Laws of Motion - Questions & Answers

Explained, Step by Step

EXERCISE-3(A)

(A) MULTIPLE CHOICE TYPE :
(Choose the correct answer from the options given below).

1. A force can change :
(a) the size or shape of a body
(b) the state of rest or of motion of a body
(c) the dimensions of a body
(d) All of the above
Step 1: Force can alter physical dimensions (size/shape).
Step 2: Force can also change the state of motion or rest.
Answer: (d) All of the above

2. The non-contact force is :
(a) force of reaction (b) force due to gravity (c) tension in string (d) force of friction
Step 1: Gravity acts on objects from a distance without physical contact.
Answer: (b) force due to gravity

3. A ball placed on a table starts rolling down when the table is tilted. This is an example of :
(a) contact force (b) non-contact force (c) both (a) and (b) (d) none of these
Step 1: The earth pulls the ball downwards via gravity.
Step 2: Gravity acts without physical connection.
Answer: (b) non-contact force

4. When a comb is rubbed on dry hair, it gets charged. This is an example of :
(a) Gravitational force (b) normal reaction force (c) magnetic force (d) electrostatic force
Step 1: Rubbing creates static electrical charges.
Answer: (d) electrostatic force

5. The nature of gravitational force is always :
(a) repulsive (b) attractive (c) depends on the bodies involved (d) None of the above
Step 1: Masses always pull each other together.
Answer: (b) attractive

6. The magnitude of a non-contact force between two bodies ......... with an increase in their separation and ..... with a decrease in their separation.
(a) increases, decreases (b) decreases, increases (c) increases, increases (d) decreases, decreases
Step 1: Non-contact forces weaken over distance.
Step 2: So force decreases as separation increases.
Step 3: Force increases as separation decreases.
Answer: (b) decreases, increases

7. On doubling the separation between two bodies, the magnitude of the non-contact force between them becomes :
(a) one-eighth (b) one-half (c) one-fourth (d) remains the same
Step 1: Force F is inversely proportional to r².
Step 2: If r becomes 2r, F becomes 1/(2r)² = 1/4 of F.
Answer: (c) one-fourth

8. A spring balance works on the principle of :
(a) gravitational force (b) magnetic force (c) restoring force (d) frictional force
Step 1: A stretched spring pulls back to its original shape.
Answer: (c) restoring force

9. A horizontal spring with two object A and B attached to its two ends are shown in the figure below. If the spring is compressed, it ..... each object and if it is stretched, it ...... each object.
(a) pushes, pulls (b) pulls, pushes (c) pushes, pushes (d) pulls, pulls
Step 1: Compression creates an outward force (pushes).
Step 2: Stretching creates an inward force (pulls).
Answer: (a) pushes, pulls

(B) VERY SHORT ANSWER TYPE :

1. Classify the following amongst contact and non-contact forces :
(a) frictional force, (b) normal reaction force, (c) force of tension in a string, (d) gravitational force, (e) electrostatic force, (f) magnetic force.
Step 1: Contact forces need physical touch (friction, normal, tension).
Step 2: Non-contact forces act at a distance (gravity, electrostatic, magnetic).
Answer: Contact force : (a), (b) and (c). Non-contact force : (d), (e) and (f).

2. Define a contact force.
Step 1: Note the requirement of physical touch for these forces.
Answer: The forces which are applied on bodies by making a physical contact with them are called contact forces.

3. Define a non-contact force.
Step 1: Note the lack of physical touch needed for these forces.
Answer: The forces experienced by bodies even without being physically in touch, are called non-contact forces.

4. Give one example in each case where : (a) the force is of contact, and (b) force is at a distance.
Step 1: Think of physical touch (friction).
Step 2: Think of action at a distance (gravity).
Answer: (a) Frictional force between a rolling ball and the ground. (b) Gravitational force between the earth and the moon.

5. State one factor on which the magnitude of a non-contact force depends. How does it depend on the factor stated by you ?
Step 1: Identify distance as the key factor.
Step 2: Inverse relationship with distance.
Answer: Distance of separation. The magnitude of force decreases as the distance increases.

6. The separation between two masses is reduced to half. How is the magnitude of gravitational force between them affected ?
Step 1: Force ∝ 1/r².
Step 2: r becomes r/2, so (r/2)² = r²/4.
Step 3: Force becomes 4 times the original.
Answer: Force will become four times.

7. Give one example in each of the following cases where a force : (a) stops a moving body. (b) moves a stationary body. (c) changes the size of a body. (d) changes the shape of a body.
Step 1: Stopping motion: Catching a ball.
Step 2: Creating motion: Pushing a book.
Step 3: Changing size: Stretching a spring.
Step 4: Changing shape: Squeezing a rubber ball.
Answer: (a) A fielder catching a moving ball. (b) Pushing a book resting on a table. (c) Stretching a spring. (d) Squeezing a rubber ball.

(C) SHORT ANSWER TYPE :

1. Explain by giving two examples each of : (a) Contact forces, and (b) Non-contact forces.
Step 1: Contact requires touch (friction, normal reaction).
Step 2: Non-contact works across space (gravity, electrostatic).
Answer: (a) Contact forces require physical contact. Examples: Frictional force, Normal reaction force. (b) Non-contact forces act from a distance without physical touch. Examples: Gravitational force, Electrostatic force.

2. State the effects of a force applied on (i) a non-rigid, and (ii) a rigid body. How does the effect of the force differ in the two cases ?
Step 1: Non-rigid bodies can deform and move.
Step 2: Rigid bodies cannot deform, they only move.
Answer: (i) On a non-rigid body, a force can change its dimensions (size or shape) and can produce motion. (ii) On a rigid body, a force only produces motion without changing its dimensions.

3. What is a frictional force ? Draw a neat labelled diagram showing a frictional force acting on a body.
Step 1: Friction opposes relative motion between surfaces.
Step 2: Diagram involves a block, push force, and opposite friction force.
Answer: When a body slides over a rough surface, a force starts acting on the body in a direction opposite to its motion, called frictional force. (Diagram: A block on a rough surface being pushed; an opposing force labelled 'Friction' acts in the opposite direction).

4. What is a normal reaction force ? Draw a neat labelled diagram showing a normal reaction force acting on a body.
Step 1: Normal reaction is the perpendicular push back from a surface.
Step 2: Balances the downward weight.
Answer: When a body exerts a force on a surface, the surface exerts an equal and opposite force on the body normal to the surface, called normal reaction force. (Diagram: A block resting on a table with Weight acting downwards and Reaction acting upwards).

(D) LONG ANSWER TYPE :

1. (a) A ball is hanging by a string from the ceiling of the roof. Draw a neat labelled diagram showing the forces acting on the ball and the string.
Step 1: Gravity pulls down (weight).
Step 2: String pulls up (tension).
Answer: Forces acting are the weight of the ball acting vertically downwards and the tension in the string acting vertically upwards.

(b) A spring is compressed against a rigid wall. Draw a neat and labelled diagram showing the forces acting on the spring.
Step 1: Hand pushes spring towards wall.
Step 2: Spring pushes back with restoring force.
Answer: The applied pushing force acts towards the wall, and the restoring force developed in the spring acts in the opposite direction away from the wall.

(c) A wooden block is placed on a table top. Name the forces acting on the block and draw a neat and labelled diagram to show the point of application and direction of these forces.
Step 1: Identify downward gravity (weight) from center.
Step 2: Identify upward support force (normal reaction) from contact surface.
Answer: Forces acting are the weight of the block acting vertically downwards from its center of gravity, and the normal reaction force exerted by the table acting vertically upwards from the surface of contact.

EXERCISE-3(B)

(A) MULTIPLE CHOICE TYPE :
(Choose the correct answer from the options given below).

