a wrong answer.

Question 72. Calculate the mass of the Earth given its radius is \( 6.38 \times 10^3 \text{ km} \), \( G = 6.67 \times 10^{-11} \), and \( g = 9.8 \text{ ms}^{-2} \).
Answer: Radius of earth \( = 6.38 \times 10^3 \text{ km} = 6.38 \times 10^6 \text{ m} \)
\( G = 6.67 \times 10^{-11} \)
Acceleration due to gravity \( = 9.8 \text{ ms}^{-2} \).
We know that
\( g = (G \times M)/R^2 \).
\( 9.8 = (6.67 \times 10^{-11} \times M)/ ( 6.38 \times 10^6 \times 6.38 \times 10^6) \)
\( 9.8 \times 6.38 \times 6.38 \times 10^{12} = 6.67 \times 10^{-11} \times M \)
\( 398.9 \times 10^{12} = 6.67 \times 10^{-11} \times M \)
\( M = 398.9 \times 10^{12}/6.67 \times 10^{-11} \)
\( M = 59 \times 10^{23} \text{ kg} \)
So mass of earth is \( 59 \times 10^{23} \text{ kg} \).
In simple words: By using the size of the Earth and how fast things fall, we can figure out that the entire Earth weighs about 6 followed by 24 zeros kilograms!

📝 Teacher's Note: This calculation shows how scientists can "weigh" the Earth without a giant scale. It's a powerful application of the gravity formula.

🎯 Exam Tip: Always convert km to m by multiplying by \( 10^3 \) before using the gravity formula. Forgetting this conversion is a very common trap.