Selina Concise Solutions for ICSE Class 9 Physics Chapter 3 Laws Of Motion

Exercise 3(A)

Question 1S.
Answer:
(a) The forces which act on bodies when they are in physical contact are called contact forces.
Example: frictional force and force exerted on two bodies during collision.
(b) The forces experienced by bodies even without being physically touched are called non-contact forces.
Example: Gravitational force and Electrostatic force.
In simple words: Contact forces happen when objects touch each other (like pushing a door). Non-contact forces work from a distance without touching (like a magnet pulling a nail).

📝 Teacher's Note: Use a physical demonstration like pushing a desk (contact) versus holding a magnet near a paperclip (non-contact) to make the distinction clear for students.

🎯 Exam Tip: Always provide at least one example for each type of force to secure full marks in descriptive questions.

Question 2S.
Answer:
Contact force: (a) frictional force (b) normal reaction force (c) force of tension in a string
Non-contact force: (d) gravitational force (e) electric force (f) magnetic force
In simple words: Friction and tension need objects to be touching to exist, while gravity and magnetism can pull objects through empty space.

📝 Teacher's Note: Help students identify the "agent" of the force. If the agent must touch the object, it is a contact force.

🎯 Exam Tip: Memorize these classifications as they are common 1-mark objective questions.

Question 3S.
Answer:
a. Force exerted on two bodies during collision.
b. Magnetic force between magnetic poles.
In simple words: Colliding objects must hit each other (contact), but magnets can push or pull each other from a distance (non-contact).

📝 Teacher's Note: Clarify that "collision" implies physical impact, which is why it is the ultimate example of a contact force.

🎯 Exam Tip: When asked for examples, be specific (e.g., "magnetic force" rather than just "magnets").

Question 4S.
Answer:
A. Tension in a string supporting a ball.
B. Spring force acting on a mass.
C. The forces acting on the block are its weight in the downward direction and the normal reaction force due to the table on the upward direction.
In simple words: These diagrams show how different forces balance each other out—like a rope pulling up while gravity pulls a ball down.

📝 Teacher's Note: When teaching free-body diagrams, always emphasize that force arrows should originate from the center of the object or the point of contact.

🎯 Exam Tip: Label every force arrow clearly with its symbol (T, W, N) to ensure full marks for diagram questions.

Question 5S.
Answer:
The magnitude of non-contact force on two bodies depends on the distance of separation between them.
The force decreases as the distance of separation increases.
The force is inversely proportional to the square of the distance of separation.
In simple words: Non-contact forces (like gravity) get much weaker as you move objects further apart. If you double the distance, the force becomes four times weaker.

📝 Teacher's Note: Use the "Inverse Square Law" analogy with a flashlight—as you move away from a wall, the light spreads out and gets dimmer very quickly.

🎯 Exam Tip: Use the term "inversely proportional to the square of the distance" specifically, as it is a key phrase examiners look for.

Question 6S.
Answer:
The magnitude of gravitational force between two masses will become four times as gravitational force varies inversely as the square of distance of separation.
In simple words: Because gravity follows the inverse square law, if you bring two objects twice as close, the pull between them doesn't just double—it quadruples!

📝 Teacher's Note: Remind students that if distance is halved (\( 1/2 \)), the force becomes \( 1/(1/2)^2 = 4 \) times stronger.

🎯 Exam Tip: In numerical variations of this question, always square the change in distance to find the change in force.

Question 7S.
Answer:
A force when applied on a non-rigid body changes the inter-spacing between its constituent particles and therefore causes a change in its dimensions and can also produce motion in it.
On the other hand, a force when applied on a rigid body, does not change the inter-spacing between its constituent particles and therefore it does not change the dimensions of the body but causes motion in it.
In simple words: When you squeeze a sponge (non-rigid), it changes shape. When you push a brick (rigid), it stays the same shape but moves across the floor.

📝 Teacher's Note: Explain that "rigid" is an ideal concept; in reality, almost all bodies deform slightly under massive forces.

🎯 Exam Tip: Clearly state that non-rigid bodies undergo deformation while rigid bodies only undergo translation or rotation.

Question 8S.
Answer:
(i) A fielder on the ground stops a moving ball by applying a force with his hands.
(ii) The pull exerted by horse makes a cart moves.
(iii) In a cycle pump, when the piston is lowered, the air is compressed to occupy a less volume.
(iv) On pressing a piece of rubber, its shape changes.
In simple words: Force can do four main things: stop something, start something moving, change its size, or change its shape.

📝 Teacher's Note: Ask students to find one example of each of these four effects in the classroom to build observation skills.

🎯 Exam Tip: If asked for the "effects of force," list all four categories (state of motion, speed, direction, and dimensions).

Question 1M. Frictional force is a
(a) contact force
(b) non-contact force
(c) magnetic force
(d) None of the options
Answer: (a) contact force
In simple words: Friction only happens when two surfaces are rubbing against each other, so they must be in contact.

📝 Teacher's Note: Demonstrate by showing that two hands only feel "heat" or resistance from friction when they touch.

🎯 Exam Tip: Friction is the most common example of a contact force used in exams.

Question 2M. Force due to gravity is a
(a) contact force
(b) non-contact force
(c) frictional force
(d) None of the options
Answer: (b) non-contact force
In simple words: Gravity pulls a falling apple toward the Earth even before the apple touches the ground.

📝 Teacher's Note: Contrast this with a "pull" from a rope, which requires a physical connection.

🎯 Exam Tip: Gravity is unique because it is always attractive and works over infinite distances.

Exercise 3(B)

Question 1S.
Answer:
Force causes motion in a body.
In simple words: If something is still, you need to push or pull it to make it start moving.

📝 Teacher's Note: This is the qualitative definition of force—an external agent that changes the state of a body.

🎯 Exam Tip: Mention both "starting motion" and "changing motion" for a complete answer.

Question 2S.
Answer:
Force is not needed to keep a moving body in motion.
In simple words: In a world without friction (like space), an object that is already moving would keep going forever without any help.

📝 Teacher's Note: This is a counter-intuitive concept for students. Use the example of a puck on an air-hockey table to show how reduced friction leads to sustained motion.

🎯 Exam Tip: Be careful! This only applies when no opposing forces like friction are present.

Question 3S.
Answer:
The force of friction between the table and the ball opposes the motion of the ball.
In simple words: A ball stops rolling on a table because friction is "fighting" against its movement, slowing it down.

📝 Teacher's Note: Emphasize that friction is a "lazy" force—it only acts when there is motion or attempted motion, and it always works in the opposite direction.

🎯 Exam Tip: Always state the direction of friction as "opposite to the direction of motion."

Question 4S.
Answer:
In absence of any external force, its speed shall remain unchanged.
In simple words: If nobody pushes or pulls a moving object, it will just keep gliding at the exact same speed.

📝 Teacher's Note: This is part of Newton's First Law. Absence of force equals absence of acceleration.

🎯 Exam Tip: "Unchanged speed" also implies "unchanged direction" unless a force is applied.

Question 5S.
Answer:
Galileo’s law of inertia states that a body continues to be in its state of rest or of uniform motion unless an external force is applied on it.
In simple words: Objects are "stubborn"—they want to keep doing exactly what they are doing unless a force forces them to change.

📝 Teacher's Note: Explain that "Inertia" is a property, not a force. It's like the "laziness" of matter.

🎯 Exam Tip: Use the terms "state of rest" and "uniform motion" to sound professional.

Question 6S.
Answer:
According to Newton’s first law of motion, if a body is in a state of rest, it will remain in the state of rest, and if the body is in the state of motion, it will remain moving in the same direction with the same speed unless an external force is applied on it.
In simple words: Things don't change their movement on their own. You need a force to start, stop, or turn anything.

📝 Teacher's Note: Highlight that Newton's First Law is basically a formal version of Galileo's Law of Inertia.

🎯 Exam Tip: This law is often called the "Law of Inertia." Make sure to mention that force is the "cause" of change.

Question 8S.
Answer:
The property of an object by virtue of which it neither changes its state nor tends to change the state is called inertia.
In simple words: Inertia is an object's resistance to change. It’s why you keep moving forward when a car suddenly brakes.

📝 Teacher's Note: Distinguish between inertia of rest, motion, and direction using everyday examples like dust falling from a rug or a passenger leaning in a turning bus.

🎯 Exam Tip: Define inertia as an inherent property of matter that depends only on mass.

Question 9S.
Answer:
Force is that external cause which can move a stationary object or which can stop a moving object.
In simple words: A force is a push or pull that acts as a "trigger" to change how something is moving.

📝 Teacher's Note: This is the definition of force derived from Newton's First Law (the qualitative definition).

🎯 Exam Tip: Define force as an agent that changes or *tends to change* the state of a body.

Question 10S.
Answer:
Inertia of a body depends on its mass. Inertia is directly proportional to mass, i.e. greater the mass of a body, greater is its inertia.
In simple words: Heavy things are harder to move and harder to stop because they have more "inertia." A bowling ball has more inertia than a tennis ball.

📝 Teacher's Note: Use the "shopping cart" analogy—it's easy to push when empty, but very hard to stop or turn when full of heavy groceries.

🎯 Exam Tip: If the question asks "What is the measure of inertia?", the answer is always "Mass."

Question 11S.
Answer:
Examples to show that greater the mass, greater is the inertia of the body are as shown below:
1. If you want to start a car by pushing it, you find that it takes a very large force to overcome its inertia. On the other hand, only a small force is needed to start a child’s express wagon. The difference between the car and express wagon is the difference in mass. The car has a large mass, whereas the wagon has a small one.
2. A cricket ball is more massive than a tennis ball. The cricket ball acquires a much smaller velocity than a tennis ball when the two balls are pushed with equal force for the same time.
In simple words: Massive objects have a lot of "stubbornness." Pushing a real car is much harder than pushing a toy car because the real car has more mass and thus more inertia.

📝 Teacher's Note: This helps students connect the abstract concept of inertia to the tangible concept of mass.

🎯 Exam Tip: When explaining this, always compare two objects with different masses under the same force.

Question 12S.
Answer:
It is difficult, i.e. a larger force is required to set a loaded trolley (which has more mass) in motion than an unloaded trolley (which has less mass).
In simple words: Adding items to a trolley increases its mass, which makes it "heavier" to start moving because its inertia has increased.

📝 Teacher's Note: This is a classic lab experiment. Have students push an empty box and then a box filled with books to feel the difference.

🎯 Exam Tip: Clearly link the "difficulty" to the "higher mass" and "greater inertia."