1. Newton's first law includes :
(a) the definition of inertia (b) the definition of force (c) both (a) and (b) (d) none of the above
Step 1: 1st law describes inertia (resistance to change).
Step 2: 1st law defines force qualitatively as the cause of change.
Answer: (c) both (a) and (b)

2. Force is a ......... quantity.
(a) scalar (b) vector (c) directionless (d) none of these
Step 1: Force has both magnitude and specific direction.
Answer: (b) vector

3. The property of an object by virtue of which it tends to retain its state of rest or of motion is called :
(a) friction (b) inertia (c) force (d) mass
Step 1: The natural resistance to change in motion is inertia.
Answer: (b) inertia

4. A student has four balls — table tennis ball, tennis ball, cricket ball and a football. The correct ascending order of inertia is :
(a) table tennis ball, tennis ball, cricket ball, football
(b) cricket ball, football, tennis ball, table tennis ball
(c) table tennis ball, football, cricket ball, tennis ball
(d) football, cricket ball, table tennis ball, tennis ball
Step 1: Inertia increases with mass.
Step 2: Order of mass is TT ball < Tennis ball < Cricket ball < Football.
Answer: (a) table tennis ball, tennis ball, cricket ball, football

5. Which one of the following is not an example of inertia of rest ?
(a) When a train suddenly starts moving forward, the passenger standing in the compartment tends to fall backwards.
(b) When a hanging carpet is beaten with a stick, dust particles start falling out of it.
(c) On shaking the branches of a tree, the fruits fall down.
(d) When a passenger jumps out of a moving train, he falls down.
Step 1: Option (d) involves a moving train, so the person is already in motion.
Step 2: This is inertia of motion, not rest.
Answer: (d) When a passenger jumps out of a moving train, he falls down.

6. An athlete often runs before taking a long jump. This is an example of :
(a) Newtons' first law (b) inertia of rest (c) inertia of motion (d) both (a) and (c)
Step 1: Running builds up forward velocity.
Step 2: Inertia of motion helps carry the athlete further (Newton's 1st law).
Answer: (d) both (a) and (c)

7. In inertia of motion, the body continues to be in a state of motion with ........ speed in ......... direction unless an external force is applied.
(a) variable, zigzag (b) same, same (c) same, variable (d) variable, same
Step 1: No external force means velocity does not change.
Step 2: Constant velocity means same speed and same direction.
Answer: (b) same, same

(B) VERY SHORT ANSWER TYPE :

1. Name the physical quantity which causes motion in a body.
Step 1: A push or pull changes motion.
Answer: Force.

2. Is force needed to keep a moving body in motion ?
Step 1: By 1st law, uniform motion needs no net force.
Answer: No.

3. A ball is moving on a perfectly smooth horizontal surface. If no force is applied on it, will its speed decrease, increase or remain unchanged ?
Step 1: Smooth surface means no friction force.
Step 2: Without force, motion remains constant.
Answer: will remain unchanged.

4. What is Galileo's law of inertia ?
Step 1: Objects resist change to their state of motion.
Answer: If a body is at rest, it remains at rest, and if it is moving, it will continue to move in the same direction with the same speed unless an external force is applied on it.

5. State Newton's first law of motion.
Step 1: Reiterate Galileo's law formally.
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 it 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.

6. Give qualitative definition of force on the basis of Newton's first law of motion.
Step 1: Force is the agent required to change states of motion.
Answer: Force is that external cause which tends to change the state of rest or the state of motion of an object.

7. Name the factor on which inertia of a body depends and state how it depends on the factor stated by you.
Step 1: Inertia is determined solely by mass.
Step 2: Direct relationship.
Answer: Mass; more the mass, more is the inertia of the body.

8. Name the two kinds of inertia.
Step 1: Resistance from standstill.
Step 2: Resistance while moving.
Answer: (1) Inertia of rest, and (2) Inertia of motion.

9. Give one example of each of the following : (a) inertia of rest or static inertia, and (b) inertia of motion or dynamic inertia.
Step 1: (a) Body wants to stay put: Train starts, you fall back.
Step 2: (b) Body wants to keep moving: Jump from moving train, you fall forward.
Answer: (a) When a train suddenly starts, passengers tend to fall backwards. (b) A passenger jumping out of a moving train falls forward.

10. An aeroplane is moving uniformly at a constant height under the action of two forces (i) upward force (lift) and (ii) downward force (weight). What is the net force on the aeroplane ?
Step 1: Uniform motion means zero acceleration.
Step 2: Zero acceleration requires zero net force.
Answer: Zero.

(C) SHORT ANSWER TYPE :

1. A ball moving on a table top eventually stops. Explain the reason.
Step 1: The table isn't perfectly smooth.
Step 2: Friction acts against the motion.
Answer: Force of friction between the ball and table top opposes the motion.

2. What is meant by the term inertia ?
Step 1: Define the inherent resistance property.
Answer: The property of an object by virtue of which it tends to retain its state of rest or of motion, is called inertia.

3. Give two examples to show that greater the mass, greater is the inertia of the body.
Step 1: Example comparing light vs heavy moving objects.
Step 2: Example comparing pushing light vs heavy stationary objects.
Answer: (1) A heavy cricket ball is harder to stop than a light tennis ball moving at the same speed. (2) A loaded trolley is harder to push than an empty one.

4. 'More the mass, more difficult it is to move the body from rest'. Explain this statement by giving an example.
Step 1: Mass is a direct measure of inertia.
Step 2: Pushing a heavier object requires overcoming higher inertia.
Answer: Mass is a measure of inertia. The heavier a body, the greater its tendency to resist any change in its state of rest. For example, pushing a stationary car requires much more effort than pushing a stationary bicycle.

5. Two equal and opposite forces act on a stationary body. Will the body move ? Give reason to your answer.
Step 1: Equal opposites cancel each other out.
Step 2: Net force = 0, so state of rest remains.
Answer: No. Net force on the body is zero, so the body will remain stationary due to inertia of rest.

6. Two equal and opposite forces act on a moving object. How is its motion affected ? Give reason.
Step 1: Forces cancel, making net force zero.
Step 2: Uniform motion continues unaffected.
Answer: Motion remains unaffected. Reason : Net force on the object is zero.

7. Why does a person fall when he jumps out from a moving train ?
Step 1: Feet hit the ground and stop immediately.
Step 2: Upper body continues moving forward due to inertia.
Answer: Because his feet come to rest immediately on touching the ground, while his upper body remains in motion due to inertia of motion.

8. Why does a coin placed on a card, drop into the tumbler when the card is rapidly flicked with the finger ?
Step 1: Finger applies force only to the card.
Step 2: Coin resists motion due to inertia of rest and drops down.
Answer: The card moves away due to the applied force, but the coin does not share this motion and stays at its place due to inertia of rest, ultimately falling into the tumbler due to gravity.

9. Why does a ball thrown vertically upwards in a moving train, come back to the thrower's hand ?
Step 1: The ball has the train's forward velocity.
Step 2: It retains this horizontal speed in the air due to inertia.
Answer: Because the ball shares the forward motion of the train due to inertia of motion even while it is in the air, moving ahead by the same distance as the person.

10. People often shake branches of a tree for getting down its fruits. Why ?
Step 1: Shaking moves the branches quickly.
Step 2: Fruits want to stay at rest (inertia) and detach.
Answer: When branches are shaken they come into motion, but the fruits remain in a state of rest due to inertia. Thus they get detached and fall down due to gravity.

(D) LONG ANSWER TYPE :

1. State and explain the law of inertia (or Newton's first law of motion).
Step 1: State the law clearly.
Step 2: Explain it defines force and inertia qualitatively.
Answer: Statement: If a body is at rest, it remains at rest and if it is moving, it will continue to move with the same speed in the same direction unless an external force is applied on it. It gives the qualitative definition of force and defines the concept of inertia.

