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

Unit I

Practice Problems 1

Question 1. Calculate the velocity of a body of mass 0.5 kg, when it has a linear momentum of 5 Ns.
Answer: Mass of body = \( m = 0.5 \text{ kg} \)
Linear momentum = \( P = 5 \text{ Ns} \)
Velocity of body = \( v = ? \)
We know \( P = mv \)

\( v = \frac{P}{m} = \frac{5}{0.5} \)
\( v = 10 \text{ ms}^{-1} \)
In simple words: Momentum is just weight multiplied by speed. To find the speed, we divide the momentum by the weight of the object.

📝 Teacher's Note: When solving momentum problems, always ensure mass is in kg and momentum in Ns or kg m/s to get velocity in m/s.

🎯 Exam Tip: Always state the formula \( P = mv \) before substituting values to ensure step marks.

Question 2. An electron of mass \( 9 \times 10^{-31} \text{ kg} \) is moving with a linear velocity of \( 6 \times 10^7 \text{ ms}^{-1} \). Calculate the linear momentum of electron.
Answer: Mass of electron = \( m = 9 \times 10^{-31} \text{ kg} \)
Velocity of electron = \( v = 6 \times 10^7 \text{ ms}^{-1} \)
Momentum of electron = \( P = ? \)
We know, \( P = mv \)
\( P = 9.1 \times 10^{-31} \times 6 \times 10^7 \)
\( P = 54 \times 10^{-24} \text{ kg ms}^{-1} \) or \( \text{Ns} \)
\( = 54 \times 10^{-24} \text{ kg ms}^{-1} \text{ Ns} \)
In simple words: Even though an electron is tiny, it moves so fast that it still has a calculable amount of momentum. We just multiply its mass and speed together.

📝 Teacher's Note: Help students practice multiplication with powers of ten. Remind them that \( 10^a \times 10^b = 10^{a+b} \).

🎯 Exam Tip: Units for momentum can be written as either \( \text{kg ms}^{-1} \) or \( \text{Ns} \); both are correct.

Question 3. A body of mass 200 g is moving with a velocity of \( 5 \text{ ms}^{-1} \). If the velocity of the body changes to \( 17 \text{ ms}^{-1} \), calculate the change in linear momentum of the body.
Answer: Mass of body = \( m = 200 \text{ g} = 0.2 \text{ kg} \)
Velocity = \( v_1 = 5 \text{ ms}^{-1} \); Velocity = \( v_2 = 17 \text{ ms}^{-1} \)
Change in linear momentum of body
\( = mv_2 - mv_1 = m (v_2 - v_1) \)
\( = 0.2(17-5) \)
\( = 2.4 \text{ Ns} \) or \( \text{kg ms}^{-1} \)
In simple words: Change in momentum is the weight of the object times the difference in its speed. It shows how much more "push" the object has after speeding up.

📝 Teacher's Note: A common mistake is using mass in grams. Always convert grams to kilograms by dividing by 1000 before starting calculations.

🎯 Exam Tip: When velocity changes, use the factored form \( m(v-u) \) to simplify your math and reduce errors.

Question 4. A motorcycle of mass 100 kg is running at \( 10 \text{ ms}^{-1} \). If its engine develops an extra linear momentum of 2000 Ns, calculate the new velocity of motorcycle.
Answer: Mass of motor cycle = \( m = 100 \text{ kg} \)
Velocity of motor cycle = \( v_1 = 10 \text{ ms}^{-1} \)
Momentum of motor cycle = \( mv_1 = 100 \times 10 = 1000 \text{ Ns} \)
When engine develops an extra linear momentum of 2000 Ns
Then total momentum of motor cycle = \( 1000 + 2000 = 3000 \text{ Ns} \)
Let \( v_2 = \) new velocity of the motor cycle.
Total momentum of motor cycle = \( mv_2 \)
\( 3000 = 100 \times v_2 \)

\( v_2 = \frac{3000}{100} = 30 \text{ ms}^{-1} \)
In simple words: The bike already had some momentum. The engine added more. We add them together to find the new total and then find the new speed.

📝 Teacher's Note: Emphasize that "extra linear momentum" is added to the "initial momentum," not replaced by it.

🎯 Exam Tip: Draw a "Before" and "After" state for momentum to keep track of the values clearly.

Practice Problems 1:

Question 1. A car initially at rest, picks up a velocity of \( 72 \text{ kmh}^{-1} \) in 20 seconds. If the mass of the car is 1000 kg, find (1) Force developed by its engine (2) Distance covered by the car.
Answer: Initial velocity = \( u = 0 \)
Final velocity = \( v = 72 \text{ kmh}^{-1} \)

\( v = 72 \times \frac{5}{18} \text{ ms}^{-1} = 20 \text{ ms}^{-1} \)
Mass of car = \( m = 1000 \text{ kg} \)
Time = \( t = 20 \text{ s} \)
(1)
\( v = u + at \)
\( 20 = 0 + a(20) \)

(Acceleration) \( a = \frac{20}{20} = 1 \text{ ms}^{-2} \)
Now, \( F = ma \)
\( F = 1000 \times 1 = 1000 \text{ N} \)
(2)
Distance covered by car = \( S = ? \)
\( S = ut + \frac{1}{2} at^2 \)
\( = 0(20) + \frac{1}{2} (1) (20)^2 = 0 + \frac{400}{2} \)
\( S = 200 \text{ m} \)
In simple words: First, we find out how much the car speeds up every second (acceleration). Then we multiply that by its weight to find the force. Finally, we use a motion formula to see how far it travelled.

📝 Teacher's Note: Converting km/h to m/s is mandatory when time is given in seconds. Multiply by \( 5/18 \) for a quick conversion.