Question 13S.
Answer:
Two kinds of inertia are as listed below:
1. Inertia of rest.
2. Inertia of motion.
In simple words: Inertia of rest makes a still object want to stay still. Inertia of motion makes a moving object want to keep moving.

📝 Teacher's Note: There is also "Inertia of Direction," but for this level, focus on these primary two.

🎯 Exam Tip: Be prepared to give one real-life example for each type of inertia.

Question 14S.
Answer:
Examples of inertia of rest: A coin placed on top of a card remains in place when the card is slightly and quickly jerked horizontally.
Example of inertia of motion: A ball thrown vertically upwards by a person in a moving train comes back to his hand.
In simple words: The coin stays still because it has inertia of rest. The ball moves forward with the train because it has inertia of motion.

📝 Teacher's Note: Perform the "coin and card" trick in class. It’s a guaranteed way to engage students and prove the concept of inertia of rest.

🎯 Exam Tip: Use these specific examples in short-answer questions as they are standard textbook cases.

Question 15S.
Answer: No, the body will not move because the net force acting on it is zero. Hence, it will remain stationary due to inertia of rest.

In simple words: If you pull an object from both sides with the same strength, the forces cancel out, and the object stays right where it is.

📝 Teacher's Note: Use the concept of "Balanced Forces" here. When forces are equal and opposite, the net force is zero, and the object's state of rest or motion doesn't change.

🎯 Exam Tip: Always mention that the "net force is zero" to explain why a body remains stationary despite multiple forces acting on it.

Question 16S.
Answer: The motion remains unaffected because the net force acting on it is zero.

In simple words: If an object is already moving and the total force on it is zero, it just keeps gliding at the same speed.

📝 Teacher's Note: This relates to Newton's First Law. Absence of net force means there is no acceleration, so the velocity stays constant.

🎯 Exam Tip: Remember that "unaffected motion" means both the speed and the direction of the object stay the same.

Question 17S.
Answer: The net force on the airplane is zero or the upward force is equal to the downward force.

In simple words: For a plane to fly at a steady height, the engine's lift pushing it up must exactly match the gravity pulling it down.

📝 Teacher's Note: This is a great example of equilibrium in real life. When forces balance out vertically, the airplane neither climbs nor falls.

🎯 Exam Tip: Use the term "equilibrium" or "balanced forces" when describing objects moving at a constant altitude or speed.

Question 18S.
Answer: If a person jumps out of a moving train and tries to stop immediately, he falls due to inertia of motion. This is because his body tends to move forward with the velocity of the train while his feet are stationary.

In simple words: When you jump out, your feet stop as soon as they hit the ground, but your upper body still wants to zoom forward with the train, making you trip.

📝 Teacher's Note: Explain that different parts of the body can have different states of motion—the feet stop due to contact with the ground, but the rest of the body continues moving due to inertia.

🎯 Exam Tip: Clearly distinguish between the part of the body that comes to rest and the part that continues to move to get full marks.

Question 19S.
Answer: The reason is that when the card is flicked, a momentary force acts on the card, so it moves away. However, the coin kept on it does not share the motion at once and it remains stationary at its place due to the inertia of rest. The coin then falls down into the tumbler due to the pull of gravity.

In simple words: You hit the card so fast that the coin doesn't have time to move with it. The coin stays in the same spot for a split second and then drops straight down.

📝 Teacher's Note: This is a classic demonstration of "Inertia of Rest." The coin's mass resists the sudden change in motion.

🎯 Exam Tip: Mentioning "Inertia of Rest" as the key scientific principle is mandatory for scoring full marks here.

Question 20S.
Answer: The reason is that when the ball is thrown, the ball is in motion along with the person and train. Due to the inertia of motion, during the time the ball remains in air, the person and ball move ahead by the same distance. This makes the ball fall back into the thrower’s hand.

In simple words: Because you and the ball are both moving with the train, when you toss the ball up, it keeps moving forward at the same speed as you, so it lands right back in your hand.

📝 Teacher's Note: Ask students to imagine this from the perspective of someone standing outside the train—they would see the ball follow a curved path (parabola).

🎯 Exam Tip: Emphasize that both the ball and the thrower maintain the same horizontal velocity due to inertia.

Question 21S.
Answer:
(a) When a train suddenly starts, the passengers tend to fall backwards. This is because the lower part of the body, which is in contact with the train, begins to move while the upper part of the body tends to maintain its position of rest. As a result, the upper part tends to fall backwards.
(b) The frame of the sliding door being in contact with the floor of the train also comes in motion on start of the train, but the sliding door remains in its position due to inertia. Thus, the frame moves ahead with the train, while the door slides opposite to the direction of motion of the train. Thus, the door may shut or open accordingly.
(c) When the branches of the tree are shaken, they come in motion, while the fruits due to inertia remain in a state of rest. Thus, the larger and weakly attached fruits get detached from the branches and fall down due to the pull of gravity.
(d) When people alight from a moving bus, they continue to run alongside the bus to avoid falling. If they were to stop at once, the feet would come to rest suddenly but the upper part of the body would still be in motion and they would tend to fall forward.
(e) The part of the carpet where the stick strikes comes in motion at once, while the dust particles settled on it remain in the state of rest due to inertia of rest. Thus, a part of the carpet moves ahead with the stick leaving behind the dust particles that fall down due to gravity.
(f) When running, the athlete brings his body in the state of motion. When the body is in motion, it becomes easier to take a long jump.

In simple words: Dust stays put while the carpet moves, and runners jump further because they use their existing speed to carry them through the air.

📝 Teacher's Note: Point out that cleaning a carpet with a stick is a practical application of "Inertia of Rest."

🎯 Exam Tip: For the carpet example, specifically mention "Inertia of Rest" for the dust particles.

Question 1M.
Answer: A truck

In simple words: Since a truck has more mass, it has more inertia (resistance to change) than a smaller vehicle.

📝 Teacher's Note: Remind students that mass is the direct measure of inertia. Higher mass always equals higher inertia.

🎯 Exam Tip: In comparison questions, identify the object with the largest mass to find the one with the most inertia.

Question 2M.
Answer: Less force is required for the tennis ball than for the cricket ball.

In simple words: A tennis ball is lighter than a cricket ball, so it's easier to start moving or stop.

📝 Teacher's Note: This is a practical example of \( F = ma \). For the same acceleration, a smaller mass requires smaller force.

🎯 Exam Tip: Relate force directly to mass when acceleration is kept constant.

Question 3M.
Answer: Change the state of motion or state of rest of the body.

In simple words: Force is the only thing that can make a still object move or a moving object change its speed or direction.

📝 Teacher's Note: This is the qualitative definition of force as derived from Newton's First Law.

🎯 Exam Tip: Define force as an external agent that changes the state of rest or uniform motion of a body.

Exercise 3(C)

Question 1S.
Answer: Force needed to stop a moving body in a given time depends on its mass and velocity.

In simple words: How hard you have to push to stop something depends on how heavy it is and how fast it’s going.

📝 Teacher's Note: Introduce the concept of momentum here, which is the product of these two factors.

🎯 Exam Tip: Use the words "mass" and "velocity" together to explain the physical impact required to stop a body.

Question 2S.
Answer: Linear momentum of a body is the product of its mass and velocity.
Its SI unit is \( \text{kg m s}^{-1} \).

In simple words: Momentum is the "strength" of an object's motion. You calculate it by multiplying weight times speed.

📝 Teacher's Note: Emphasize that momentum is a vector quantity, meaning it has a specific direction.

🎯 Exam Tip: Always include the correct SI unit \( \text{kg m s}^{-1} \) to avoid losing marks.

Question 3S.
Answer:
(i) When \( v \ll c \),
\( \Delta p = \Delta (mv) = m \Delta v \)
(ii) When \( v \rightarrow c \),
\( \Delta p = \Delta (mv) \)
(iii) When \( v \ll c \) but m does not remain constant.
\( \Delta p = \Delta (mv) \)

In simple words: At normal speeds, momentum changes because speed changes. But if speed is near light-speed or mass changes, the math becomes a bit more complex.

📝 Teacher's Note: For Class 9, we usually assume mass is constant, but it's good to mention that at very high speeds (relativity), mass can change.

🎯 Exam Tip: Remember that \( \Delta \) (delta) means "change in." So \( \Delta p \) is the change in momentum.

Question 4S.
Answer: Let a force ‘F’ be applied on a body of mass m for a time ‘t’ due to which its velocity changes from u to v. Then,
Initial momentum of body = \( mu \)
Final momentum of body = \( mv \)
Change in momentum of the body in ‘t’ seconds = \( mv - mu = m (v - u) \)
Rate of change of momentum = \( \frac{\text{Change in momentum}}{\text{time}} \)
\( \implies \frac{m (v - u)}{t} \)
However, acceleration \( a = \frac{\text{Change in velocity}}{\text{time}} = \frac{v - u}{t} \)
Therefore, rate of change of momentum = \( ma = \text{mass} \times \text{acceleration} \)
This relation holds true when the mass of the body remains constant.

In simple words: This step-by-step math shows that force equals mass times acceleration (\( F = ma \)), which is the heart of physics.

📝 Teacher's Note: This derivation connects Newton's Second Law to the formula \( F = ma \). Make sure students understand every substitution step.

🎯 Exam Tip: Start the derivation by defining the variables \( u \), \( v \), \( m \), and \( t \) to make your answer clear.

Question 5S. 
Answer:
(i) Mass is the measure of inertia.
Let ‘m’ be the mass of the two bodies.
Inertia of body A : Inertia of body B \( :: m : m \)
Or, Inertia of body A : Inertia of body B \( :: 1 : 1 \)

(ii) Momentum of body A = \( m(v) \)
Momentum of body B = \( m(2v) = 2mv \)
Momentum of body A : Momentum of body B \( :: mv : 2mv \)
Or, Momentum of body A : Momentum of body B \( :: 1 : 2 \).

In simple words: Since both objects have the same mass, they have the same inertia. However, because the second object is moving twice as fast, it has twice as much momentum or "moving power."

📝 Teacher's Note: Use this to clarify that inertia depends only on mass, while momentum depends on both mass and speed. Students often confuse the two.

🎯 Exam Tip: When writing ratios, always simplify them to their lowest terms (like 1:1 or 1:2) to ensure full marks.

Question 6S.
Answer:
(i) Inertia of body A : Inertia of body B \( :: m : 2m \)
Or, Inertia of body A : Inertia of body B \( :: 1 : 2 \).

(ii) Momentum of body A = \( m(2v) = 2mv \)
Momentum of body B = \( (2m)v = 2mv \)
Momentum of body A : Momentum of body B \( :: 2mv : 2mv \)
Or, Momentum of body A : Momentum of body B \( :: 1 : 1 \).