EXERCISE-3(C)

(A) MULTIPLE CHOICE TYPE :
(Choose the correct answer from the options given below).

1. The force needed to stop a moving body in a definite time depends on :
(a) mass of the body (b) velocity of the body (c) both (a) and (b) (d) none of these
Step 1: Force depends on momentum change.
Step 2: Momentum is mass × velocity.
Answer: (c) both (a) and (b)

2. Linear momentum is a :
(a) scalar quantity (b) vector quantity (c) directionless quantity (d) none of these
Step 1: Momentum = mass × velocity.
Step 2: Velocity has direction, making momentum a vector.
Answer: (b) vector quantity

3. Newton's first law of motion defines force :
(a) quantitatively (b) qualitatively (c) both (a) and (b) (d) none of these
Step 1: 1st law explains what force *is* conceptually.
Step 2: 2nd law gives the math (quantitative).
Answer: (b) qualitatively

4. According to Newton's second law, the acceleration produced in a body is :
(a) inversely proportional to the force applied on it
(b) directly proportional to the force applied on it
(c) not dependent on force
(d) none of these
Step 1: F = ma, so a = F/m.
Step 2: 'a' increases when 'F' increases.
Answer: (b) directly proportional to the force applied on it

5. The correct form of Newton's second law is :
(a) F = Δp / Δt (b) F = m (Δv / Δt) (c) F = v (Δm / Δt) (d) F = mv
Step 1: Second law is rate of change of momentum.
Answer: (a) F = Δp / Δt

6. For a given force F, the graph plotted for acceleration against mass is :
(a) [Straight line from origin] (b) [Horizontal line] (c) [Straight line downwards] (d) [Curve downwards (hyperbola)]
Step 1: a = F/m.
Step 2: Inverse relationship creates a hyperbolic curve.
Answer: (d) [Curve downwards (hyperbola)]

7. The acceleration produced in a body by a force of a given magnitude depends on :
(a) size of the body (b) shape of the body (c) mass of the body (d) none of these
Step 1: Formula a = F/m shows dependence on mass.
Answer: (c) mass of the body

8. 1 Newton is equal to :
(a) 10³ dyne (b) 10⁸ dyne (c) 10¹⁰ dyne (d) 10⁵ dyne
Step 1: 1 N = 1 kg × 1 m/s² = 1000 g × 100 cm/s².
Step 2: 1000 × 100 = 100000 = 10⁵.
Answer: (d) 10⁵ dyne

9. The force required to produce an acceleration of 5 m s⁻² in a body of mass 2 kg is :
(a) 2.5 N (b) 20 N (c) 10 N (d) 25 N
Step 1: F = ma.
Step 2: F = 2 kg × 5 m/s² = 10 N.
Answer: (c) 10 N

10. According to Newton's second law of motion, the change in momentum takes place in a direction :
(a) opposite to the direction of force applied (b) perpendicular to the direction of force applied (c) similar to the direction of force applied (d) none of these
Step 1: Force and momentum change are vectors pointing the same way.
Answer: (c) similar to the direction of force applied

11. For relation F = mΔv/Δt to hold true, the condition(s) necessary is/are :
(a) velocities are equal to the velocity of light
(b) velocities are much smaller than the velocity of light
(c) mass remains constant
(d) both (b) and (c)
Step 1: Mass must be constant to pull it out of the Δ operator.
Step 2: Mass is only constant if velocity << speed of light.
Answer: (d) both (b) and (c)

12. Two balls of masses 4m and 8m are in motion with velocities 2v and v respectively. The ratio of their momentum would be :
(a) 2 : 1 (b) 1 : 2 (c) 1 : 1 (d) 1 : 4
Step 1: Momentum 1 = 4m × 2v = 8mv.
Step 2: Momentum 2 = 8m × v = 8mv.
Step 3: Ratio is 8mv / 8mv = 1:1.
Answer: (c) 1 : 1

(B) VERY SHORT ANSWER TYPE :

1. Define linear momentum and state its S.I. unit.
Step 1: Product of mass and velocity.
Step 2: Units are kg and m/s.
Answer: The linear momentum of a body is the product of its mass and velocity. S.I. unit is kg m s⁻¹.

2. A body of mass m moving with a velocity v is acted upon by a force. Write expression for change in momentum in each of the following cases : (i) when v << c, (ii) when v → c, and (iii) when v << c but m does not remain constant. Here c is the speed of light.
Step 1: (i) Constant mass allows pulling m outside Δ.
Step 2: (ii) & (iii) Mass is variable, keep it inside Δ.
Answer: (i) m Δv, (ii) Δ(mv), (iii) Δ(mv)

3. Two bodies A and B of same mass are moving with velocities v and 2v respectively. Compare their (i) inertia, (ii) momentum.
Step 1: Inertia depends only on mass (m:m = 1:1).
Step 2: Momentum is mv. Ratio is mv : m(2v) = 1:2.
Answer: (i) 1 : 1 (ii) 1 : 2

4. Two balls A and B of masses m and 2m are in motion with velocities 2v and v respectively. Compare : (i) their inertia, (ii) their momentum, and (iii) the force needed to stop them in the same time.
Step 1: Inertia depends on mass (m:2m = 1:2).
Step 2: Momentum is m(2v) vs 2m(v) = 2mv:2mv = 1:1.
Step 3: Force = momentum change / time. Same momentum means 1:1 ratio.
Answer: (i) 1 : 2 (ii) 1 : 1 (iii) 1 : 1

5. Name the S.I. unit of force and define it.
Step 1: Unit is Newton.
Step 2: Definition comes from F = ma (1 kg and 1 m/s²).
Answer: Newton. One newton is the force which when acts on a body of mass 1 kg, produces an acceleration of 1 m s⁻².

6. What is the C.G.S. unit of force ? How is it defined?
Step 1: Unit is Dyne.
Step 2: Definition from 1 gram and 1 cm/s².
Answer: Dyne. One dyne is the force which when acts on a body of mass 1 g, produces an acceleration of 1 cm s⁻².

7. Name the S.I. and C.G.S. units of force. How are they related ?
Step 1: S.I. is Newton, C.G.S. is dyne.
Step 2: Factor is 10⁵.
Answer: S.I. unit is Newton, C.G.S. unit is dyne. 1 Newton = 10⁵ dyne.

(C) SHORT ANSWER TYPE :

1. Name the two factors on which the force needed to stop a moving body in a given time depends.
Step 1: Force depends on rate of change of momentum.
Step 2: Momentum consists of mass and velocity.
Answer: (i) Mass of the moving body, and (ii) Velocity of the moving body.

2. Show that the rate of change of momentum = mass × acceleration. Under what condition does this relation hold ?
Step 1: Rate of change of p = Δ(mv)/Δt.
Step 2: If mass is constant, it becomes m(Δv/Δt).
Step 3: Since Δv/Δt = a, we get m × a.
Answer: Rate of change of momentum = Δp/Δt = Δ(mv)/Δt. If mass remains constant, this equals m(Δv/Δt) = m × a (since Δv/Δt = acceleration). This holds true when velocity is much smaller than the speed of light and mass remains constant.

3. State Newton's second law of motion. What information do you get from it ?
Step 1: Define relationship between momentum change and force.
Step 2: It provides mathematical calculation of force.
Answer: The rate of change of momentum of a body is directly proportional to the force applied on it and the change in momentum takes place in the direction in which the force is applied. It gives the quantitative value of force.

4. How does Newton's second law of motion differ from first law of motion ?
Step 1: 1st law is descriptive.
Step 2: 2nd law provides a formula.
Answer: The first law defines force qualitatively and defines inertia, whereas the second law gives the quantitative measure of force relating it to mass and acceleration.