🎯 Exam Tip: "Initially at rest" is a keyword meaning \( u = 0 \). Identifying these hidden values is crucial.

Question 2. A golfer hits a ball at rest, such that the contact between the ball and golf stick is for 0.1 s. If the golf ball covers a linear distance of 400 m in 2 s, find the magnitude of force applied. Mass of golf ball is 50 g.
Answer: Distance covered by ball = 400 m
Time taken = 2 s

\( \therefore \text{Uniform velocity of the ball} = \frac{\text{Distance}}{\text{Time}} = \frac{400}{2} = 200 \text{ ms}^{-1} \)

Now initial velocity of the ball = \( u = 0 \)
Time for which force acts on the ball = \( t = 0.1 \text{ s} \)
Final velocity of the ball after the force stops acting on it = \( v = 200 \text{ ms}^{-1} \)
Using, \( v = u + at \)
We have, \( 200 = 0 + a (0.1) \)

\( a = \frac{200}{0.1} = 2000 \text{ ms}^{-2} \)
Mass of the ball = \( m = 50 \text{ g} = 0.05 \text{ kg} \)
Force acting on the ball = \( F = ma \)
\( F = 0.05 \times 2000 \)
\( F = 100 \text{N} \)
In simple words: The ball was hit so hard it reached a huge speed almost instantly. We calculated that speed from its long flight, then figured out the force needed to reach that speed in just a tiny fraction of a second.

📝 Teacher's Note: This problem distinguishes between the time the force acts (0.1s) and the time the ball travels (2s). Force calculations only use the time the force is actually being applied.

🎯 Exam Tip: Always check units. 50g must be 0.05kg. Using 50 will give an answer 1000 times too large!

Practice Problems 2:

Question 1. A car of mass 800 kg, moving at \( 54 \text{ kmh}^{-1} \) is brought to rest over a distance of 15 m. Find the retarding force developed by the brakes of the car.
Answer: Mass of car = \( m = 800 \text{ kg} \)
Initial velocity of car = \( u = 54 \text{ kmh}^{-1} \)

\( u = 54 \times \frac{5}{18} \text{ ms}^{-1} = 15 \text{ ms}^{-1} \)
Final velocity of car = \( v = 0 \)
Distance covered = \( S = 15 \text{ m} \)
Retarding force = \( F = ? \)
\( v^2 - u^2 = 2aS \)
\( (0)^2 - (15)^2 = 2a(15) \)
\( 30 a = -225 \)
\( a = -7.5 \text{ ms}^{-2} \)
\( F = ma = (800) (-7.5) \)
\( F = -6000 \text{N} \)
In simple words: Retarding force is just "braking force." We find out how much the car slows down (negative acceleration) and use that to find the force applied by the brakes.

📝 Teacher's Note: The negative sign in acceleration and force indicates that they are acting opposite to the direction of motion.

🎯 Exam Tip: If the question asks for "retarding force," you can state the final answer as a positive value, noting that "retardation" implies the direction.

Question 2. A cricket player holds a cricket ball of mass 100 g by moving his hands backward by 0.75 m. If the initial velocity of the ball is \( 108 \text{ kmh}^{-1} \), find the retarding force applied by the player.
Answer: Mass of ball \( m = 100 \text{ g} = 0.1 \text{ kg} \)
Distance covered = \( S = 0.75 \text{ m} \)
Initial velocity of ball = \( u = 108 \text{ kmh}^{-1} \)

\( u = 108 \times \frac{5}{18} \text{ ms}^{-1} = 30 \text{ ms}^{-1} \)
Final velocity of ball = \( v = 0 \)
Retarding force = \( F = ? \)
\( v^2 - u^2 = 2aS \)
\( (0)^2 - (30)^2 = 2a (0.75) \)
\( 1.5 a = -900 \)
Retarding force = \( F = ma \)
\( F = 0.1 \times (-600) \)
\( F = -60 \text{ N} \)
In simple words: When a player catches a ball, they pull their hands back to stop it over a longer distance. This reduces the force hitting their hands. Here, the force used to stop the fast ball was 60 Newtons.

📝 Teacher's Note: Relate this to Newton's Second Law: increasing the distance (and thus time) taken to stop the ball reduces the impact force.

🎯 Exam Tip: Be careful with the math for \( 2 \times 0.75 = 1.5 \). Mental math errors often happen with decimals.

Practice Problems 3:

Question 1. A force of 600 dynes acts on a glass ball of mass 200 g for 12 s. If initially the ball is at rest, find (1) Final velocity (2) Distance covered.
Answer: \( F = 600 \text{ dyne} = 600 \text{ g cms}^{-2} \)
\( m = 200 \text{ g} \)
\( \text{Acceleration } a = \frac{F}{m} = \frac{600 \text{ g cms}^{-2}}{200 \text{ g}} \)
\( a = 3 \text{ cms}^{-2} \)
Also time = \( t = 12 \text{ s} \)
Initial velocity = \( u = 0 \)
(1)
Final velocity = \( v = ? \)
\( v = u + at \)
\( v = 0 + 3 (12) \)
\( v = 36 \text{ cms}^{-1} \)
(2)
Distance covered = \( S = ? \)
\( v^2 - u^2 = 2as \)
\( (36)^2 - (0)^2 = 2(3)s \)

\( S = \frac{36 \times 36}{6} = 36 \times 6 = 216 \text{ cm} \)
In simple words: This problem uses the CGS system (grams and centimeters). We find the acceleration first, then use it to find the final speed and total distance the ball rolled.

📝 Teacher's Note: In CGS, 1 Dyne = \( 1 \text{ g cm/s}^2 \). It's important to keep all units in CGS to avoid having to convert to SI mid-problem.