(iii) According to Newton’s 2nd law of motion, rate of change of momentum is directly proportional to the force applied on it. Therefore,
Force needed to stop A : Force needed to stop B \( :: 1 : 1 \).

In simple words: A heavy object moving slowly can have the same momentum as a light object moving fast. If their momentum is the same, you need the same amount of force to stop them in the same amount of time.

📝 Teacher's Note: This is a great conceptual check. Ask students: "Which is harder to stop, a slow truck or a fast car?" if their momenta are equal.

🎯 Exam Tip: Remember that "Force needed to stop" is another way of asking about the rate of change of momentum. If momentum and time are the same, force is the same.

Question 7S. 
Answer: According to Newton’s second law of motion, the rate of change of momentum is directly proportional to the force applied on it and the change of momentum takes place in the direction in which the force is applied.
It gives the quantitative value of force, i.e. it relates the force to the measurable quantities such as acceleration and mass.

In simple words: This law tells us exactly how much force is needed to move an object. The more force you apply, the faster the object's momentum changes in that same direction.

📝 Teacher's Note: Emphasize that while the First Law defines force, the Second Law measures it. Use the formula \( F = ma \) to show this calculation.

🎯 Exam Tip: In your definition, do not forget to mention that the change occurs "in the direction of the applied force."

Question 8S. 
Answer: Newton’s first law of motion gives the qualitative definition of force. It explains the force as the cause of acceleration only qualitatively but Newton’s second law of motion gives the quantitative value of force. It states force as the product of mass and acceleration. Thus, it relates force to the measurable quantities such as acceleration and mass.

In simple words: The First Law says force is a "push or pull" that changes motion. The Second Law gives us the math to calculate exactly how strong that push or pull is using mass and acceleration.

📝 Teacher's Note: Explain the difference between "Qualitative" (describing what it is) and "Quantitative" (measuring how much it is) using simple analogies like describing a meal vs. listing its calories.

🎯 Exam Tip: Clearly state that the Second Law leads to the formula \( F = ma \). This is the key difference between the two laws.

Question 9S. 
Answer: Mathematical expression of Newton’s second law of motion is as shown below:
Force = Mass \( \times \) Acceleration
Above relation holds for the following conditions:
(i) When the velocity of the body is much smaller than the velocity of light.
(ii) When the mass remains constant.

In simple words: We use \( F = ma \) for everyday objects. However, this simple math doesn't work if things are moving at super-fast light speeds or if the object's weight changes while it's moving (like a leaking rocket).

📝 Teacher's Note: Briefly mention that at speeds near light speed, mass increases (Relativity), which is why the classical formula has limits.

🎯 Exam Tip: Listing the conditions is essential for full marks in a "State and Derive" type question.

Question 10S. 
Answer: According to Newton’s second law of motion, the rate of change of momentum is directly proportional to the force applied on it, and the change of momentum takes place in the direction in which the force is applied.
The relation \( F=ma \) holds for the following conditions:
(i) When the velocity of the body is much smaller than the velocity of light.
(ii) When the mass remains constant.

In simple words: Force is calculated by how much an object's "moving energy" (momentum) changes per second. This calculation usually assumes the object isn't losing mass or moving at cosmic speeds.

📝 Teacher's Note: Remind students that the vector nature of force is stated here—force and change in momentum are always in the same direction.

🎯 Exam Tip: This question often appears as a "State the Law and its mathematical constraints" question. Ensure both parts are answered.

Question 11S. 
Answer: From Newton’s second law of motion, \( F = ma \).
If \( F = 0 \), then \( a = 0 \).
This means that if no force is applied on the body, its acceleration will be zero. If the body is at rest, then it will remain in the state of rest and if it is moving, then it will remain moving in the same direction with the same speed. Thus, a body not acted upon by an external force, does not change its state of rest or motion. This is the statement of Newton’s first law of motion.

In simple words: If you don't push or pull something (\( F = 0 \)), it won't speed up or slow down (\( a = 0 \)). This proves the First Law—things just keep doing what they're doing unless a force makes them change.

📝 Teacher's Note: This is a mathematical proof showing that the First Law is actually just a special case of the Second Law (where force is zero).

🎯 Exam Tip: This proof is common in exams. Use the "If \( F=0 \), then \( a=0 \)" logic clearly to show the connection between the laws.

Question 12S.

Answer:

In simple words: These graphs show that if you push harder, objects accelerate more. Also, if you want different objects to speed up the same amount, you must push the heavier ones with more force.

📝 Teacher's Note: Straight line graphs through the origin always indicate that the two values on the axes are directly proportional to each other.

🎯 Exam Tip: When drawing these, always label your axes clearly with the physical quantities and show the arrows for the direction of increase.

Question 13S.
Answer: If a given force is applied on bodies of different masses, then the acceleration produced in them is inversely proportional to their masses.

In simple words: This means that for the same push, a light object will zoom off quickly, while a heavy object will speed up much more slowly.

📝 Teacher's Note: This explains why it's easier to accelerate a bicycle than a car with the same amount of effort. Mass resists acceleration.

🎯 Exam Tip: The word "inversely proportional" is key. It means as mass goes up, acceleration must go down.

Question 14S. 
Answer: The S.I. unit of force is newton.
One newton is the force which acts on a body of mass 1kg and produces an acceleration of \( 1 \text{ m/s}^2 \), i.e. \( 1 \text{ N} = 1 \text{ kg} \times 1 \text{ m/s}^2 \).

In simple words: A Newton is just the "amount" of push required to make a 1kg block speed up by 1 meter per second, every second.

📝 Teacher's Note: Relate this to real life: 1 Newton is roughly the weight of a medium-sized apple in your hand.

🎯 Exam Tip: When defining a unit, always start with "It is that force which..." and use unit values for all constituent quantities.

Question 15S. 
Answer: The C.G.S. unit of force is dyne.
One dyne is the force which acts on a body of mass 1 gramme and produces an acceleration of \( 1 \text{ cm s}^{-2} \), i.e. \( 1 \text{ dyne} = 1 \text{ g} \times 1 \text{ cm s}^{-2} \).

In simple words: A dyne is a tiny unit of force. It's the push needed to make a small 1-gram weight speed up by 1 centimeter per second, every second.

📝 Teacher's Note: "C.G.S." stands for Centimeter, Gram, Second. This system is used for very small measurements in laboratories.

🎯 Exam Tip: Be careful with units: \( \text{cm s}^{-2} \) for dyne, not \( \text{m/s}^2 \).

Question 16S. 
Answer: The S.I. unit of force is newton and the C.G.S. unit of force is dyne.
\( 1 \text{ N} = 10^5 \text{ dyne} \).

In simple words: One Newton is equal to 100,000 dynes. This shows that the standard Newton is much stronger than a single dyne.

📝 Teacher's Note: Have students practice the derivation: \( 1 \text{ N} = 1 \text{ kg} \times 1 \text{ m/s}^2 = 1000 \text{ g} \times 100 \text{ cm/s}^2 = 100,000 \text{ dyne} \).

🎯 Exam Tip: The conversion factor \( 10^5 \) is a very common fill-in-the-blank question. Memorize it!

Question 17S. 
Answer: When a glass vessel falls from a height on a hard floor, it comes to rest almost instantaneously, i.e. in a very short time, so the floor exerts a large force on the vessel and it breaks. However, if it falls on a carpet, then the time duration, in which the vessel comes to rest, increases, so the carpet exerts less force on the vessel and it does not break.

In simple words: A soft carpet acts like a cushion. It takes longer to stop the glass, which makes the "impact" much gentler. On a hard floor, the stop is instant and violent.

📝 Teacher's Note: This is a perfect example of Impulse. By increasing the time of contact (\( t \)), the average force (\( F \)) decreases for the same change in momentum.

🎯 Exam Tip: Explain this using the logic that Force is inversely proportional to time (\( F \propto 1/t \)) for a fixed change in momentum.

Question 18S
Answer: (a) We pull our hands back while catching a fast moving cricket ball, because by doing so, we increase the time of catch, i.e. increase the time to bring about a given change in momentum, and hence, the rate of change of momentum decreases. Thus, a small force is exerted on our hands by the ball.

In simple words: Moving your hands back gives the ball more time to slow down. Because the slowdown happens over a longer time, it doesn't sting your hands as much.

📝 Teacher's Note: Relate this to why cars have airbags—they prolong the crash time to reduce the impact force on the passenger.

🎯 Exam Tip: Use the phrase "rate of change of momentum" to explain why the force is reduced. This is the scientific term examiners look for.

Question 1M.
Answer: \( Mv \)
In simple words: Momentum is just how heavy something is times how fast it's moving.

📝 Teacher's Note: \( P = mv \) is the fundamental formula for linear momentum.

🎯 Exam Tip: Remember that momentum is a vector quantity and its unit is \( \text{kg m/s} \).

Question 2M.
Answer: \( \text{N s} \)
In simple words: This is the unit for Impulse, which is force times time.

📝 Teacher's Note: Impulse is also equal to the change in momentum.

🎯 Exam Tip: \( 1 \text{ N s} = 1 \text{ kg m/s} \). They are the same unit in different forms.

Question 3M.
Answer: \( F = \frac{\Delta p}{\Delta t} \)
In simple words: This formula says force is just how fast momentum is changing.

📝 Teacher's Note: This is the most accurate statement of Newton's Second Law.

🎯 Exam Tip: The delta symbol (\( \Delta \)) represents "change in."

Question 4M.
Answer: Mass of the body
In simple words: Inertia depends entirely on how much "stuff" (mass) is in the object.

📝 Teacher's Note: Emphasize that speed or volume does not affect inertia; only mass does.

🎯 Exam Tip: If mass doubles, inertia doubles. It's a simple direct relationship.

Question 1N.
Answer:
Mass of the body, \( m = 5\text{kg} \)
Velocity, \( v = 2 \text{ m/s} \)
Linear momentum \( = mv = (5)(2) \text{ kg m/s} \)
\( = 10 \text{ kg m/s} \)
In simple words: To find the "moving power," just multiply the weight (5kg) by the speed (2m/s) to get 10.

📝 Teacher's Note: A simple plug-and-chug numerical to practice the formula \( p=mv \).

🎯 Exam Tip: Always state the formula you are using before doing the calculation.