5. Write the mathematical form of Newton's second law of motion. State condition if any.
Step 1: General formula is F = Δp / Δt.
Step 2: This is universally true.
Answer: F = Δp / Δt. The condition is that this general form is true regardless of whether mass is constant or variable.

6. State Newton's second law of motion. Under what condition does it take the form F = ma ?
Step 1: State momentum relation.
Step 2: Condition is constant mass.
Answer: Newton's second law: The rate of change of momentum of a body is directly proportional to the applied force. It takes the form F = ma when the mass of the body remains constant.

7. How can Newton's first law of motion be obtained from the second law of motion ?
Step 1: Use F = ma.
Step 2: If F = 0, then a = 0 (constant velocity or rest).
Answer: From F = ma, if no force is applied (F = 0), then acceleration a = 0. This means the body remains at rest or moves with uniform velocity, which is the statement of the first law.

8. Why does a glass vessel break when it falls on a hard floor, but it does not break when it falls on a carpet ?
Step 1: Hard floor brings it to rest instantly, meaning tiny Δt.
Step 2: Force F = Δp/Δt. Tiny Δt means huge Force (breaking it).
Step 3: Carpet yields, increasing Δt and lowering Force.
Answer: On a hard floor, it comes to rest almost instantaneously, meaning the rate of change of momentum is very high, exerting a large force that breaks it. On a carpet, the time duration to come to rest increases, so less force is exerted.

9. Use Newton's second law of motion to explain the following :
(a) A cricketer pulls his hands back while catching a fast moving cricket ball.
(b) An athlete prefers to land on sand instead of hard floor while taking a high jump.
Step 1: Both scenarios aim to increase the stopping time (Δt).
Step 2: By F = Δp/Δt, larger time means smaller impact force.
Answer: (a) Pulling hands back increases the time to stop the ball, decreasing the rate of change of momentum and thus reducing the force on the hands. (b) Sand yields and increases the time to bring the athlete to rest, lowering the force exerted on his feet.

(D) LONG ANSWER TYPE :

1. Draw graphs to show the dependence of (i) acceleration on force for a constant mass, and (ii) force on mass for a constant acceleration.
Step 1: (i) a = F/m (direct proportion).
Step 2: (ii) F = ma (direct proportion).
Answer: (i) A straight line graph starting from the origin (a ∝ F). (ii) A straight line graph starting from the origin (F ∝ m).

2. How does the acceleration produced by a given force depend on mass of the body ? Draw a graph to show it.
Step 1: a = F/m.
Step 2: Inverse relationship.
Answer: Acceleration is inversely proportional to the mass of the body (a ∝ 1/m). The graph of acceleration against mass is a curve (hyperbola).

3. How does acceleration vary with the force applied for a given body ? Show it with the help of a graph.
Step 1: a = F/m.
Step 2: Direct relationship.
Answer: Acceleration is directly proportional to the force applied (a ∝ F). The graph is a straight line passing through the origin.

(E) NUMERICALS :

1. A body of mass 5 kg is moving with velocity 2 m s⁻¹. Calculate its linear momentum.
Step 1: m = 5 kg, v = 2 m/s.
Step 2: p = m × v = 5 × 2.
Answer: 10 kg m s⁻¹

2. The linear momentum of a ball of mass 50 g is 0.5 kg m s⁻¹. Find its velocity.
Step 1: m = 50 g = 0.05 kg, p = 0.5 kg m/s.
Step 2: v = p / m = 0.5 / 0.05.
Answer: 10 m s⁻¹

3. A force of 15 N acts on a body of mass 2 kg. Calculate the acceleration produced.
Step 1: F = 15 N, m = 2 kg.
Step 2: a = F / m = 15 / 2.
Answer: 7.5 m s⁻²

4. A force of 10 N acts on a body of mass 5 kg. Find the acceleration produced.
Step 1: F = 10 N, m = 5 kg.
Step 2: a = F / m = 10 / 5.
Answer: 2 m s⁻²

5. Calculate the magnitude of force which when applied on a body of mass 0.5 kg produces an acceleration of 5 m s⁻².
Step 1: m = 0.5 kg, a = 5 m/s².
Step 2: F = m × a = 0.5 × 5.
Answer: 2.5 N

6. A force of 10 N acts on a body of mass 2 kg for 3 s, initially at rest. Calculate : (i) the velocity acquired by the body, and (ii) change in momentum of the body.
Step 1: F = 10 N, m = 2 kg, t = 3 s, u = 0.
Step 2: a = F / m = 10 / 2 = 5 m/s².
Step 3: (i) v = u + at = 0 + 5(3) = 15 m/s.
Step 4: (ii) Δp = m(v - u) = 2(15 - 0) = 30.
Answer: (i) 15 m s⁻¹, (ii) 30 kg m s⁻¹

7. A force acts for 10 s on a stationary body of mass 100 kg after which the force ceases to act. The body moves through a distance of 100 m in the next 5 s. Calculate : (i) the velocity acquired by the body, (ii) the acceleration produced by the force, and (iii) the magnitude of the force.
Step 1: (i) Body moves 100 m in 5 s at constant velocity. v = d / t = 100 / 5 = 20 m/s.
Step 2: (ii) Initial u = 0. Acceleration a = (v - u) / t = (20 - 0) / 10 = 2 m/s².
Step 3: (iii) Force F = ma = 100 × 2 = 200 N.
Answer: (i) 20 m s⁻¹, (ii) 2 m s⁻², (iii) 200 N

8. Fig. 3.11 shows the velocity-time graph of a particle of mass 100 g moving in a straight line. Calculate the force acting on the particle.
Step 1: m = 100 g = 0.1 kg.
Step 2: From graph, velocity changes 0 to 20 m/s in 5 s.
Step 3: a = (20 - 0) / 5 = 4 m/s².
Step 4: F = ma = 0.1 × 4 = 0.4 N.
Answer: 0.4 N

9. A force causes an acceleration of 10 m s⁻² in a body of mass 500 g. What acceleration will be caused by the same force in a body of mass 5 kg ?
Step 1: F = m₁a₁ = 0.5 kg × 10 m/s² = 5 N.
Step 2: a₂ = F / m₂ = 5 / 5 = 1 m/s².
Answer: 1 m s⁻²

10. A cricket ball of mass 150 g moving at a speed of 25 ms⁻¹ is brought to rest by a player in 0.03 s. Find the average force applied by the player.
Step 1: m = 0.15 kg, u = 25 m/s, v = 0, t = 0.03 s.
Step 2: a = (v - u) / t = (0 - 25) / 0.03 = -833.33 m/s².
Step 3: F = ma = 0.15 × (-833.33) = -125 N.
Answer: -125 N

11. A force acts for 0.1 s on a body of mass 2.0 kg initially at rest. The force is then withdrawn and the body moves with a velocity of 2 m s⁻¹. Find the magnitude of force.
Step 1: m = 2 kg, u = 0, v = 2 m/s, t = 0.1 s.
Step 2: a = (v - u) / t = (2 - 0) / 0.1 = 20 m/s².
Step 3: F = ma = 2 × 20 = 40 N.
Answer: 40 N

12. A body of mass 500 g, initially at rest, is acted upon by a force which causes it to move a distance of 4 m in 2 s. Calculate the force applied.
Step 1: m = 0.5 kg, u = 0, s = 4 m, t = 2 s.
Step 2: s = ut + 0.5at² ➔ 4 = 0 + 0.5×a×(2)² ➔ 4 = 2a ➔ a = 2 m/s².
Step 3: F = ma = 0.5 × 2 = 1 N.
Answer: 1 N