🎯 Exam Tip: If the question is given in Dynes and Grams, keep your answer in cm/s and cm unless specifically asked for SI units.

Question 2. A bullet of mass 30 g, and moving with a velocity x hits a wooden target with a force of 187.5 N. If the bullet penetrates 80 cm, find the value of x
Answer: Mass of bullet = \( m = 30 \text{ g} = \frac{30}{1000} \text{ kg} = 0.03 \text{ kg} \)
Force = \( F = 187.5 \text{ N} \)
Acceleration = \( a = \frac{F}{m} \) [because of retardation]
\( a = -\frac{187.5}{0.03} \)
\( a = -\frac{18750}{3} = -6250 \text{ ms}^{-2} \)
Initial velocity = \( u = x \)
Final velocity = \( v = 0 \)
Distance covered by bullet = \( S = 80 \text{ cm} = \frac{80}{100} \text{ m} \)
\( S = 0.8 \text{m} \)
\( v^2 - u^2 = 2as \)
\( (0)^2 - (x)^2 = 2 \times (-6250) \times 0.8 \)
\( -x^2 = -10000 \)
\( x^2 = 10000 \)
\( x = \sqrt{10000} = 100 \text{ ms}^{-1} \)
In simple words: The bullet stops inside the wood, so we know the final speed is zero. We use the force and mass to find the "braking" power of the wood and work backwards to find how fast the bullet was going at the start.

📝 Teacher's Note: Point out the negative sign for acceleration again. The wooden target provides resistance, which is a retarding force.

🎯 Exam Tip: Always convert cm to m in SI problems. \( 80 \text{ cm} = 0.8 \text{ m} \).

Question 3. A car of mass 1000 kg develops a force of 500 N over a distance of 49 m. If initially the car is at rest find (1) Final velocity (2) time for which it accelerates.
Answer: Mass of car = \( m = 1000 \text{ kg} \)
Force = \( F = 500 \text{ N} \)

\( \text{Acceleration of car } a = \frac{F}{m} = \frac{500}{1000} = 0.5 \text{ ms}^{-2} \)

Initially velocity = \( u = 0 \)
Distance covered = \( S = 49 \text{ m} \)
Final velocity = \( v = ? \)
Time = \( t = ? \)
(1)
\( v^2 - u^2 = 2a S \)
\( v^2 - (0)^2 = 2 (0.5) (49) \)
\( v^2 = 49 \)
\( v = 7 \text{ ms}^{-1} \)
(2)
\( v = u + at \)
\( 7 = 0 + 0.5t \)
\( 0.5 t = 7 \)

\( t = \frac{7}{0.5} = 14 \text{ s} \)
In simple words: The car gets a steady push. We find its acceleration, then use its travel distance to find the final speed, and finally calculate how long it took to reach that speed.

📝 Teacher's Note: If students get stuck with \( v = \sqrt{49} \), remind them of basic squares. Square roots in physics problems are often perfect squares like 49, 100, or 144.

🎯 Exam Tip: Solve parts in order. Part (1) gives you the final velocity needed to solve Part (2).

Exercise 1

Question 1. (a) Define force.
(b) State four effects which a force can bring about. Give two examples in each case.
Answer: (a) Force : It is defined as an external cause which changes or tends to change the state of rest or uniform motion of a body in a straight line.
(b) Effects of force : A force can bring about the following effects.

1. Force can set a stationary body into motion.
For example :
1. A player can set a ball (at rest) in motion by hitting it with suitable material like hockey.
2. A magnet can move an iron nail.

2. Force can stop the moving bodies.
For example :
1. A speeding car is stopped by the force of friction of brakes.
2. A rolling football stops because of the force of friction from the ground.

3. Force can change the speed or direction of a moving body.
For example :
1. A stone projected vertically upwards changes its speed and direction of motion because of the force of gravity.
2. A moving bicycle starts running faster, when more force is applied on its peddles.

4. Force can change the dimensions of a body.
For example :
1. Lenght of a rubber band increases, when stretching force is applied on it.
2. We clay can be moulded in any shape by applying a force with hands.
In simple words: Force is a push or pull. It can make things move, stop them, turn them, or even squish and stretch them.

📝 Teacher's Note: Use classroom demos like stretching a spring or pushing a toy car to illustrate these effects visually.

🎯 Exam Tip: When listing effects, always pair them with an example to show a complete understanding.

Question 2. What do you understand by the following terms?
(a) Contact forces (b) Non-contact forces
Answer: (a) The forces which act on bodies when they are in actual contact, are called the contact forces. Two examples are as below:
1. Force of friction
2. Force of tension.

(b) The forces which act on bodies without being physically touched, are called the non-contact forces or the forces at a distance. The gravitational force, electrostatic force and magnetic force are the non-contact forces.
In simple words: Contact forces need a "touch," like pushing a box. Non-contact forces work from a distance, like a magnet pulling a nail or Earth pulling an apple down.

📝 Teacher's Note: Explain that non-contact forces act through "fields" (gravitational, magnetic, etc.).

🎯 Exam Tip: "Force at a distance" is a good synonym to use for non-contact forces.

Question 3. (a) What do you understand by the term inertia?
(b) What are its kinds?
(c) Give two examples of each kind, stated in (b).
Answer: (a) Inertia. “Is the property of a body due to which it cannot change its state (rest or of uniform motion) itself.” Untill some external force is applied on it.
(b) The three kinds of inertia are:
1. Inertia of rest,
2. Inertia of motion,
3. Inertia of direction.

(1) Inertia of rest: The tendency of a body to continue in its state of rest, even when some external unbalanced force is applied on it, is called inertia of rest.