Question 2N.
Answer:
Linear momentum \( = 0.5 \text{ kg m/s} \)
Mass, \( m = 50 \text{ g} = 0.05 \text{ kg} \)
Velocity \( = \text{Linear momentum/mass} \)
\( = \frac{0.5}{0.05} \text{ m/s} \)
\( = 10 \text{ m/s} \)
In simple words: We work backwards here. If we know the power and the weight, we divide to find out how fast it was going.

📝 Teacher's Note: Converting 50g to 0.05kg is a common step where students make mistakes. Remind them to always use SI units (kg).

🎯 Exam Tip: Check your units first! Don't divide grams into kg-based momentum directly.

Question 3N.
Answer:
Force, \( F = 15 \text{ N} \)
Mass, \( m = 2\text{kg} \)
Acceleration, \( a = F/m \) [ From Newton’s second law]
Or, \( a = (15/2) \text{ ms}^{-2} \)
Or, \( a = 7.5 \text{ ms}^{-2} \)
In simple words: If you push a 2kg block with a force of 15, it will speed up by 7.5 meters per second every second.

📝 Teacher's Note: This is a direct application of \( F=ma \). Acceleration is the "result" of applying force to a mass.

🎯 Exam Tip: Use the unit \( \text{ms}^{-2} \) for acceleration to be technically correct.

Question 4N.
Answer:
Force, \( F = 10 \text{ N} \)
Mass, \( m = 5\text{kg} \)
Acceleration, \( a = F/m \) [ From Newton’s second law]
Or, \( a = (10/5) \text{ ms}^{-2} \)
Or, \( a = 2 \text{ ms}^{-2} \)
In simple words: Pushing a 5kg weight with 10 units of force makes it gain 2 units of speed every second.

📝 Teacher's Note: Compare this to 3N: even with less force, a heavier mass accelerates much more slowly.

🎯 Exam Tip: Always show the division step clearly in numerical problems.

Question 5N.
Answer:
Mass, \( m = 0.5 \text{ kg} \).
Acceleration, \( a = 5 \text{ ms}^{-2} \)
Force, \( F = ma \) [ From Newton’s second law]
Or, \( F = (0.5) (5) \text{ N} = 2.5 \text{ N} \).
In simple words: To make a half-kg ball speed up by 5 every second, you only need 2.5 units of force.

📝 Teacher's Note: Emphasize that force and acceleration always act in the same direction.

🎯 Exam Tip: Write the unit 'N' at the end of your answer for force.

Question 6N.
Answer:
Force, \( F = 10 \text{ N} \)
Mass, \( m = 2 \text{ kg} \)
Time, \( t = 3 \text{ s} \)
Initial velocity, \( u = 0 \text{ m/s} \).

(i) Let \( v \) be the final velocity acquired.
From Newton’s second law,
\( F = ma \).
Or, \( a = F/m = 10/2 = 5 \text{ ms}^{-2} \).
From the 1st equation of motion,
\( a = (v - u)/t \)
Or, \( v = at + u \).
Or, \( v = (5)(3) + 0 = 15 \text{ m/s} \).

(ii) Change in momentum = Final momentum – initial momentum
\( \Delta p = mv - mu \).
Or, \( \Delta p = m (v - u) \).
Or, \( \Delta p = 2 ( 15 - 0) = 30 \text{ kg m/s} \).
In simple words: We push a 2kg block for 3 seconds. First, we find it gains 5 speed every second, so after 3 seconds, it's going at 15. Its moving power changed by 30 units.

📝 Teacher's Note: This problem combines dynamics (\( F=ma \)) with kinematics (\( v=u+at \)). It's a very common exam format.

🎯 Exam Tip: Solve step-by-step. Find acceleration first, then use it to find velocity.

Question 7N.
Answer:
Mass, \( m = 100 \text{ kg} \)
Distance moved, \( s = 100 \text{ m} \)
Initial velocity, \( u = 0 \)

(i) Because the body moves through a distance of 100 m in 5 s,
Velocity of the body = Distance moved / time taken
Velocity \( = (100/5) = 20 \text{ m/s} \)

(ii) From Newton’s third equation of motion,
\( v^2 - u^2 = 2as \).
Or, \( a = (v^2 - u^2) /2s \).
Or, \( a = [ (20^2 - 0^2)/ 2(100) ] \text{ ms}^{-2} \).
Or, \( a = 2 \text{ ms}^{-2} \).

(iii) Force, \( F = ma \)
Or, \( F = (100) (2) \text{ N} \).
Or, \( F = 200 \text{ N} \).
In simple words: A giant 100kg object moves 100 meters. We find it ends up going 20m/s. Using the math of distance and speed, we find it was gaining 2 speed every second. That required a 200 Newton push.

📝 Teacher's Note: Remind students that \( u = 0 \) whenever a body starts from rest.

🎯 Exam Tip: Double check if "time taken" is provided for the whole distance or just a segment. In part (i), we assume constant velocity for that calculation.

Question 8N.
Answer:
Slope of a velocity-time graph gives the value of acceleration.
Here, slope \( = 20/5 = 4 \text{ m/s}^2 \).
Or, acceleration, \( a = 4 \text{ m/s}^2 \).
Force = Mass \( \times \) Acceleration.
Given mass, \( m = 100 \text{ g} = 0.1 \text{ kg} \).
Force \( = (0.1) (4) = 0.4 \text{ N} \).
In simple words: Looking at a graph, we see speed going from 0 to 20 in 5 seconds. That means it gains 4 speed every second. For a small 100g weight, that only needs 0.4 units of force.

📝 Teacher's Note: Interpreting graphs is a vital skill. Slope = Change in Y / Change in X.

🎯 Exam Tip: Mention "Slope of v-t graph = acceleration" to get a bonus mark for explanation.

Question 9N.
Answer:
Mass, \( m = 2 \text{ kg} \)
Initial velocity, \( u = 0 \)
Final velocity, \( v = 2 \text{ m/s} \)
Time, \( t = 0.1 \text{ s} \)
Acceleration = Change in velocity/time
Or, \( a = (v - u) /t \)
Or, \( a = (2 - 0)/ 0.1 = 20 \text{ ms}^{-2} \).
Force = Mass x Acceleration
Or, \( F = (2) (20) = 40 \text{ N} \).
In simple words: We pushed a block to 2 speed in just a blink of an eye (0.1 seconds). Because we did it so fast, it had a huge acceleration of 20, requiring a 40 Newton push.

📝 Teacher's Note: When the time is very small, the force needed is very large. This is the concept of impact.

🎯 Exam Tip: Be careful with decimals like 0.1 in the denominator—it moves the decimal of the numerator to the right.

Question 10N.
Answer:
Mass, \( m = 100\text{g} = 0.1 \text{ kg} \).
Initial velocity, \( u = 30 \text{ m/s} \).
Final velocity, \( v = 0 \).
Time, \( t = 0.03 \text{ s} \).
Acceleration = Change in velocity/time.
Or, \( a = (v - u)/t \).
Or, \( a = (0 - 30)/ 0.03 = -1000 \text{ ms}^{-2} \).
Here, negative sign indicates retardation.
Now, Force = Mass x Acceleration
Or, \( F = (0.1) (1000) = 100 \text{ N} \).
In simple words: A ball hitting something and stopping in an instant is like a mini-explosion of force. It slowed down by 1000 speed units per second!

📝 Teacher's Note: This explains why catching a hard ball with "soft hands" (increasing time) is so important.

🎯 Exam Tip: Always note that a "negative acceleration" is called "retardation" or "deceleration."

Question 11N.
Answer:
Mass, \( m = 480 \text{ kg} \).
Initial velocity, \( u = 54 \text{ km/hr} = 15 \text{ m/s} \).
Final velocity, \( v = 0 \).
Time, \( t = 10 \text{ s} \).
Acceleration = Change in velocity/time.
Or, \( a = (v - u)/t \).
Or, \( a = (0 - 15)/10 = -1.5 \text{ ms}^{-2} \).
Here, negative sign indicates retardation.
Now, Force = Mass x Acceleration
Or, \( F = (480) (1.5) = 720 \text{ N} \).
In simple words: A heavy car stops over 10 seconds. We convert its highway speed to meters per second first. It slowed down gently by 1.5 units each second, requiring 720 units of braking force.

📝 Teacher's Note: To convert km/hr to m/s, multiply by \( 5/18 \).

🎯 Exam Tip: KM/HR to M/S conversion is a standard trap. Do it first before any other step.

Question 12N.
Answer:
Mass, \( m = 50 \text{ gm} = 0.05 \text{ kg} \).
Initial velocity, \( u = 100 \text{ m/s} \).
Final velocity, \( v = 0 \).
Distance, \( s = 2\text{cm} = 0.02 \text{ m} \).

(i) Initial momentum \( = mu = (0.05) (100) = 5 \text{ kg m/s} \).
(ii) Final momentum \( = mv = (0.05) (0) = 0 \text{ kg m/s} \).

(iii) Acceleration, \( a = (v^2 - u^2)/2s \).
Or, \( a = (0^2 - 100^2)/ 2(0.02) \).
Or, \( a = - 2.5 \times 10^5 \text{ ms}^{-2} \).
Therefore, retardation is \( 2.5 \times 10^5 \text{ ms}^{-2} \).

(iv) Force, \( F = ma \)
Or, \( F = (0.05 \text{ kg}) (2.5 \times 10^5 \text{ ms}^{-2}) \)
Or, \( F = 12500 \text{ N} \)
In simple words: A bullet stops inside wood in just 2cm. Its moving power goes from 5 to zero instantly. This creates a massive force of 12,500 Newtons, which is why bullets are so destructive.

📝 Teacher's Note: This problem shows how much force is packed into a tiny, fast-moving bullet.

🎯 Exam Tip: 2cm is a tiny distance; make sure to convert it to 0.02m to get the correct large force value.

Question 13N.
Answer:
Let the force be \( F \).
Force \( F \) causes an acceleration, \( a = 10 \text{ m/s}^2 \) in a body of mass, \( m = 500 \text{ g} \) or \( 0.5 \text{ kg} \)
Thus, \( F = ma \)
Or, \( F = (0.5) (10) = 5 \text{ N} \)
Let \( a’ \) be the acceleration which force \( F (=5\text{N}) \) cause on a body of mass, \( m’ = 5 \text{ kg} \).
Then, \( a’ = F/m’ \).
Or, \( a’ = (5/5) \text{ ms}^{-2} \).
Or, \( a’ = 1 \text{ ms}^{-2} \).
In simple words: We find a secret force by using a small weight. Then we use that same secret push on a 10-times heavier weight. Since it's 10 times heavier, it only speeds up 1/10th as much.

📝 Teacher's Note: This numerical beautifully demonstrates the inverse proportionality between mass and acceleration for a fixed force.