13. A car of mass 480 kg moving at a speed of 54 km h⁻¹ is stopped by applying brakes in 10 s. Calculate the force applied by the brakes.
Step 1: m = 480 kg, u = 54 km/h = 15 m/s, v = 0, t = 10 s.
Step 2: a = (v - u) / t = (0 - 15) / 10 = -1.5 m/s².
Step 3: F = m|a| = 480 × 1.5 = 720 N.
Answer: 720 N

14. A car is moving with a uniform velocity of 30 m s⁻¹. It is stopped in 2 s by applying a force of 1500 N through its brakes. Calculate : (a) the change in momentum of car, (b) the retardation produced in car, and (c) the mass of the car.
Step 1: u = 30 m/s, v = 0, t = 2 s, F = 1500 N.
Step 2: (b) retardation = (u - v) / t = 30 / 2 = 15 m/s².
Step 3: (c) mass m = F / a = 1500 / 15 = 100 kg.
Step 4: (a) Δp = m(u - v) = 100 × 30 = 3000 kg m/s.
Answer: (a) 3000 kg m s⁻¹, (b) 15 m s⁻², (c) 100 kg

15. A bullet of mass 50 g moving with an initial velocity of 100 m s⁻¹, strikes a wooden block and comes to rest after penetrating a distance 2 cm in it. Calculate : (i) initial momentum of the bullet, (ii) final momentum of the bullet, (iii) retardation caused by the wooden block, and (iv) resistive force exerted by the wooden block
Step 1: m = 0.05 kg, u = 100 m/s, v = 0, s = 0.02 m.
Step 2: (i) Initial p = 0.05 × 100 = 5 kg m/s.
Step 3: (ii) Final p = 0.05 × 0 = 0.
Step 4: (iii) v² = u² + 2as ➔ 0 = 10000 + 2(a)(0.02) ➔ 0.04a = -10000 ➔ a = -2.5 × 10⁵ m/s².
Step 5: (iv) F = ma = 0.05 × 2.5 × 10⁵ = 12500 N.
Answer: (i) 5 kg m s⁻¹, (ii) zero, (iii) 2.5 × 10⁵ m s⁻², (iv) 12500 N

EXERCISE-3(D)

(A) MULTIPLE CHOICE TYPE :
(Choose the correct answer from the options given below).

1. Newton's third law :
(a) defines the force qualitatively (b) defines the force quantitatively (c) explains the way a force acts on a body (d) gives the direction of force.
Step 1: 3rd law describes the interaction (action-reaction) between two bodies.
Answer: (c) explains the way a force acts on a body

2. The motion of a boat moving away from the shore is an example of :
(a) Newton's first law (b) Newton's second law (c) Newton's third law (d) not related with Newton's law
Step 1: Stepping creates backward action on boat.
Step 2: Boat reacts by pushing you forward, and boat moves backward.
Answer: (c) Newton's third law

3. In an interaction of two bodies A and B, which of the following statements is correct ?
(a) F_AB = F_BA (b) F_BA ≥ F_AB (c) F_AB = - F_BA (d) F_AB > F_BA
Step 1: Forces are equal in magnitude.
Step 2: Forces are opposite in direction (denoted by negative sign).
Answer: (c) F_AB = - F_BA

4. A footballer hits a football with a force of 10 N. What force does the footballer experience ?
(a) 5 N (b) 10 N (c) 20 N (d) 15 N
Step 1: Action = Reaction in magnitude.
Answer: (b) 10 N

5. Out of the following statements, which one is correct ?
(a) the sum of action and reaction on a body is zero.
(b) action and reaction act on different bodies in opposite direction.
(c) action and reaction are always in opposite direction on the same body.
(d) action and reaction act on different bodies in the same direction.
Step 1: Action and reaction always act on two different interacting bodies.
Answer: (b) action and reaction act on different bodies in opposite direction.

6. Which of the following are examples of Newton's third law of motion ?
(i) While catching a ball, the cricketer withdraws his hands along with the ball.
(ii) Athletes often land on sand after taking a high jump
(iii) Motion of a man on ground
(a) (iii) (b) (ii) (c) (i) (d) all of the above
Step 1: Walking requires pushing ground backward (action) to get forward reaction.
Answer: (a) (iii)

7. Why is it necessary to wear seat belts while driving according to Newton's law of motion ?
(a) To increase the inertia of the passenger.
(b) To balance the action and reaction forces acting between the passenger and car.
(c) To prevent passengers from being thrown forward due to their inertia when the car stops suddenly.
(d) To increase the momentum of passengers during a collision by applying an external force.
Step 1: Passenger tends to keep moving (inertia of motion) when car stops.
Answer: (c) To prevent passengers from being thrown forward due to their inertia when the car stops suddenly.

8. A student is riding a bicycle on a level road. Apply Newton's third law of motion and identify the force which is equal and opposite to the backward push of the rear wheel on the road.
(a) The force exerted by him on pedals.
(b) The resistance of the air and force of friction.
(c) The forward push of the road on the rear wheel.
(d) all of the above.
Step 1: Action = backward push by wheel on road.
Step 2: Reaction = forward push by road on wheel.
Answer: (c) The forward push of the road on the rear wheel.

9. Out of the following, which statement does not relate to Newton's laws of motion ?
(a) If a body A exerts a force on body B, then body B exerts an equal and opposite force on body A.
(b) The force needed to produce a given acceleration in a body is inversely proportional to the mass of the body.
(c) The rate of change of momentum of a body is directly proportional to the external force acting on the body and takes place in the direction of the force.
(d) A body continues in a state of rest or of uniform motion in a straight line unless acted upon by an external force.
Step 1: By F = ma, Force needed is *directly* proportional to mass, not inversely.
Answer: (b) The force needed to produce a given acceleration in a body is inversely proportional to the mass of the body.

(B) VERY SHORT ANSWER TYPE :

1. State Newton's third law of motion.
Step 1: State relationship between action and reaction.
Answer: To every action there is always an equal and opposite reaction.

2. Name and state the action and reaction in the following cases :
(a) firing a bullet from a gun,
(b) hammering a nail,
(c) a book lying on a table,
(d) a moving rocket,
(e) a person walking on the floor,
(f) a moving train colliding with a stationary train.
Step 1: Identify the two interacting bodies in each case.
Answer: (a) Action: Force on bullet. Reaction: Recoil force on gun. (b) Action: Hammer hits nail. Reaction: Nail pushes hammer. (c) Action: Weight of book on table. Reaction: Upward normal force of table on book. (d) Action: Rocket expels gases downward. Reaction: Gases push rocket upward. (e) Action: Foot pushes ground backward. Reaction: Ground pushes foot forward. (f) Action: Moving train exerts force on stationary train. Reaction: Stationary train exerts equal and opposite force on moving train.

3. 'The action and reaction both act simultaneously.' Is this statement true ?
Step 1: They exist as a pair at the exact same time.
Answer: Yes.

4. 'The action and reaction are equal in magnitude'. Is this statement true ?
Step 1: Magnitudes of interacting forces are always identical.
Answer: Yes.

5. Comment on the statement 'the sum of action and reaction on a body is zero'. [Hint : The statement is wrong]
Step 1: Action and reaction act on two *different* bodies.
Step 2: You cannot sum forces acting on separate objects.
Answer: The statement is wrong because action and reaction act on two different interacting bodies, so they cannot be summed up to find the net force on a single body.

(C) SHORT ANSWER TYPE :

1. State the usefulness of Newton's third law of motion.
Step 1: It explains how forces exist in pairs.
Step 2: Helps explain locomotion (walking, rocket launch).
Answer: It explains how forces always exist in pairs and helps understand the interaction between bodies, such as walking, swimming, or rocket propulsion.