(2) Inertia of motion: The tendency of a body to continue in its state of motion, in a straight line, even when some external unbalanced force acts on it, is called inertia of motion.

(3) Inertia of direction: The tendency of a body by which it is unable to change its direction of motion, even when some external unbalanced force acts on it, is called inertia of direction.

(c) Examples of inertia of rest:
1. Imagine a pile of books placed on a sheet of paper. If the paper is suddenly pulled with a jerk, the books are left behind, because of the inertia of rest.
2. When a carpet is suddenly jerked, the dust flies off. Because on the sudden movement, the carpet moves, but dust on account of the inertia of rest, is left behind.

Examples of inertia of motion:
1. A man standing in a moving bus falls forward as soon as tiie bus stops, due to the inertia of motion of upper part of his body.
2. Before taking a long jump, a boy runs a certain distance, because in doing so he picks up the inertia of motion, which helps him in taking a longer leap.

Examples of inertia of direction:
1. It is a common experience that passengers tend to fall sideways, when a speeding bus takes a sharp turn. It is because, when the bus is moving along straight line in a particular direction suddenly takes a sharp turn, the passengers on account of inertia of direction continue along their direction and hence fall sideways.
2. The sparks produced during sharpening of a knife against a grinding wheel leave the rim of the wheel tangentially because of inertia of direction.
In simple words: Inertia is "laziness." Objects want to keep doing what they are already doing. If they are still, they want to stay still. If they are moving, they want to keep moving in the same direction.

📝 Teacher's Note: The "Bus" examples are the most relatable. Explain how our feet move with the bus (due to friction) but our bodies stay put (due to inertia).

🎯 Exam Tip: Be precise with terminology: "tendency of a body" is a key phrase in the definition of inertia.

Question 4. What do you understand by the term momentum?
Answer: Momentum : The instantaneous force which body possesses due to combined effect of mass and velocity is called momentum of the body. Mathematically, momentum is the product of mass and velocity of body.
\( P = mv \)
where P represents the momentum of the body.
In simple words: Momentum is the "strength" of a moving object. A heavy truck has more momentum than a toy car, even at the same speed.

📝 Teacher's Note: Use the analogy of catching a tennis ball vs. a bowling ball at the same speed to explain momentum intuitively.

🎯 Exam Tip: State both the verbal definition and the mathematical formula \( P = mv \).

Question 5. State two factors which determine the momentum of a body.
Answer: Factors which determine the momentum of a body are :
1. Mass of the moving body : Larger the mass of a body, larger will be its momentum.
2. Velocity of the body : Larger the velocity of the body, larger will be its momentum.
In simple words: How heavy an object is and how fast it moves are the two things that decide its momentum.

📝 Teacher's Note: Show the direct proportionality: \( P \propto m \) and \( P \propto v \).

🎯 Exam Tip: If asked to explain, mention that doubling either mass or velocity will double the momentum.

Question 6. State units of momentum in (1) CGS system (2) SI system.
Answer: 1. In CGS system, unit of momentum is \( \text{g cms}^{-1} \).
2. In SI system, unit of momentum is \( \text{kgms}^{-1} \).
In simple words: Units are the "labels" we use. In science, we use grams/centimeters for small things and kilograms/meters for standard things.

📝 Teacher's Note: Remind students that unit of Force is Newton, but Momentum is \( \text{kg m/s} \). Don't mix them up.

🎯 Exam Tip: Ensure the superscripts (\( ^{-1} \)) are written clearly to avoid confusion with division signs.

Question 7. Define Newton’s second law of motion.
Answer: Newton’s second law of motion states that rate of change of momentum is directly proportional to force applied and takes place in the direction of force.
In simple words: The harder you push something, the faster its speed changes. This change always happens in the same direction you pushed it.

📝 Teacher's Note: This is the law that gives us the most important formula in physics: \( F = ma \).

🎯 Exam Tip: The "rate of change" is the key phrase here. It implies time is a factor.

Question 8. Prove mathematically \( F = ma \).
Answer: Derivation of \( F = ma \) from Newton’s Second Law of Motion:
Newton introduced the concept of momentum and say “The momentum of a moving body is defined as the product of its mass and velocity.”
Thus, \( p = mv \), where \( p = \) momentum of body
\( m = \) mass of body
\( v = \) velocity of body
Suppose the velocity of body of mass \( m \) changes from \( u \) to \( v \) in time \( t \).
Initial momentum, \( p_1 = mu \)
and final momentum, \( p_2 = mv \)
the change in momentum, \( (p_2-p_1) \) takes place in time \( t \). Then according to Newton’s second law of motion, the magnitude of force F is:

\( \frac{P_2 - P_1}{t} \propto F \), or \( F = \frac{km(v - u)}{t} \)

where \( k = \) constant of proportionality

Now, \( a = \frac{(v - u)}{t} \), where \( a = \) acceleration of body

\( \therefore F = kma \)
\( \implies F = ma \) (\( \because k = 1 \), constant)
This relation holds good when mass of the body remains constant.
In simple words: Force equals mass times acceleration. This proof shows that force is what causes an object's momentum to change over time.

📝 Teacher's Note: Explain that \( k \) is chosen as 1 based on the definition of a Newton (the force that gives 1kg mass an accel of \( 1 \text{ m/s}^2 \)).

🎯 Exam Tip: This derivation is very common. Practice the steps: change in momentum \( \rightarrow \) rate of change \( \rightarrow \) substitution of \( a \).