🎯 Exam Tip: Treat this as a two-part problem. Solve for F first, then use that F for the second mass.

Question 14N.
Answer:
Initial velocity, \( u = 30 \text{ m/s} \)
Final velocity, \( v = 0 \)
Time, \( t = 2\text{s} \)
Force, \( F = 1500 \text{ N} \)

Here, \( a = (v - u)/t = (0 - 30)/ 2 = - 15 \text{ ms}^{-2} \). Here, negative sign indicates retardation.
Now, \( F = ma \).
Or, \( m = F/a = (1500/ 15) = 100 \text{ kg} \).

(a) Change in momentum = Final momentum – Initial momentum
Or, \( \Delta p = m (v - u) \)
Or, \( \Delta p = 100 (0 - 30) \)
Or, \( \Delta p = 3000 \text{ kg m/s} \)

(b) Acceleration, \( a = (v - u)/t \).
Or, \( a = (0 - 30)/ 2 = - 15 \text{ ms}^{-2} \),
Here, negative sign indicates retardation.
Thus, retardation \( = 15 \text{ ms}^{-2} \).

(c) From Newton’s second law of motion,
\( F = ma \)
Or, \( m = F/a = (1500/ 15) = 100 \text{ kg} \).
In simple words: We know the braking force and how fast the car stopped. We use that to calculate that the car must weigh 100kg. Its total "moving power" changed by 3000 units.

📝 Teacher's Note: The magnitude of change in momentum is what's usually asked for, so you can drop the negative sign in final answers like 3000.

🎯 Exam Tip: Be consistent with signs. If acceleration is -15, force could also be considered -1500 (opposing motion), giving a positive mass.

Exercise 3(D)

Question 1S.
Answer:
Newton’s third law explains how a force acts on an object.
In simple words: Forces never come alone. If you touch something, it touches you back!

📝 Teacher's Note: Use the term "Interaction" to describe forces between two objects.

🎯 Exam Tip: Focus on the phrase "forces act in pairs."

Question 2S.
Answer:
According to Newton’s third law of motion, to every action there is always an equal and opposite reaction. The action and reaction act simultaneously on two different bodies.
In simple words: If you push a wall, the wall pushes you back with the exact same strength at the exact same time.

📝 Teacher's Note: Clarify that "action" and "reaction" are just names for the two forces in the interaction; either one can be called the action.

🎯 Exam Tip: Mentioning "different bodies" is crucial. Action and reaction never act on the same body, which is why they don't cancel out.

Question 3S.
Answer:
Law of action and reaction: In an interaction of two bodies A and B, the magnitude of action, i.e. the force \( F_{AB} \) applied by the body B on the body A, is equal in magnitude to the reaction, i.e., the force \( F_{BA} \) applied by the body A on the body B, but they are in directions opposite to each other.
Examples:
1. When a book is placed on a table, it does not move downwards. It implies that the resultant force on the book is zero, which is possible only if the table exerts an upward force of reaction on the book, equal to the weight of the book.
2. While moving on the ground, we exert a force by our feet to push the ground backwards; the ground exerts a force of the same magnitude on our feet forward, which makes it easier for us to move.
Explanation: In the above stated example, there are two objects and two forces. In the first example, the weight of the book acts downwards (action) and the force of the table acts upwards (reaction).
In the second example, our feet exerts a force on the ground (action) and the ground exerts an equal and opposite force (reaction) on our feet.
In simple words: Every time there is a push, there is a push back. That push back is what lets us walk forward or keeps a book from falling through a table.

📝 Teacher's Note: Have students stand on skateboards and push off a wall to see the "reaction" move them backwards.

🎯 Exam Tip: When giving examples, clearly identify which force is the action and which is the reaction.

Question 4S.
Answer:
(a) Action: Force exerted on the bullet.
Reaction: Recoil experienced by the gun.
(b) Action: The force exerted by the hammer on the nail.
Reaction: The force applied by the nail on the hammer.
(c) Action: Weight of the book acting downwards.
Reaction: Force acted by the table upwards.
(d) Action: Force exerted by the rocket on the gases backwards.
Reaction: Force exerted by outgoing gases on the rocket in forward direction.
(e) Action: Force exerted by the feet on the ground in backward direction.
Reaction: Force exerted by the ground on feet in forward direction.
(f) Action: Force exerted by a moving train on a stationary train.
Reaction: Force exerted by a stationary train on a moving train.
In simple words: In every situation—shooting a gun, hammering a nail, or a rocket flying—there is always a pair of forces working against each other.

📝 Teacher's Note: This is a great list for "identifying action and reaction" quiz questions.

🎯 Exam Tip: Ensure that your action-reaction pairs involve two distinct objects (e.g., foot and ground, bullet and gun).

Question 5S.
Answer:
When a rocket moves in space, it pushes gases outside, i.e. the rocket applies force on the gases in the backward direction. As a reaction, the gases put equal amount of force on the rocket in the opposite direction and the rocket moves in the forward direction.
In simple words: A rocket works by "throwing" fuel backwards. As it throws fuel out one end, the fuel pushes the rocket forward at the other end.

📝 Teacher's Note: Emphasize that rockets don't need "air" to push against; they push against their own exhaust gases.

🎯 Exam Tip: Use the words "backward direction" for gas and "forward direction" for rocket to show the "opposite" part of the law.

Question 6S.
Answer:
When a man fires a bullet from a gun, a force \( F \) is exerted on the bullet (action), and the gun experiences an equal and opposite recoil (reaction) and hence gets recoiled.
In simple words: The gun pushes the bullet out fast. Because of the push back, the gun kicks into your shoulder.

📝 Teacher's Note: The "recoil" speed is much less than the bullet's speed because the gun is much heavier (\( F = ma \)).

🎯 Exam Tip: Be sure to include both "action" and "reaction" labels in your diagram for full credit.

Question 7S.
Answer: When a man exerts a force (action) on the boat by stepping into it, its force of reaction makes him step out of the boat, and the boat tends to leave the shore due to the force exerted by the man (i.e. action).

Question 8S.
Answer: Couple two spring balances A and B as shown in the figure. When we pull the balance B, both the balances show the same reading indicating that both the action and reaction forces are equal and opposite. In this case, the pull of either of the two spring balances can be regarded as action and that of the other balance as the reaction.
In simple words: When you connect two spring scales and pull them, they both show the same number. This happens because the force you pull with is perfectly matched by a force pulling back.

📝 Teacher's Note: This experiment is the best way to prove that action and reaction are always exactly equal. Ensure students notice that both scales must be identical and zeroed properly before the experiment.

🎯 Exam Tip: Always remember that action and reaction act on two different bodies. In this experiment, Scale A pulls Scale B, and Scale B pulls Scale A.

Question 9S.
Answer: To move a boat, the boatman pushes (action) the water backwards with his oar. In this response, the water exerts an equal and opposite force (reaction) in the forward direction on the boat due to which the boat moves ahead.
In simple words: To move forward in water, you have to push the water behind you. The water then pushes you forward with the same amount of strength.

📝 Teacher's Note: Explain that the water is the medium that provides the reaction force. If there was no water (vacuum), the oars wouldn't be able to push the boat forward.

🎯 Exam Tip: Use the keywords "action", "reaction", and "equal and opposite" to score full marks in this descriptive question.

Question 10S.
Answer: A person pushing a wall hard (action) by his palm, experiences a force (reaction) exerted by the wall on his palm in the opposite direction; thus, he is liable to fall backwards.
In simple words: If you push a wall, the wall pushes you back just as hard. If you push hard enough without bracing yourself, the wall's "push back" can knock you over.

📝 Teacher's Note: Ask students to stand on a skateboard and push a wall. They will move away from the wall, providing a physical proof of the reaction force.

🎯 Exam Tip: Specify that the falling backwards is caused by the reaction force acting on the person's hands.

Question 11S.
Answer: Yes, action and reaction act simultaneously.
In simple words: Action and reaction happen at the exact same moment. One doesn't wait for the other to finish.

📝 Teacher's Note: This is a common point of confusion. There is zero time delay between the action and the reaction force.

🎯 Exam Tip: Use the word "simultaneous" to describe the timing of Newton's third law forces.

Question 12S.
Answer: Yes, action and reaction are equal in magnitude.
In simple words: If you hit a table with 10 units of force, the table hits your hand back with exactly 10 units of force.

📝 Teacher's Note: Magnitude refers to the size or amount of the force. The strength is identical, even if the effects (like damage) are different due to different materials.

🎯 Exam Tip: Mention "equal in magnitude" and "opposite in direction" together for a complete description.

Question 13S.
Answer: When a falling ball strikes the ground, it exerts a force on the ground. The ground exerts a force back at the ball in the opposite direction. This is the reason the ball rises upwards.
In simple words: A ball bounces because it hits the floor, and the floor hits it back, pushing it up into the air.

📝 Teacher's Note: The elasticity of the ball helps convert the kinetic energy into potential energy and back, but the upward push itself comes from the ground's reaction force.

🎯 Exam Tip: Identify the ground as the source of the upward reaction force.

Question 14S.
Answer: The given statement is wrong.
Reason: According to Newton’s third law of motion, the action and reaction act simultaneously on different bodies. Hence they do not cancel each other.
In simple words: Forces only cancel out if they pull on the same object. Since action and reaction pull on different objects, they can't cancel each other out.

📝 Teacher's Note: Use the example of a horse pulling a cart. The horse pulls the cart (Action), and the cart pulls the horse (Reaction). Since these forces are on different objects, the cart still moves.

🎯 Exam Tip: This is a favorite conceptual question. The key phrase is "act on different bodies."

Question 1M.
Answer: Explains the way the force acts on a body.
In simple words: Newton's third law tells us that forces always happen in pairs between two things.

📝 Teacher's Note: This is the qualitative definition of force interaction.

🎯 Exam Tip: Distinguish between the First Law (definition of force), Second Law (measurement of force), and Third Law (nature of force interaction).

Question 2M.
Answer: Different bodies in opposite directions
In simple words: Action and reaction always push or pull on two separate things and in exactly opposite ways.

📝 Teacher's Note: Emphasize that the forces never act on the same object.

🎯 Exam Tip: Look for "different bodies" in the options for any question about action-reaction pairs.

Question 1N.
Answer: The wall exerts an equal force of 10 N on the boy in the opposite direction, i.e. west.
In simple words: If a boy pushes a wall with 10 N toward the east, the wall pushes him back with exactly 10 N toward the west.

📝 Teacher's Note: This numerical demonstrates that reaction is exactly equal to action in value but opposite in vector direction.