2. State and explain the law of action and reaction, by giving two examples.
Step 1: Define 3rd law.
Step 2: Provide examples like pushing a wall and book on table.
Answer: To every action, there is an equal and opposite reaction acting on two different bodies. Example 1: Pushing a wall (action) makes you feel a backward push (reaction). Example 2: A book on a table exerts its weight downwards (action), and the table pushes it upwards with an equal force (reaction).

3. Explain the motion of a rocket with the help of Newton's third law.
Step 1: Rocket expels exhaust gas downwards.
Step 2: Gas pushes rocket upwards.
Answer: The rocket expels hot gases downwards at high velocity (action). These gases exert an equal and opposite reaction force upwards on the rocket, making it move forward.

4. When a shot is fired from a gun, the gun gets recoiled. Explain.
Step 1: Gun exerts forward force on bullet.
Step 2: Bullet exerts backward force on gun.
Answer: Firing the shot exerts a forward force (action) on the bullet. Simultaneously, the bullet exerts an equal and opposite backward force (reaction) on the gun, causing it to recoil.

5. When you step ashore from a stationary boat, it tends to leave the shore. Explain.
Step 1: You push backward on the boat to step forward.
Step 2: The boat moves backward due to this action.
Answer: You exert a force (action) backwards on the boat to step forward. The boat experiences this force and moves away from the shore due to the reaction pushing you forward.

6. When two spring balances joined at their free ends, are pulled apart, both show the same reading. Explain.
Step 1: Pulling one spring balance pulls the other.
Step 2: By 3rd law, interacting forces are equal.
Answer: Pulling one spring balance exerts a force (action) on the other. By Newton's third law, the other balance exerts an equal and opposite force (reaction), so both register the same magnitude of force.

7. To move a boat ahead in water, the boatman has to push the water backwards by his oar. Explain.
Step 1: Pushing water backwards is the action.
Step 2: Water pushing boat forwards is the reaction.
Answer: By pushing water backwards (action), the water exerts an equal and opposite forward force (reaction) on the boat, causing it to move ahead.

8. A person pushing a wall hard is liable to fall back. Give reason.
Step 1: Pushing wall applies action.
Step 2: Wall applies equal reaction, pushing person backward.
Answer: The person applies an action force on the wall. The rigid wall applies an equal and opposite reaction force on the person, which can cause him to lose balance and fall back.

9. A light ball falling on ground, after striking the ground rises upwards. Explain the reason.
Step 1: Ball hits ground (action).
Step 2: Ground pushes ball up (reaction).
Answer: When it strikes the ground, it exerts an action force on it. The ground immediately exerts an equal and opposite reaction force upwards on the ball, making it bounce.

(D) NUMERICALS :

1. A boy pushes a wall with a force of 10 N towards east. What force is exerted by the wall on the boy ?
Step 1: Action = 10 N East.
Step 2: Reaction = equal and opposite = 10 N West.
Answer: 10 N towards west

2. In Fig. 3.17, a block of weight 15 N is hanging from a rigid support by a string. What force is exerted by (a) block on the string, (b) string on the block. Name them and show them in the diagram.
Step 1: (a) Block pulls string down by its weight = 15 N.
Step 2: (b) String pulls block up with tension = 15 N.
Answer: (a) 15 N downwards (weight), (b) 15 N upwards (tension).

EXERCISE-3(E)

(A) MULTIPLE CHOICE TYPE :
(Choose the correct answer from the options given below).

1. The gravitational force between two bodies is :
(a) always repulsive (b) always attractive (c) attractive only at large distances (d) repulsive only at large distances.
Step 1: Gravity is purely a pulling force.
Answer: (b) always attractive

2. The gravitational force of attraction between two particles is because of :
(a) their masses (b) their shapes (c) their sizes (d) none of these
Step 1: Masses create the gravitational field.
Answer: (a) their masses

3. The law of gravitation was given by :
(a) Thomas Alva Edison (b) Alexander Graham Bell (c) Sir Isaac Newton (d) Albert Einstein
Step 1: Newton formulated the Universal Law of Gravitation.
Answer: (c) Sir Isaac Newton

4. The force of attraction acting between two bodies is :
(a) directly proportional to the product of their masses
(b) directly proportional to the square of the distance between them.
(c) inversely proportional to the square of the distance between them.
(d) both (a) and (c)
Step 1: F ∝ m₁m₂.
Step 2: F ∝ 1/r².
Answer: (d) both (a) and (c)

5. The value of gravitational constant G depends on :
(a) the nature of the particle (b) the temperature of the particle (c) medium (d) none of these
Step 1: G is a universal constant, independent of conditions.
Answer: (d) none of these

6. The value of gravitational constant G changes with the change in :
(a) height (b) place (c) distance (d) it remains unchanged
Step 1: Universal constant G never changes anywhere.
Answer: (d) it remains unchanged

7. The gravitational force between two masses is :
(a) directly proportional to the product of the masses.
(b) inversely proportional to the square of the separation between them.
(c) significant between heavenly bodies, but is insignificant between atomic bodies
(d) all of the above
Step 1: All three statements correctly describe gravity properties.
Answer: (d) all of the above

8. The gravitational force of attraction between the two bodies of masses 60 kg and 40 kg separated by a distance 10 m is : (where G = 6.7 × 10⁻¹¹ Nm² kg⁻²).
(a) 1.60 × 10⁻⁹ N (b) 2.8 × 10⁻¹¹ N (c) 1.0 × 10⁻¹¹ N (d) 1.6 × 10⁻¹¹ N
Step 1: F = G(m₁m₂)/r² = 6.7×10⁻¹¹ × 60 × 40 / 10².
Step 2: = 6.7×10⁻¹¹ × 2400 / 100 = 1.608 × 10⁻⁹ N.
Answer: (a) 1.60 × 10⁻⁹ N

9. The value of g on a planet depends on :
(a) the mass of the planet (b) the radius of the planet (c) both (a) and (b) (d) composition of the planet
Step 1: Formula g = GM/R².
Step 2: It depends on mass (M) and radius (R) of planet.
Answer: (c) both (a) and (b)

10. A body is projected vertically upward with an initial velocity u. If acceleration due to gravity is g, the time for which it remains in air, is :
(a) u / g (b) ug (c) 2u / g (d) u / 2g
Step 1: Time to go up = u/g.
Step 2: Total time = time up + time down = 2u/g.
Answer: (c) 2u / g

11. An object falling freely from rest reaches ground in 2 s. If acceleration due to gravity is 9.8 m s⁻², the velocity of the object on reaching the ground will be:
(a) 9.8 m s⁻¹ (b) 4.9 m s⁻¹ (c) 19.6 m s⁻¹ (d) zero.
Step 1: v = u + gt = 0 + 9.8 × 2 = 19.6 m/s.
Answer: (c) 19.6 m s⁻¹

12. A body of mass 25 kg is taken from the earth to the moon. If the value of g on moon is 1.6 ms⁻², then the weight of the body on moon is :
(a) 50 N (b) 100 N (c) 40 N (d) 60 N
Step 1: Weight = mg.
Step 2: W = 25 × 1.6 = 40 N.
Answer: (c) 40 N

13. The relationship between weight and mass is :
(a) W = m (b) W = mG (c) W = mg (d) m = Wg
Step 1: Weight is force of gravity.
Step 2: F = ma becomes W = mg.
Answer: (c) W = mg

14. The S.I. unit of weight is :
(a) kg (b) Joule (c) Newton (d) dyne
Step 1: Weight is a force, so S.I. unit is Newton.
Answer: (c) Newton

15. The mass of a body :
(a) changes with a change of place
(b) changes with direction
(c) changes when the velocity of the body is close to the velocity of light
(d) none of these
Step 1: Mass is constant unless speed approaches speed of light (relativistic effects).
Answer: (c) changes when the velocity of the body is close to the velocity of light

16. Identify the correct statement from the following :
(a) Mass is a force and is measured in Newton.
(b) The weight of a body is the same everywhere in the universe.
(c) Mass remains constant, but weight changes with gravity.
(d) Weight is measured using a physical balance.
Step 1: Mass is constant matter. Weight is gravity pull (changes with g).
Answer: (c) Mass remains constant, but weight changes with gravity.