Question 9. Define absolute units of force in CGS as well as SI system.
Answer: Absolute unit of force in CGS system is dyne and in SI system is Newton (N).
One dyne : When the body of mass 1 gram moves with an acceleration of \( 1 \text{ cms}^{-2} \), then the force acting on the body is called one dyne.
\( 1 \text{ dyne} = 1 \text{ g cms}^{-2} \)
One Newton : When a body of mass 1 kg moves with an acceleration of \( 1 \text{ ms}^{-2} \), then force acting on the body is said to be one newton.
OR
That force is said to be one newton, which producers an acceleration of \( 1 \text{ ms}^{-2} \) in a body of mass 1 kg.
\( 1 \text{ N} = 1 \text{ kg ms}^{-2} \)
In simple words: A Newton is the "standard" push needed to speed up 1 kg by 1 meter per second every second. A dyne is the tiny version using grams and centimeters.

📝 Teacher's Note: Relate the "Newton" to the weight of a small apple (approx. 100g) to give students a physical sense of the unit.

🎯 Exam Tip: Definitions of "1 Newton" and "1 Dyne" are frequently asked. Memorize them verbatim.

Question 10. Derive the relation between newton and dyne.
Answer: Relation between newton and dyne :
\( 1 \text{ N} = 1 \text{ kg} \times \text{ms}^{-2} = 1000 \text{ g} \times 100 \text{ cm s}^{-2} \)
\( 1 \text{ N} = 100000 \text{ g cm s}^{-2} \)
\( 1 \text{ N} = 10^5 \text{ g cm s}^{-2} \)
\( 1 \text{ N} = 10^5 \text{ dynes} \) (\( \because 1 \text{ dyne} = 1 \text{ g cm s}^{-2} \))
In simple words: One Newton is equal to 100,000 dynes. It's like saying one dollar is 100 cents, but on a much larger scale!

📝 Teacher's Note: Show the conversion of kg to g (\( 10^3 \)) and m to cm (\( 10^2 \)) explicitly to show how we get \( 10^5 \).

🎯 Exam Tip: The conversion factor \( 10^5 \) is the critical part of the answer.

Question 11. State Newton’s third law of motion and give two examples.
Answer: Newton’s third law of motion states “to every action there is an equal and opposite reaction.” It is useful for rocket propulsion.

Examples :
(1) Consider a book lying on the table. Its weight (w) acts vertically downward (Action on the table) and book does not fall. That means table is exerting equal force on the book, but in opposite direction [normal reaction R] called reaction. Thus, action and reaction are equal and opposite.
(2) When we swim in water, we push the water backward [Action] and water in turn exerts equal force on us but in opposite direction (Reaction) i.e. there are two different objects man and water and two forces equal in magnitude and opposite in direction.
In simple words: You can't touch something without it touching you back just as hard. If you push a wall, the wall pushes you back with the same force.

📝 Teacher's Note: Clarify that action and reaction forces *always* act on two different bodies. That is why they don't cancel each other out.

🎯 Exam Tip: "Equal and opposite" are the two most important words in this definition.

Question 12. Explain the following :
1. Why do we jerk wet clothes before spreading them on a line?
2. Why does dust fly off, when carpet is hit with a stick?
3. Why do fruits fall off the branches in the strong wind?
4. Why does a pillion rider fall forward, when the driver of a two-wheeler suddenly applies the brakes?
5. Why does a boatman push the bank backward with a long bamboo pole, on launching his boat in water?
6. Why is it difficult to walk on marshy ground?
7. Why is it dangerous to jump out of a moving vehicle? How can the danger be minimised?
8. Why does a boat-man push water backward with the oars, while rowing a boat?
Answer:
1. When clothes are suddenly jerked, the dust flies off.
2. Because on the sudden movement, the clothes moves, but dust on account of the inertia of rest, is left behind.
3. When carpet is beaten with a stick, then carpet comes in motion but dust particles present on them try to remain at rest because of inertia of rest and hence the dust fly off. (hi) Strong winds slake the branches of a tree, laden with fruits, vigorously. As a result, branches come in motion but fruits try to remaining at rest due to inertia of rest and hence get detached from the tree and fall off.
4. When the driver of a two wheeler suddenly applies the brakes, then lower part of pillion rider comes to rest but his upper part remain in motion due to inertia of motion. As a result, pillion rider falls forward.
5. On launching his boat in water, a boatman push the bank backward with a long bamboo pole. As a result bank offers equal and opposite reaction and hence the boat move.
6. It becomes difficult to walk on marshy ground because when we push the marshy ground with our feet, the ground yields. So it does not react back with same force.
7. It is dangerous to jump out of a moving vehicle. Because when we jump out of a moving vehicles, then our feet will suddenly come to rest, while the rest of the body will be in the state of motion and hence, one can fall down and get seriously injured. We can minimise this danger by running along with the moving bus and in the same direction in which the bus is moving.
8. A boatman push water backwards with the oars. As a reaction, water pushes the boat in forward direction with the same force.
In simple words: Most of these are explained by "laziness" (inertia)—objects wanting to stay still or keep moving. The boat examples are explained by "pushing back" (action-reaction).

📝 Teacher's Note: Use these reasoning questions to differentiate between Newton's First Law (Inertia) and Third Law (Action/Reaction).

🎯 Exam Tip: In your answer, clearly identify which law of motion explains the situation (e.g., "Due to inertia of rest...").