🎯 Exam Tip: Always specify the direction (e.g., west) if the action direction is given.

Question 2N.
Answer:
(a) A block exerts 15 N force (weight) on the string downwards.
(b) The string exerts an equal force of 15 N on the block in the opposite direction, i.e. upward direction (tension).
In simple words: The weight pulls the string down, and the string pulls the weight up with the same force to keep it from falling.

📝 Teacher's Note: This upward force in a string is called "Tension." It is an internal molecular reaction to the weight being applied.

🎯 Exam Tip: In diagrams, always draw the tension arrow pointing away from the mass along the string.

Exercise 3(E)

Question 1S.
Answer: Newton’s law of gravitation: Every particle in the universe attracts every other particle with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them, and the direction of the force is along the line joining the masses.
In simple words: Everything in the universe pulls on everything else. The heavier the things are, the stronger the pull. The further apart they are, the weaker the pull.

📝 Teacher's Note: This is a "Universal" law because it applies to everything from two tiny atoms to two giant galaxies.

🎯 Exam Tip: Don't forget to mention that the force acts along the line joining the centers of the two objects.

Question 2S.
Answer: Gravitational force is always attractive.
In simple words: Gravity never pushes things away; it only pulls things together.

📝 Teacher's Note: Contrast this with magnetic or electrical forces, which can both pull (attract) and push (repel).

🎯 Exam Tip: This is a unique property of gravity—it has no "negative" or "repulsive" form.

Question 3S.
Answer: \[ F \propto \frac{Gm_1m_2}{d^2} \]
Here G is a constant of proportionality called the universal gravitational constant.
In simple words: This math formula calculates exactly how strong gravity is. It uses the masses of the two objects and the distance between them.

📝 Teacher's Note: Help students understand that 'G' is a constant that remains the same everywhere in the universe, unlike 'g' (acceleration due to gravity) which changes.

🎯 Exam Tip: Memorize the formula correctly. Mass is multiplied at the top, and distance squared is at the bottom.

Question 4S.
Answer: The gravitational force of attraction between two masses is inversely proportional to the square of distance between them.
In simple words: If you move objects twice as far apart, the gravitational pull doesn't just get half as weak; it gets four times weaker!

📝 Teacher's Note: This is called the "Inverse Square Law." It is a fundamental concept in physics that also applies to light and sound intensity.

🎯 Exam Tip: Use the term "inversely proportional to the square of distance" to get full marks.

Question 5S.
Answer: If the distance between the masses becomes half, the force reduces to one-fourth.
In simple words: Wait, actually the law says if distance is halved, the force increases by four times! (Check formula: \( F \propto 1/d^2 \), if \( d = 1/2 \), then \( F = 1/(1/4) = 4 \)). The PDF text says "reduces to one-fourth" which is a common typo in study guides—be careful!

📝 Teacher's Note: Use this to teach students how to substitute values into formulas to check if a statement is true. Correct the textbook error during class.

🎯 Exam Tip: In calculations, always substitute the change in distance and then square it to find the new force.

Question 6S.
Answer: The gravitational constant is defined as the force of attraction between two bodies of unit mass separated by a unit distance.
In simple words: If you have two 1-kg blocks exactly 1 meter apart, the tiny pull between them is exactly equal to the number G.

📝 Teacher's Note: This is the conceptual way to define 'G'. It gives students a physical idea of what the constant represents.

🎯 Exam Tip: Use the words "unit mass" (1 kg) and "unit distance" (1 m) in your definition.

Question 7S.
Answer: The value of G in the S.I. system is \( 6.67 \times 10^{-11} \text{ Nm}^2\text{kg}^{-2} \).
In simple words: G is an incredibly small number. This is why you don't feel a gravitational pull from your friends or the furniture—the force is too small to notice.

📝 Teacher's Note: Point out the negative exponent to show how small the number is. This helps explain why gravity is considered a "weak" force.

🎯 Exam Tip: Memorize the units (\( \text{Nm}^2\text{kg}^{-2} \)) carefully. They are derived directly from the formula.

Question 8S.
Answer: The gravitational force of attraction is significant to explain the motion of heavenly bodies, e.g. motion of planets around the Sun, motion of the Moon around the Earth etc.
In simple words: Gravity is the "invisible rope" that keeps the moon orbiting Earth and the planets orbiting the Sun.

📝 Teacher's Note: Connect this to centripetal force. Gravity provides the pull needed to keep objects in circular orbits.

🎯 Exam Tip: Give examples of both planetary and lunar orbits to show the significance.

Question 9S.
Answer: The force with which the Earth attracts a body towards its centre is called the force due to gravity.
In simple words: Gravity is the reason things fall down. It's the Earth's massive pull pulling everything toward its center.

📝 Teacher's Note: Distinguish between "Gravitation" (pull between any two masses) and "Gravity" (specifically the pull from Earth or a planet).

🎯 Exam Tip: Mention that the pull is always directed "towards the centre of the Earth."

Question 10S.
Answer: The force due to gravity on a body of mass m kept on the surface of Earth (mass=M and radius=R) is equal to the force of attraction between the Earth and that body.
\[ F = \frac{GMm}{R^2} \]
In simple words: This formula calculates your weight on Earth by using the Earth's mass and how far you are from its center (the radius).

📝 Teacher's Note: This is a specific application of the universal law. Here, 'M' is Earth's mass and 'R' is Earth's radius.

🎯 Exam Tip: Remember that 'R' is the distance from the center of the Earth to the object on the surface.

Question 11S.
Answer: The rate at which the velocity of a freely falling body increases is called acceleration due to gravity. Its S.I. unit is \( \text{m/s}^2 \).
In simple words: When you drop something, it goes faster and faster as it falls. The speed it gains every second is called 'g'.

📝 Teacher's Note: This acceleration is the same for all objects, regardless of their weight, if there is no air resistance.

🎯 Exam Tip: Use the correct unit \( \text{m/s}^2 \) for any type of acceleration.

Question 12S.
Answer: The average value of ‘g’ on the Earth’s surface is \( 9.8 \text{ m/s}^2 \).
In simple words: On Earth, a falling object gains about 9.8 meters per second of speed for every second it falls.

📝 Teacher's Note: Explain that this is an "average" because Earth isn't a perfect sphere—'g' is slightly higher at the poles and lower at the equator.

🎯 Exam Tip: Use 9.8 or 10 in numericals, depending on what the question paper specifies.

Question 13S.
Answer: Let \( g \) be the acceleration due to gravity on the Earth’s surface (mass = M and radius = R).
Then, \( g = \frac{GM}{R^2} \).
In simple words: This shows that gravity on Earth depends only on the Earth's own size and weight, not on how heavy you are.

📝 Teacher's Note: This is a key derivation. It proves that the small 'm' (mass of the object) cancels out, which is why a feather and a hammer fall at the same rate in a vacuum.

🎯 Exam Tip: Remember that \( g \) is inversely proportional to the square of the radius (\( R^2 \)).

Question 14S.
Answer: Acceleration due to gravity (g) is directly proportional to universal gravitational constant (G).
In simple words: If the universe's general gravity constant G were stronger, the gravity pulling you to Earth (g) would also be stronger.

📝 Teacher's Note: This is a direct relationship from the formula \( g = G(M/R^2) \). If G doubles, g doubles.

🎯 Exam Tip: Understand the difference between the constant G and the variable g.

Question 15S.
Answer:
\( h = ut + \frac{1}{2}gt^2 \)
\( h = \frac{1}{2}gt^2 \)
In simple words: This formula tells you how far an object falls (\( h \)) if you just drop it (starting at speed 0) and let it fall for a certain time (\( t \)).

📝 Teacher's Note: In the second version, \( u \) (initial velocity) is 0 because the object is dropped from rest.

🎯 Exam Tip: Always check if the object was thrown (\( u > 0 \)) or dropped (\( u = 0 \)) before picking the formula.

Question 16S.
Answer: If a body is thrown vertically up with an initial velocity u to a height h, then there will be retardation (\( a = -g \)).
At the highest point, the final velocity \( v = 0 \).
Thus, from the third equation,
\( v^2 = u^2 - 2gh \)
\( \implies 0 = u^2 - 2gh \)
\( \implies h_{max} = \frac{u^2}{2g} \)
In simple words: If you throw a ball up, it slows down until it stops at the very top. This formula calculates how high it will go based on how fast you threw it.

📝 Teacher's Note: "Retardation" means slowing down. Here, gravity is working against the upward motion, which is why we use \( -g \).

🎯 Exam Tip: Remember that at the maximum height, the velocity is always zero for a split second.

Question 17S.
Answer: Mass: The mass of a body is the quantity of matter it contains.
Weight: The weight of a body is the force with which the Earth attracts it.
In simple words: Mass is how much "stuff" is in you. Weight is how hard the Earth pulls on that stuff.

📝 Teacher's Note: This is one of the most important distinctions in physics. Use a "Moon" example: your mass stays the same on the Moon, but your weight changes.

🎯 Exam Tip: Define mass as "quantity of matter" and weight as "force of attraction."

Question 18S.
Answer: Mass is a scalar quantity, but weight is a vector quantity. Mass is the measure of the quantity of matter contained in a body, but weight is the measure of force with which the Earth attracts the body. Mass of a body is always constant but weight varies from place to place.
In simple words: Mass doesn't have a direction and never changes. Weight always pulls down and changes depending on where you are (like on a high mountain or another planet).

📝 Teacher's Note: Scalar means just a number. Vector means a number plus a direction (down, in the case of weight).

🎯 Exam Tip: Remember that mass is "constant" and weight is "variable."

Question 19S.
Answer: The S.I. unit of mass is kg and that of weight is newton.

Question 20S.
Answer: \( W = mg \)
At the centre of Earth, \( g = 0 \).
Therefore, \( W = 0 \).
In simple words: If you were at the exact center of the Earth, the planet would pull you in every direction at once, so you would weigh nothing!

📝 Teacher's Note: This is a great conceptual question. The gravitational pull from all parts of the Earth cancels out at the center.

🎯 Exam Tip: Acceleration due to gravity is zero at the center of the Earth.

Question 21S.
Answer: Mass of a body is always constant.
In simple words: Whether you are on Earth, the Moon, or floating in space, the amount of "stuff" that makes you up never changes.

📝 Teacher's Note: This is true for classical physics. Mass is an inherent property of the object.

🎯 Exam Tip: Mass is independent of position and gravity.

Question 22S.
Answer: \( 1 \text{ kgf} = 9.8 \text{ N} \).
One kilogramme force is the force due to gravity on a mass of 1 kilogramme.
In simple words: 1 kgf is just the weight of a 1 kg bag of sugar. On Earth, that weight is equal to about 9.8 Newtons.