(B) VERY SHORT ANSWER TYPE :

1. State Newton's law of gravitation.
Step 1: Define relationship of masses and distance.
Answer: Every particle in the universe attracts every other mass 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.

2. State whether the gravitational force between two masses is attractive or repulsive ?
Step 1: Gravity always pulls.
Answer: Always attractive.

3. Write an expression for the gravitational force of attraction between two bodies of masses m₁ and m₂ separated by a distance r.
Step 1: Use universal gravitation formula.
Answer: F = G (m₁m₂) / r²

4. How is the gravitational force between two masses affected if the separation between them is doubled?
Step 1: Inverse square law (1/2² = 1/4).
Answer: Force reduces to one-fourth.

5. Define gravitational constant G.
Step 1: Relate it to unit masses and unit distance.
Answer: Gravitational constant G is numerically equal to the magnitude of force of attraction between two masses each of 1 kg placed at a separation of 1 m.

6. Write the numerical value of gravitational constant G with its S.I. unit.
Step 1: Recall the constant value.
Answer: 6.67 × 10⁻¹¹ N m² kg⁻²

7. Define the term acceleration due to gravity ? Write its S.I. unit.
Step 1: Define it as velocity rate change for falling body.
Answer: The rate at which the velocity of a freely falling body increases is called acceleration due to gravity. Its S.I. unit is m s⁻².

8. Write down the average value of g on the earth's surface.
Step 1: Recall average constant.
Answer: 9.8 m s⁻²

9. How are g and G related ?
Step 1: Formula connecting both.
Answer: g = GM / R²

10. A body falls freely under gravity from rest and reaches the ground in time t. Write an expression for the height fallen by the body.
Step 1: Use h = ut + 0.5gt² with u=0.
Answer: h = 1/2 gt²

11. A body is thrown vertically upwards with an initial velocity u. Write an expression for the maximum height attained by the body.
Step 1: Use v² = u² - 2gh with v=0.
Answer: h = u² / 2g

12. State the S.I. units of (a) mass and (b) weight.
Step 1: Mass is matter, Weight is force.
Answer: (a) kg, (b) Newton

13. The value of g at the centre of the earth is zero. What will be the weight of a body of mass m kg at the centre of the earth ?
Step 1: W = mg. If g=0, W=0.
Answer: Zero.

14. Which of the following quantity does not change by change of place of a body : mass or weight ?
Step 1: Mass is constant regardless of gravity.
Answer: Mass.

(C) SHORT ANSWER TYPE :

1. How does the gravitational force of attraction between two masses depend on the distance between them ?
Step 1: State the inverse square relation.
Answer: It is inversely proportional to the square of the distance between them.

2. List applications of law of gravitation ?
Step 1: Recall uses in astronomy and earth phenomena.
Answer: Determining the mass and size of celestial bodies, explaining the formation of ocean tides, launch and motion of satellites, and discovery of new planets.

3. What do you understand by the term force due to gravity ?
Step 1: Define Earth's pull.
Answer: The force with which the earth attracts a body towards its centre is called the force due to gravity on the body.

4. Write an expression for the force due to gravity on a body of mass m and explain the meaning of the symbols used in it.
Step 1: Write formula and define terms.
Answer: F = GMm / R². Here G is universal gravitational constant, M is mass of earth, m is mass of the body, and R is radius of the earth.

5. How is the acceleration due to gravity on the surface of the earth related to its mass and radius ?
Step 1: Use g = GM/R² to explain direct and inverse relations.
Answer: g = GM / R², so it is directly proportional to the mass of the earth and inversely proportional to the square of its radius.

6. Define the terms mass and weight.
Step 1: Mass = matter. Weight = gravity force.
Answer: Mass of a body is the quantity of matter it contains. Weight of a body is the force with which the earth attracts it.

7. Distinguish between mass and weight.
Step 1: Point out scalar vs vector, and constant vs variable.
Answer: Mass is a constant scalar quantity measured in kg, while weight is a variable vector quantity measured in Newton.

8. Explain the meaning of the following statement '1 kgf = 9.8 N'.
Step 1: Relate kgf to the force needed to hold 1 kg.
Answer: It means that a force of 9.8 Newtons is required to hold a mass of 1 kg on our palm against gravity.

9. Why are seat belts important in vehicles? Which law of motion explains their importance?
Step 1: Prevents forward momentum when stopping (inertia).
Answer: Seat belts prevent passengers from being thrown forward when brakes are applied. It is explained by Newton's first law of motion (inertia of motion).

(D) NUMERICALS :

1. The force of attraction between two bodies at a certain separation is 10 N. What will be the force of attraction between them if the separation is reduced to half ?
Step 1: F = 10 N.
Step 2: Distance reduced to half (r/2). Force ∝ 1/r².
Step 3: Force becomes 4 × F = 4 × 10 = 40 N.
Answer: 40 N

2. Write the approximate weight of a body of mass 5 kg. What assumption have you made ?
Step 1: Mass = 5 kg.
Step 2: Assume g = 10 m/s² for approximation.
Step 3: Weight = mg = 5 × 10 = 50 N.
Answer: 50 N (Assumption : g = 10 m s⁻²)

3. Calculate the weight of a body of mass 10 kg in (a) kgf and (b) newton. Take g = 9.8 m s⁻².
Step 1: m = 10 kg, g = 9.8 m/s².
Step 2: (a) Weight in kgf is numerically equal to mass in kg = 10 kgf.
Step 3: (b) Weight in N = mg = 10 × 9.8 = 98 N.
Answer: (a) 10 kgf, (b) 98 newton.

4. State the magnitude and direction of the force of gravity acting on a body of mass 5 kg. Take g = 9.8 m s⁻².
Step 1: m = 5 kg, g = 9.8 m/s².
Step 2: F = mg = 5 × 9.8 = 49 N.
Step 3: Gravity always pulls towards Earth's center (downwards).
Answer: Force of gravity on the body = 49 newton vertically downwards.

5. The weight of a body is 2.0 N. What is the mass of the body ? (g = 10 m s⁻²)
Step 1: W = 2.0 N, g = 10 m/s².
Step 2: m = W / g = 2.0 / 10 = 0.2 kg.
Answer: 0.2 kg

6. The weight of a body on earth is 98 N where the acceleration due to gravity is 9.8 m s⁻². What will be its (a) mass and (b) weight on moon where the acceleration due to gravity is 1.6 m s⁻² ?
Step 1: Earth weight = 98 N, g_earth = 9.8 ➔ m = 98/9.8 = 10 kg.
Step 2: (a) Mass is constant everywhere = 10 kg.
Step 3: (b) Weight on moon = m × g_moon = 10 × 1.6 = 16 N.
Answer: (a) 10 kg, (b) 16 N

7. A man weighs 600 N on earth. What would be his approximate weight on moon ? Give reason for your answer ?
Step 1: Moon gravity is approximately 1/6th of Earth gravity.
Step 2: Weight on moon = Earth weight / 6 = 600 / 6 = 100 N.
Answer: 100 N. Reason : The value of g on moon = 1/6th the value of g on earth.