UNIT II - EXERCISE 2

(A) Objective Questions

Question 1. The mass of earth is \( 6 \times 10^{24} \text{ kg} \) and radius of earth is \( 6.4 \times 10^6 \text{ m} \). The magnitude of force between the mass of 1 kg and the earth is :
(a) 770 N
(b) 9.810 N
(c) 830 N
(d) 9.790 N
Answer: (a) 9.770 N
Explanation :
Mass of earth = \( m_1 = 6 \times 10^{24} \text{ kg} \)
Mass of the body = \( m_2 = 1 \text{ kg} \)
Radius of earth = \( r = 6.4 \times 10^6 \text{ m} \)
Universal gravitational constant = \( G = 6.67 \times 10^{-11} \text{ Nm}^2 \text{ kg}^{-2} \)
Force = \( F = ? \)
\( F = \frac{G m_1 m_2}{r^2} = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1}{(6.4 \times 10^6)^2} \)
\( = \frac{6.67 \times 6 \times 10^{-11-12+24}}{6.4 \times 6.4} \)
\( F = \frac{6.67 \times 6 \times 10}{6.4 \times 6.4} \)
\( F = 9.770 \text{ N} \)
In simple words: This is the formula for gravity. It shows that Earth pulls on a 1 kg block with a force of about 9.8 Newtons.

📝 Teacher's Note: This force is exactly what we call the "weight" of the 1 kg object on Earth's surface.

🎯 Exam Tip: Learn the value of \( G \) (\( 6.67 \times 10^{-11} \)) as it is rarely provided in the question itself.

Question 1=2. A man is walking from east to west on a rough surface. The force on the man is directed :
(a) from west to east
(b) from east to west
(c) along the north
(d) along the west
Answer: (a) from west to east
Explanation: Force of friction always acts in a direction opposite to direction of motion.
In simple words: Friction is a "resistance" force. If you move West, friction tries to pull you back East.

📝 Teacher's Note: This is a tricky concept. While friction *helps* us walk forward, the frictional force *on* our feet is opposite to the direction of our stride.

🎯 Exam Tip: Remember: Friction = Opposition. It always points against the movement.

Question 3. Inertia is the property of a body by virtue of which the body is :
(a) unable to change by itself the state of rest
(b) unable to change by itself the state of uniform motion
(c) unable to change by itself the direction of motion
(d) unable to change by itself the state of rest or uniform motion.
Answer: (d) unable to change by itself the state of rest or uniform motion.
In simple words: Inertia means an object cannot change what it is doing (staying still or moving) unless something else pushes or pulls it.

📝 Teacher's Note: Choice (d) is the most complete answer because it covers all aspects of Newton's First Law.

🎯 Exam Tip: Always look for the most comprehensive definition in MCQs.

Question 4. The impulse of a body is equal to :
(a) rate of change of its momentum
(b) change in its momentum
(c) the product of force applied on it and the time of application of force.
(d) both (b) and (c).
Answer: (d) both (b) and (c).
In simple words: Impulse is like a "quick hit." It is calculated by Force times Time, and it results in a change in the object's momentum.

📝 Teacher's Note: Impulse (\( J = F \Delta t = \Delta p \)) links force and time to the resulting change in motion.

🎯 Exam Tip: Impulse and Momentum have the same units (\( \text{Ns} \) or \( \text{kg m/s} \)).

Question 5. A force acts on a body of mass 3 kg such that its velocity changes from \( 4 \text{ ms}^{-1} \) to \( 10 \text{ ms}^{-1} \). The change in momentum of the body is
(a) \( 42 \text{ kgms}^{-1} \)
(b) \( 2 \text{ kgms}^{-1} \)
(c) \( 18 \text{ kgms}^{-1} \)
(d) \( 14 \text{ kgms}^{-1} \)
Answer: (c) 18 kgms⁻¹
Explanation :
Mass = \( m = 3 \text{ kg} \)
Velocity \( v_1 = 4 \text{ ms}^{-1} \)
Velocity \( v_2 = 10 \text{ ms}^{-1} \)
Change in momentum of body
\( = mv_2 - mv_1 = m (v_2 - v_1) = 3 (10 - 4) \)
\( = 3 \times 6 = 18 \text{ kgms}^{-1} \)
In simple words: Subtract the start speed from the end speed and multiply by the weight. \( 3 \text{ kg} \times 6 \text{ m/s difference} = 18 \).

📝 Teacher's Note: Simple subtraction of velocities before multiplication is the easiest path to the answer.

🎯 Exam Tip: Check your subtraction twice; \( 10 - 4 \) is a very easy place to make a silly mistake under exam pressure.

Question 6. Action-reaction forces
(a) act on the same body
(b) act on different bodies
(c) act along different lines
(d) act in same direction
Answer: (b) act on different bodies
In simple words: If you kick a ball, the action is on the ball, but the reaction is on your foot. Two different things!

📝 Teacher's Note: This is why action and reaction don't cancel out. Forces only cancel if they act on the same single object.

🎯 Exam Tip: This is the most important characteristic of Newton's Third Law to remember for conceptual questions.

Question 7. Which of the following are vector quantities?
(a) Momentum
(b) Velocity
(c) Force
(d) All of the options
Answer: (d) All of the options
Explanation : All those quantities which have magnitude as well as direction are known as vector quantities.
In simple words: For all these three, it matters "which way" they are going. Direction is a key part of their identity.

📝 Teacher's Note: Contrast these with scalars like mass or time, where "direction" makes no sense.

🎯 Exam Tip: If you find two correct answers in a list like this, "All of the options" is almost certainly the right choice.

Question 8. A woman drawing water from a village well, falls backward, when the rope snaps. This is on account of
(a) Newton’s third law of motion
(b) Newton’s law of gravitation
(c) Newton’s second law of motion
(d) Newton’s first law of motion
Answer: (a) Newton’s third law of motion
Explanation : To every action, there is an equal and opposite reaction.
In simple words: The woman was pulling the rope (Action). The rope was pulling her (Reaction). When the rope broke, the "Action" vanished, and the "Reaction" pushed her backward.

📝 Teacher's Note: Use the "tug-of-war" analogy. When one side lets go, the other side falls due to the sudden loss of the opposing force.