📝 Teacher's Note: kgf stands for Kilogram-Force. It's a non-SI unit used to describe weight in terms of mass, making it easier for everyday use.

🎯 Exam Tip: Use the conversion factor 9.8 to turn kgf into Newtons.

Question 1M.
Answer: Always attractive
In simple words: Gravity only pulls things together; it never pushes them apart.

📝 Teacher's Note: This differentiates gravity from other forces like electrostatic force which can repel.

🎯 Exam Tip: Gravity is "strictly attractive" in nature.

Question 2M.
Answer: \( 6.7 \times 10^{-11} \text{ Nm}^2 \text{kg}^{-2} \)
In simple words: This is the universal value for 'G'. It's very small because gravity is a weak force.

📝 Teacher's Note: Ensure students can read scientific notation correctly (the -11 means 10 zeros before the 6).

🎯 Exam Tip: Memorize the value and the units together.

Question 3M.
Answer: \( 6.7 \times 10^{-11} \text{ N} \)
In simple words: This is the force between two 1kg bodies 1 meter apart.

📝 Teacher's Note: This is just a numerical check based on the definition of 'G'.

🎯 Exam Tip: Force is measured in Newtons (N).

Question 4M.
Answer: \( \frac{2u}{g} \)
METHOD: Let ‘t’ be the time in which the body reaches its maximum height.
Initial velocity = u.
Final velocity (at the highest point) = 0.
Acceleration due to gravity = g (negative sign indicates the body is moving against gravity).
Using the first equation of motion,
\( v = u + gt \).
We get,
\( 0 = u - gt \)
Or \( t = u/g \)
Now total time for which the ball remains in air = Time of ascent + Time of descent
Because time of ascent = Time of descent,
Total time taken = \( u/g + u/g = 2u/g \)
In simple words: The time a ball spends in the air is double the time it takes to reach the top. You just divide the starting speed by gravity and multiply by two.

📝 Teacher's Note: This assumes there is no air resistance. In a real-world scenario, the time of descent might be slightly different due to drag.

🎯 Exam Tip: Time of flight is always double the time of ascent.

Question 5M.
Answer: \( 19.6 \text{ m s}^{-1} \)
METHOD: Given, \( u = 0 \)
\( g = 9.8 \text{ m/s}^2 \)
Time \( t = 2\text{s} \)
Let ‘v’ be the velocity of object on reaching the ground.
Using the first equation of motion,
\( v = u + gt \)
We get,
\( v = 0 + (9.8) (2) \)
Or, \( v = 19.6 \text{ m/s} \).
In simple words: If you drop a ball, it gains 9.8 speed every second. After 2 seconds, it will be going exactly \( 9.8 + 9.8 = 19.6 \text{ meters per second} \).

📝 Teacher's Note: This is a standard free-fall numerical. It helps students understand how 'g' works as a multiplier for time.

🎯 Exam Tip: Always state your "Givens" (like \( u=0 \)) at the start of the numerical.

Question 1N.
Answer: Given the force of attraction between two bodies = 10 N
Now, \( F = G\frac{Mm}{R^2} \).
If the new distance \( R' = R/2 \), then let \( F' \) be the force acting between the bodies. Then:
\( F' = G\frac{Mm}{(R/2)^2} = 4G\frac{Mm}{R^2} \)
\( \implies F' = 4F \)
\( \implies F' = 4 \times 10 = 40 \text{ N} \)
In simple words: If you bring two objects twice as close to each other, the pull between them becomes four times stronger.

📝 Teacher's Note: This is the numerical application of the inverse square law. Remind students that squaring a fraction like \( 1/2 \) gives \( 1/4 \), which "flips" to the top to become 4.

🎯 Exam Tip: Be careful with the squaring part! Halving distance doesn't double force; it quadruples it.

Question 2N.
Answer: Weight = \( mg \)
\( W = (5) (9.8) = 50 \text{ N} \).
Assumption: Value of acceleration due to gravity = \( 9.8 \text{ m/s}^2 \).
In simple words: A 5kg bag of rice weighs about 50 Newtons on Earth. You just multiply the mass by gravity.

📝 Teacher's Note: This is the simplest weight calculation. Use it to introduce the unit Newton (N).

🎯 Exam Tip: Always mention your assumption (like \( g = 9.8 \)) if the question doesn't provide it.

Question 3N.
Answer: Mass = 10 kg
(i) Weight (in kgf) = \( 10 \times 1 \text{ kgf} = 10 \text{ kgf} \)
[1 kgf = 9.8 N]
(ii) Weight (in newton) = \( 10 \times 9.8 = 98 \text{ N} \).
In simple words: A 10kg mass weighs 10 kgf. To find Newtons, multiply by 9.8.

📝 Teacher's Note: This teaches students how to convert between the practical unit (kgf) and the scientific unit (Newton).

🎯 Exam Tip: The numerical value of weight in kgf is exactly the same as the mass in kg.

Question 4N.
Answer: Mass = 5 kg.
\( g = 9.8 \text{ m/s}^2 \).
Let F be the force of gravity,
\( F = mg \).
\( F = (5) (9.8) = 49 \text{ N} \).
Force of gravity always acts downwards.
In simple words: The force of Earth's gravity on a 5kg object is 49 Newtons, and it's always pulling straight down toward the ground.

📝 Teacher's Note: Remind students that force is a vector, so they should always specify the direction ("downwards").

🎯 Exam Tip: Always include the direction for force questions if applicable.

Question 5N.
Answer: Weight, \( W = 2.0 \text{ N} \)
\( g = 9.8 \text{ m/s}^2 \)
Let ‘m’ be the mass of the body.
\( W = mg \)
Or, \( m = W/g = (2/9.8) \text{ kg} = 0.2 \text{ kg} \).
In simple words: Mass is the amount of matter in an object. You find it by dividing the object's weight by the strength of gravity pulling on it.

📝 Teacher's Note: When solving weight-to-mass problems, ensure students understand that mass is an inherent property and doesn't change with location, whereas weight does.

🎯 Exam Tip: Always include the units (\( \text{kg} \)) in your final answer to avoid losing marks for incomplete responses.

Question 6N.
Answer: Weight of the body on Earth = \( 98 \text{ N} \).
Acceleration due to gravity on Earth = \( 9.8 \text{ m/s}^2 \).
Let ‘m’ be the mass of the body on Earth.
\( m = W/g \)
\( m = (98/9.8) = 10 \text{ kg} \)
Thus, the mass of the body is \( 10 \text{ kg} \), which always remains constant.
(a) Mass on moon = mass on Earth = \( 10 \text{ kg} \)
(b) Let weight on moon is \( W’ \).
\( W’ = \text{mass} \times \text{acceleration due to gravity on the Moon} \).
[Given, acceleration due to gravity on the Moon = \( 1.6 \text{ m/s}^2 \)]
\( W’ = 10 \times 1.6 = 16 \text{ N} \).
In simple words: An object's mass stays exactly the same whether it's on Earth or the Moon. However, because the Moon's gravity is weaker, the object will weigh much less there.

📝 Teacher's Note: This question tests the fundamental distinction between mass and weight. A good analogy is that mass is the number of "atoms" in a box, which doesn't change just because you move the box.

🎯 Exam Tip: Remember that mass is a constant; if you find the mass on Earth, it is the same everywhere else in the universe.

Question 7N.
Answer: Man’s weight on Earth = \( 600 \text{ N} \)
Man’s weight on the Moon = \( (1/6) \) man’s weight on Earth;
Because the acceleration due to gravity on the Moon is \( 1/6\text{th} \) that of Earth and \( w = mg \).
Therefore, man’s weight on Moon = \( (600/6) = 100 \text{ N} \).
In simple words: Moon's gravity is six times weaker than Earth's. So, whatever you weigh here, you'll weigh six times less on the Moon.

📝 Teacher's Note: Use this example to explain why astronauts can jump so high on the Moon—their muscles are pushing a body that "feels" much lighter.

🎯 Exam Tip: When given a weight on Earth and asked for weight on the Moon, the standard ratio to use is \( 1/6 \).

Question 8N.
Answer: Mass, \( m = 10.5 \text{ kg} \)
\( G = 10 \text{ m/s}^2 \)
(a) Force, \( F = mg \)
\( F = (10.5) (10) = 105 \text{ N} \)
(b) Weight, \( w = mg \)
\( w = (10.5) (10) = 105 \text{ N} \)
In simple words: The force of gravity and the weight of an object are the same thing—they both describe the pull of a planet on an object.

📝 Teacher's Note: Clarify for students that weight is technically the force due to gravity. The calculation is identical.

🎯 Exam Tip: Note the value of \( g \) provided in the question. If it's \( 10 \text{ m/s}^2 \), use that instead of \( 9.8 \text{ m/s}^2 \).

Question 9N.
Answer: Let ‘S’ be the height.
Time taken, \( t = 3\text{s} \); \( g = 9.8 \text{ m/s}^2 \)
Initial velocity, \( u = 0 \) (because the body starts from rest)
(a) Using the second equation of motion,
\( S = ut + (1/2) gt^2 \)
We get,
\( S = 0 + (1/2) (9.8) (3) (3) \)
\( S = 44.1 \text{ m} \)
(b) Let ‘v’ be the velocity with which the ball strikes the ground.
Using the third equation of motion,
\( v^2 - u^2 = 2gs \)
or, \( v^2 - 0^2 = 2(9.8) (44.1) \)
or, \( v^2 = 864.36 \)
or, \( v = 29.4 \text{ m/s} \)
In simple words: If you drop a ball, it speeds up as it falls. We can use the time it takes to fall to calculate both how high it started and how fast it was going when it hit the ground.

📝 Teacher's Note: Remind students that for "dropped" objects, the initial velocity (\( u \)) is always zero.

🎯 Exam Tip: Be careful with squaring the time (\( t^2 \)) in the formula. It's a common area for calculation errors.

Question 10N.
Answer: Mass, \( m = 5\text{kg} \)
Force, \( F = mg \)
\( F = (5) (9.8) = 49 \text{ N} \)
Assumption: Value of acceleration due to gravity is \( 9.8 \text{ m/s}^2 \).
In simple words: The gravitational pull on a \( 5\text{kg} \) block on Earth is \( 49 \text{ Newtons} \).

📝 Teacher's Note: This is a direct application of Newton's Law of Gravitation simplified for the Earth's surface.

🎯 Exam Tip: State your assumed value of \( g \) if it is not explicitly given in the question.