🎯 Exam Tip: Any situation involving a sudden recoil or backward movement is usually an example of the Third Law.

Question 9. When you kick a stone, you get hurt. Due to which property this happens?
(a) Inertia of stone
(b) Velocity of the kick
(c) Momentum of the kick
(d) Reaction of the stone.
Answer: (d) Reaction of the stone.
Explanation : The property of a body by which it is unable to change its state of rest or of uniform motion by itself, even when some external force is applied on it, is called inertia.
In simple words: You hurt your foot because the stone pushes back on your foot just as hard as you kicked it.

📝 Teacher's Note: While the explanation text mentions inertia, the *reason* you feel pain is the reaction force applied *by* the stone *on* your foot.

🎯 Exam Tip: Pain or impact felt by the "attacker" is always a result of the reaction force.

(B) Subjective Questions

Question 1. State Newton’s law of gravitation.
Answer: This law states that the force of attraction acting between two particles is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
\( i.e. F \propto m_1 m_2 \)
And \( F \propto \frac{1}{r^2} \)
Combining (i) and (ii), we have
\( F \propto \frac{m_1 m_2}{r^2} \)
or \( F = \frac{G m_1 m_2}{r^2} \)
Where G is the constant of proportionality and is known as the universal gravitational constant. Its value at all places in this universe is \( 6.67 \times 10^{-11} \text{ Nm}^2 \text{ kg}^{-2} \).
In simple words: Everything in the universe pulls on everything else. Heavier things pull harder, and things that are far away pull much weaker.

📝 Teacher's Note: Emphasize the word "Universal." It means this rule works for two apples, two planets, or two galaxies identically.

🎯 Exam Tip: Don't forget the "square" of the distance. It is an inverse-square law, which is a very common point of error.

Question 2. How is acceleration due to gravity related to (1) mass of a planet (2) distance of body from the center of earth?
Answer: Acceleration due to gravity : The acceleration of a freely falling body, under the action of gravity of earth, is called acceleration due to gravity.
Consider a body of mass ‘m’ on the surface of earth such that it falls towards it with an acceleration ‘g’.
Let, M = mass of the earth
d = distance between center of earth and center of the body
F = Force acting on the body
\( F = mg \) ...(i)
\( F = G \frac{Mm}{d^2} \) ...(ii)
Comparing (i) and (ii)
\( mg = G \frac{Mm}{d^2} \)
\( g = \frac{GM}{d^2} \)
From this equation, it is clear that
(1) \( g \propto M \)
Acceleration due to gravity on the earth is directly proportional to the mass of that earth.
(2) \( g \propto \frac{1}{d^2} \)
Acceleration due to gravity is inversely proportional to square of the distance of the body from centre of the earth.
In simple words: A bigger planet has stronger gravity. Also, the higher up you go (away from the center), the weaker the gravity gets.

📝 Teacher's Note: Note that the small 'm' (mass of the object) cancels out. This proves that all objects fall with the same acceleration regardless of their own weight.

🎯 Exam Tip: The relation \( g = GM/R^2 \) is the fundamental link between mass, radius, and gravity. Master this derivation!

Question 3. (a) What do you understand by the term mass?
(b) State two important characteristics of mass.
(c) State units of mass in CGS and SI systems.
(d ) Name the device used for measuring mass.
Answer: (a) Mass: The quantity of matter contained in a body is known as its mass.
(b) Characteristics of mass :
1. It is independent of the position and surrounding of a body.
2. It is a scalar quantity.
3. It remains constant at all places, provided the velocity of the body is not too high.
(c) CGS unit of mass is gram (g). SI unit of mass is kilogram (kg).
(d) Physical balance is used to measure the mass of a body.
In simple words: Mass is how much "stuff" is inside you. It doesn't change even if you go to the moon or float in space.

📝 Teacher's Note: Mass is an intrinsic property. Contrast this with weight to help students understand the difference clearly.

🎯 Exam Tip: Remember: Mass = Physical Balance; Weight = Spring Balance.

Question 4. (a) What do you understand by the term weight?
(b) State two important characteristics of weight
(c) State the units of weight in CGS and SI system.
(d) Name the device used for measuring weight
Answer: (a) Weight: Weight of a body is defined as the force with which the earth attracts it.
(b) Characteristics of weight :
1. It depends upon the position and surroundings of body.
2. It is a vector quantity.
3. If changes from place to place on the surface of earth due to the change in the value of acceleration due to gravity (g).
(c) In CGS system, unit of weight is dyne. In SI system, unit of weight is newton (N).
(d) Spring balance is used to measure the weight of the body.
In simple words: Weight is the "pull" of Earth on you. If you go to a planet with weaker gravity, your weight changes, even though your body (mass) is the same.

📝 Teacher's Note: Since weight is a force, its units are the same as force (Newtons). It always acts towards the center of the planet.

🎯 Exam Tip: If a problem gives mass, you must multiply by 'g' to find the weight. \( W = mg \).

Question 5. State four differences between mass and weight.
Answer:

In simple words: Mass is the amount of stuff; Weight is the strength of gravity's pull. Mass stays same; Weight changes depending on where you are.

📝 Teacher's Note: This table is the most common 3-mark question in this unit. Ensure students know at least 4 clear points.

🎯 Exam Tip: Mentioning the SI unit and the measuring device are the two easiest points to score marks.

Question 6. Does a body weigh same at all places of the Earth? Give a reason for your answer.
Answer: No, A body does not weigh same at all places of the earth because value of acceleration due to gravity (g) is different at different places on the surface of earth.
In simple words: Earth is slightly flattened at the poles, so you are closer to the center there. Gravity pulls slightly harder at the poles than at the equator.

📝 Teacher's Note: The Earth is an oblate spheroid. Radius is smaller at poles, so \( g \) is larger, and thus weight is larger.

🎯 Exam Tip: The keyword "acceleration due to gravity (g)" is essential in your reasoning.

Question 7. Why is gold not weighed by a spring balance?
Answer: Spring balance measures the weight of a body. Weight of a body changes from place to place on the surface of earth due to change in value of acceleration due to gravity. That is why gold is not weighed by a spring balance.
In simple words: A spring balance measures "pull," which changes based on location. Since gold is very expensive, we must measure its actual "amount" (mass) using a beam balance so the deal is fair everywhere.

📝 Teacher's Note: Traders would buy gold at the equator and sell it at the poles for a profit if they used spring balances!

🎯 Exam Tip: Explain that a beam balance compares two masses and is independent of local gravity.

Question 8. A man sits in a machine which generates acceleration five times more than acceleration due to gravity. If the mass of man is 80 kg, what is his weight? Take \( g = 10 \text{ ms}^{-2} \).
Answer: Mass of the man = \( m = 80 \text{ kg} \)
Acceleration due to gravity = \( g = 10 \text{ ms}^{-2} \)
Acceleration generated by the machine = \( a = 5 g = 5 \times 10 = 50 \text{ ms}^{-2} \)

Case-I: When machine accelerates downward
Effective weight of the man = \( mg - ma = m(g - a) \)
\( = 80 (50-10) = 3200 \text{ N} \)

Case-II : When the machine accelerates upward
Effective weight of man = \( mg + ma = m (g + a) \)
\( = 80 (10 + 50) = 80 \times 60 = 4800 \text{ N} \)
In simple words: When the machine zooms up, you feel much heavier (4800 N). When it drops down fast, your effective weight changes too.

📝 Teacher's Note: Use the "elevator" analogy. You feel heavier when an elevator starts moving up and lighter when it starts moving down.

🎯 Exam Tip: Always analyze the direction of acceleration. Upward = Add, Downward = Subtract.

Question 9. A man weighs 800 N the at the equator. How does the weight of man change at the following places?
(a) At poles
(b) 100 km up in space
(c) 10 km down in a mine.
Answer: Weight of the man at equator = 800 N
(a) At poles : As value of g increases at poles as compared to that at equator, so weight of the body at poles will be more than 800 N.
(b) As the value of g decreases with increase in height, so weight of the man 100 km up in the space is less than 800 N.
(c) As value of g decreases with increase in depth, so weight of the man 10 km down in a mine is less than 800 N.
In simple words: You are heaviest at the surface (poles), and you lose weight as you move away from the surface, whether you go up into space or down into a hole.

📝 Teacher's Note: It's a common misconception that gravity increases as you go down. Actually, the mass of earth "above" you starts pulling you up, cancelling out some of the pull from below.

🎯 Exam Tip: Gravity is maximum at the surface (at the poles specifically) and decreases with both height and depth.

Question 10. How is weight affected in the following cases, when initially the body is weighed in Delhi with a spring balance?
1. Body is taken to Moscow.
2. Body is taken to Ceylon.
3. Body is taken to sea level.
4. Body is taken to a high mountain.
5. Body is taken deep inside a mine.
Answer:
1. When body is taken to Moscow, then weight of the body weighs slightly more than Delhi.
2. When body is taken to Ceylon, then weight of the body is slightly more than at Delhi.
3. When body is taken to sea level, then its weight remains same.
4. As the value of g increases with increase in height of body from surface of the earth, so when a body is taken to a high mountain, its weight decreases.
5. As the value of g decreases with depth, so when body is taken deep inside a mine, its weight decreases.
In simple words: Moving towards the poles (Moscow) increases weight. Going higher (Mountains) or deeper (Mines) decreases weight.

📝 Teacher's Note: Ceylon is closer to the equator than Delhi, so weight should technically decrease. The provided solution text says "more," likely a local typo in the source material. Explain the latitude effect to students.

🎯 Exam Tip: Distance from the center of the Earth is the determining factor for these geographical questions.

Question 11. Describe, briefly how can you calculate the value of ‘g’ with a simple pendulum.
Answer: Simple pendulum : A simple pendulum consists of a small heavy mass in the form a bob suspended by a light inelastic string.
The pendulum is suspended from a suitable support (the thread may be held firmly between two halves of a cork held by a clamp and stand). The pendulum is allowed to oscillate and its time period for one oscillation is noted with the help of a stopwatch by observing a large number of oscillations. The length of the pendulum is changed several times and the time period is determined in each case. A graph is plotted between l, the length of the pendulum, and \( T^2 \), the square of time period.
We know, time period of a simple pendulum \( = T = 2\pi\sqrt{\frac{l}{g}} \)
where \( l = \) length of the pendulum and \( g = \) acceleration due to gravity.

\( T = 2\pi\sqrt{\frac{l}{g}} \)
Squaring both sides,
\( T^2 = 4\pi^2 \left( \frac{l}{g} \right) \implies g = 4\pi^2 \left( \frac{l}{T^2} \right) \)

By substituting the value of \( \frac{l}{T^2} \) in this equation, we can calculate the value of acceleration due to gravity.
In simple words: We time how long it takes for a weight on a string to swing. By changing the string length and timing it again, we can use a special formula to figure out exactly how strong Earth's gravity is.

📝 Teacher's Note: The slope of the \( l \) vs \( T^2 \) graph is exactly \( g / 4\pi^2 \). This is a standard lab experiment for Class 9/10.

🎯 Exam Tip: In the derivation, "Squaring both sides" is the most critical step. Don't skip it!