Question 11N.
Answer: Given, maximum height reached, \( s = 20 \text{ m} \)
Acceleration due to gravity, \( g = 10 \text{ m/s}^2 \)
(a) Let ‘u’ be the initial velocity.
At the highest point, velocity \( = 0 \)
Using the third equation of motion,
\( v^2 - u^2 = 2gs \)
or, \( 0 - u^2 = 2 (-10) (20) \text{ m/s} \)
or, \( u^2 = (400) \text{ m/s} \) [Negative sign indicates that the motion is against gravity]
or, \( u = 20 \text{ m/s} \)
(b) Let \( v’ \) be the final velocity of the ball on reaching the ground.
Considering the motion from the highest point to ground,
Velocity at highest point \( = 0 = \) Initial velocity for downward journey of the ball.
Distance travelled, \( s = 20\text{m} \)
Using the third equation of motion,
\( v^2 - u^2 = 2gs \)
or, \( v^2 - 0 = 2 (10) (20) \text{ m/s} \)
or, \( v^2 = 400 \text{ m/s} \)
or, \( v = 20 \text{ m/s} \)
(c) Now total time for which the ball remains in air, \( t = 2u/g \).
Or, \( t = 2 (20)/(10) \).
Or, \( t = 4\text{s} \).
In simple words: When you throw a ball up, it takes the same amount of time to go up as it does to come down. Its speed when it reaches your hand is the same as the speed you threw it with.

📝 Teacher's Note: Emphasize the symmetry of motion under gravity—ascent and descent are mirror images of each other.

🎯 Exam Tip: At maximum height, final velocity is always zero. Use this fact to solve for initial velocity or height.

Question 12N.
Answer: Initial velocity \( u = 0 \)
Final velocity \( = 20 \text{ m/s} \)
\( g = 10 \text{ m/s}^2 \)
Let ‘h’ be the height of the tower.
Using the third equation of motion,
\( v^2 - u^2 = 2gs \)
or, \( (20)^2 - 0 = 2 (10) h \)
or, \( h = 20 \text{ m} \)
In simple words: If a ball dropped from a tower reaches a speed of \( 20 \text{ m/s} \), the tower must be \( 20 \text{ meters} \) high.

📝 Teacher's Note: This demonstrates how to find the displacement using velocity when time is not provided.

🎯 Exam Tip: Ensure you use the correct kinematic equation based on the variables provided (\( u, v, g \)).

Question 13N.
Answer: Total time of journey \( = 6 \text{ s} \)
\( g = 10 \text{ m/s}^2 \)
(i) Let ‘H’ be the greatest height.
Time of ascent, \( t = 6/2 = 3 \text{ s} \),
For ascent, initial velocity, \( u = 0 \)
Using the second equation of motion,
\( H = ut + (1/2) gt^2 \)
\( H = 0 + (1/2) (10) (3)^2 \)
\( H = 45 \text{ m} \)
(ii) Let \( u’ \) be the initial velocity.
Final velocity, \( v = 0 \)
Using the third equation of motion,
\( v^2 - u^2 = 2gH \)
or, \( v^2 - 0 = 2(10) (45) \)
or, \( v^2 = 900 \)
or, \( v = 30 \text{ m/s} \)
In simple words: By knowing how long a ball stayed in the air, we can find out how high it went and how fast it was moving when it was thrown.

📝 Teacher's Note: Total time of flight is twice the time taken to reach the maximum height.

🎯 Exam Tip: Divide the total flight time by \( 2 \) to find the time taken for the ascent or descent phase.

Question 14N.
Answer: Initial velocity, \( u = 20 \text{ m/s} \)
Time, \( t = 2\text{s} \)
\( g = 10 \text{ m/s}^2 \)
Maximum height reached in \( 2\text{s} \), \( H = (1/2) gt^2 \)
Or, \( H = (1/2) (10) (2)^2 \)
Or, \( H = 20 \text{ m} \)
In simple words: Throwing a ball at \( 20 \text{ m/s} \) will send it up exactly \( 20 \text{ meters} \).

📝 Teacher's Note: This is a simplified calculation for height reached when the object comes to rest at the top.

🎯 Exam Tip: If acceleration is against motion (going up), use a negative sign in the general equations, or use the derived peak height formula.

Question 15N.
Answer: (a) Height, \( s = 80\text{m} \)
\( g = 10 \text{ m/s}^2 \)
Using the second equation of motion,
\( S = ut + (1/2) gt^2 \)
Or, \( 80 = 0 + (1/2) (10) (t)^2 \)
Or, \( (t)^2 = 16 \)
Or, \( t = 4\text{s} \)
(b) Let ‘v’ be the velocity on reaching the ground.
Using the third equation of motion,
\( v^2 - u^2 = 2gH \)
or, \( v^2 - 0 = 2(10) (80) \)
or, \( v^2 = 1600 \)
or, \( v = 40 \text{ m/s} \)
In simple words: Dropping a ball from an \( 80 \text{ meter} \) building takes \( 4 \text{ seconds} \), and it will be moving at \( 40 \text{ m/s} \) when it hits the ground.

📝 Teacher's Note: These problems help students visualize gravity as a steady acceleration of \( 10 \text{ m/s} \) every single second.

🎯 Exam Tip: Always show the square root step clearly when solving for time or velocity.

Question 16N.
Answer: Given time \( t = 2.5 \), \( g = 9.8 \text{ m/s}^2 \)
Height, \( H = (1/2) gt^2 \)
Or, \( H = (1/2) (9.8) (2.5)^2 \)
Or, \( H = 30.6 \text{ m} \)
In simple words: If you drop something and it takes \( 2.5 \text{ seconds} \) to land, you dropped it from about \( 30 \text{ meters} \) high.

📝 Teacher's Note: Direct substitution into the distance formula for free fall.

🎯 Exam Tip: Use the precise value of \( g \) (\( 9.8 \)) if provided, otherwise \( 10 \) is usually acceptable.

Question 17N.
Answer: Initial velocity, \( u = 49 \text{ m/s} \)
\( g = 9.8 \text{ m/s}^2 \)
(i) Let H be the maximum height attained.
At the highest point, velocity \( = 0 \).
Using the third equation of motion,
\( v^2 - u^2 = 2gH \)
or, \( 0 - 49^2 = 2(-9.8) (H) \)
or, \( H = (49^2)/ 19.6 \)
or, \( H = 122.5 \text{ m} \)
(ii) Total time of flight is given by \( t = 2u/g \)
Or, \( t = 2(49)/ 9.8 \)
Or, \( t = 10 \text{ s} \)
In simple words: A fast-thrown ball (\( 49 \text{ m/s} \)) will fly for \( 10 \text{ seconds} \) and reach a peak of \( 122.5 \text{ meters} \).

📝 Teacher's Note: This problem emphasizes calculating both vertical distance and time duration for projectile motion.

🎯 Exam Tip: Time of flight is the total duration until the object returns to its starting level.

Question 18N.
Answer: Initial velocity \( u = 0 \)
Time \( t = 4 \text{ s} \)
\( g = 10 \text{ m/s}^2 \)
Let ‘H’ be the height of the tower.
Using the second equation of motion,
\( H = ut + (1/2) gt^2 \)
Or, \( H = 0 + (1/2)(10)(4)^2 \)
Or, \( H = 80 \text{ m} \)
In simple words: A \( 4 \text{ second} \) fall means the object started from an \( 80 \text{ meter} \) height.

📝 Teacher's Note: Another check on the relationship between time and displacement in free fall.

🎯 Exam Tip: "Dropped" always implies initial velocity \( u = 0 \).

Question 19N.
Answer: (i) Time \( t = 20 \text{ s} \)
\( g = 10 \text{ m/s}^2 \)
Let ‘D’ be the depth of the well.
Using the second equation of motion,
\( D = ut + (1/2) gt^2 \)
\( D = 0 + (1/2)(10)(20)^2 \)
\( D = 2000 \text{ m} \)
(ii) Speed of sound \( = 330 \text{ m/s} \)
Depth of well \( = 2000 \text{ m} \)
Time taken to hear the echo after the pebble reaches the water surface = Depth/speed
\( = (2000/330) \text{ s} \)
\( = 6.1 \text{ s} \)
Time taken for pebble to reach the water surface \( = 20 \text{ s} \).
Therefore, the total time taken to hear the echo after the pebble is dropped \( = 20 + 6.1 = 26.1 \text{ s} \).
In simple words: If you drop a stone into a deep well, you first wait for it to fall (\( 20 \text{s} \)), and then you wait for the sound of the splash to travel back up to your ears (\( 6 \text{s} \)).

📝 Teacher's Note: This is a sophisticated problem that combines free-fall motion with constant speed sound wave travel.

🎯 Exam Tip: Total time for an echo includes both the falling time of the object and the upward travel time of the sound.

Question 20N.
Answer: Let x be the height of the tower.
Let h be the distance from the top of the tower to the highest point as shown in the diagram.
Initial velocity \( u = 19.6 \text{ m/s} \)
\( g = 9.8 \text{ m/s}^2 \)
At the highest point, velocity \( = 0 \)
Using the third equation of motion,
\( v^2 - u^2 = 2gh \)
Or, \( - (19.6)^2 = 2 (-9.8) h \)
Or, \( h = 19.6 \text{ m} \)
If the ball takes time \( t_1 \) to go to the highest point from the top of building, then for the upward journey from the relation, \( v = u - gt \),
\( 0 = 19.6 - (9.8) (t_1) \)
Or, \( t_1 = 2\text{s} \)
(ii) Let us consider the motion for the part (x+h)
Time taken to travel from highest point to the ground \( = (5 - 2) = 3\text{s} \)
Using the equation \( s = ut + (1/2) gt^2 \)
We get,
\( (x + h) = 0 + (1/2) (9.8) (3)^2 \)
Or, \( (x + 19.6) = 44.1 \text{ m} \)
Or, \( x = 44.1 - 19.6 = 24.5 \text{ m} \)
Thus, height of the tower \( = 24.5 \text{ m} \)
(iii) Let v be the velocity of the ball on reaching the ground.
Using the relation, \( v = u + gt \)
We get:
\( v = 0 + (9.8) (3) \)
Or, \( v = 29.4 \text{ m/s} \)
In simple words: This problem tracks a ball from the top of a building, up to its peak, and then all the way down to the ground. We use the split time to find out how tall the building is.

📝 Teacher's Note: Breaking complex motion into upward and downward segments makes it much easier to solve.

🎯 Exam Tip: Draw a clear diagram for multi-stage motion problems to track heights and velocities correctly.