8. What is the (a) force of gravity and (b) weight of a block of mass 10.5 kg ? Take g = 10 m s⁻².
Step 1: m = 10.5 kg, g = 10 m/s².
Step 2: (a) Force of gravity = mg = 10.5 × 10 = 105 N.
Step 3: (b) Weight is simply the force of gravity = 105 N.
Answer: (a) 105 N, (b) 105 N

9. A ball is released from a height and it reaches the ground in 3 s. If g = 9.8 m s⁻², find : (a) the height from which the ball was released, (b) the velocity with which the ball will strike the ground.
Step 1: u = 0 (released), t = 3 s, g = 9.8 m/s².
Step 2: (a) h = ut + 0.5gt² = 0 + 0.5(9.8)(3)² = 4.9 × 9 = 44.1 m.
Step 3: (b) v = u + gt = 0 + 9.8 × 3 = 29.4 m/s.
Answer: (a) 44.1 m, (b) 29.4 m s⁻¹

10. What force, in newton, your muscles need to apply to hold a mass of 5 kg in your hand ? State the assumption.
Step 1: Muscle force must equal weight to hold it steady.
Step 2: Assume g = 9.8 m/s².
Step 3: Force = mg = 5 × 9.8 = 49 N.
Answer: 49 N. Assumption : g = 9.8 N kg⁻¹

11. A ball is thrown vertically upwards. It goes to a height 20 m and then returns to the ground. Taking acceleration due to gravity g to be 10 m s⁻², find : (a) the initial velocity of the ball (b) the final velocity of the ball on reaching the ground and (c) the total time of journey of the ball.
Step 1: h = 20 m, g = 10 m/s², v at top = 0.
Step 2: (a) v² = u² - 2gh ➔ 0 = u² - 2(10)(20) ➔ u² = 400 ➔ u = 20 m/s.
Step 3: (b) Final velocity on return equals initial velocity = 20 m/s.
Step 4: (c) Total time = 2u/g = 2(20)/10 = 4 s.
Answer: (a) 20 m s⁻¹ (b) 20 m s⁻¹ (c) 4 s

12. A body is dropped from the top of a tower. It acquires a velocity 20 m s⁻¹ on reaching the ground. Calculate the height of the tower. (Take g = 10 m s⁻²)
Step 1: u = 0, v = 20 m/s, g = 10 m/s².
Step 2: v² = u² + 2gh ➔ 400 = 0 + 20h ➔ h = 400/20 = 20 m.
Answer: 20 m

13. A ball is thrown vertically upwards. It returns 6 s later. Calculate : (i) the greatest height reached by the ball, and (ii) the initial velocity of the ball. (Take g = 10 m s⁻²)
Step 1: Total time = 6 s, so time to reach top = 3 s.
Step 2: v at top = 0, g = 10.
Step 3: (ii) v = u - gt ➔ 0 = u - 10(3) ➔ u = 30 m/s.
Step 4: (i) h = u² / 2g = 900 / 20 = 45 m.
Answer: (i) 45 m, (ii) 30 m s⁻¹

14. A pebble is thrown vertically upwards with a speed of 20 m s⁻¹. How high will it be after 2 s ? (Take g = 10 m s⁻²)
Step 1: u = 20 m/s, t = 2 s, g = 10 m/s².
Step 2: h = ut - 0.5gt² = 20(2) - 0.5(10)(2)².
Step 3: h = 40 - 20 = 20 m.
Answer: 20 m

15. (a) How long will a stone take to fall to the ground from the top of a building 80 m high and (b) what will be the velocity of the stone on reaching the ground ? (Take g = 10 m s⁻²)
Step 1: h = 80 m, u = 0, g = 10 m/s².
Step 2: (a) h = 0.5gt² ➔ 80 = 5t² ➔ t² = 16 ➔ t = 4 s.
Step 3: (b) v = u + gt = 0 + 10(4) = 40 m/s.
Answer: (a) 4 s, (b) 40 m s⁻¹

16. A body falls from the top of a building and reaches the ground 2.5 s later. How high is the building ? (Take g = 9.8 m s⁻²)
Step 1: u = 0, t = 2.5 s, g = 9.8 m/s².
Step 2: h = 0.5gt² = 0.5 × 9.8 × (2.5)².
Step 3: h = 4.9 × 6.25 = 30.625 m.
Answer: 30.6 m

17. A ball is thrown vertically upwards with an initial velocity of 49 m s⁻¹. Calculate : (i) the maximum height attained, (ii) the time taken by it before it reaches the ground again. (Take g = 9.8 m s⁻²).
Step 1: u = 49 m/s, g = 9.8, v at top = 0.
Step 2: (i) h = u² / 2g = 2401 / 19.6 = 122.5 m.
Step 3: (ii) Total time = 2u / g = 2(49) / 9.8 = 10 s.
Answer: (i) 122.5 m, (ii) 10 s

18. A stone is dropped freely from the top of a tower and it reaches the ground in 4 s. Taking g = 10 m s⁻², calculate the height of the tower.
Step 1: u = 0, t = 4 s, g = 10 m/s².
Step 2: h = 0.5gt² = 0.5 × 10 × 16 = 80 m.
Answer: 80 m

19. A pebble is dropped freely in a well from its top. It takes 20 s for the pebble to reach the water surface in the well. Taking g = 10 m s⁻² and speed of sound = 330 m s⁻¹, find : (i) the depth of water surface, and (ii) the time when echo is heard after the pebble is dropped.
Step 1: t1 = 20 s, u = 0, g = 10 m/s².
Step 2: (i) Depth h = 0.5g(t1)² = 0.5(10)(400) = 2000 m.
Step 3: (ii) Time for sound t2 = h/v = 2000/330 ≈ 6.06 s.
Step 4: Total time = 20 + 6.06 = 26.06 s.
Answer: (i) 2000 m (ii) 26.1 s

20. A ball is thrown vertically upwards from the top of a tower with an initial velocity of 19.6 m s⁻¹. The ball reaches the ground after 5 s. Calculate : (i) the height of the tower, (ii) the velocity of ball on reaching the ground. Take g = 9.8 m s⁻².
Step 1: u = -19.6 m/s (upward is negative relative to downward gravity), t = 5 s, g = 9.8 m/s².
Step 2: (i) h = ut + 0.5gt² = -19.6(5) + 0.5(9.8)(25) = -98 + 122.5 = 24.5 m.
Step 3: (ii) v = u + gt = -19.6 + 9.8(5) = -19.6 + 49 = 29.4 m/s.
Answer: (i) 24.5 m (ii) 29.4 m s⁻¹

Competency Focussed Questions

(E) ASSERTION – REASON TYPE QUESTIONS :
(Choose the answer based on the codes given below.)
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are True and R is not the correct explanation of A
(c) Assertion is false but reason is true
(d) Assertion is true but reason is false

(i) Assertion (A) : The magnitude of non-contact force between two bodies decreases with the increase in distance of separation between them.
Reason (R) : The magnitude of non-contact force between two bodies varies inversely as the square of distance of separation between them.
Step 1: Force decreases as distance increases.
Step 2: Inverse square law confirms this relation exactly.
Answer: (a)

(ii) Assertion (A) : Inertia is an inherent property of each body by virtue of which it has a tendency to resist the change in its state of rest only.
Reason (R) : Force is that external cause which tends to change the state of rest or the state of motion of an object.
Step 1: Inertia resists change in BOTH rest and motion, so Assertion is false.
Step 2: Reason correctly defines force.
Answer: (c)

(iii) Assertion (A) : A cricket ball acquires much smaller velocity than a tennis ball when the two balls are pushed with equal force for the same duration.
Reason (R) : Mass is a measure of inertia.
Step 1: F = ma. Higher mass (cricket ball) means lower acceleration for the same force, thus smaller velocity.
Step 2: Mass is indeed the measure of inertia. Reason explains Assertion.
Answer: (a)

(iv) Assertion (A) : While catching a ball, a cricketer withdraws his hands alongwith the ball.
Reason (R) : To every action, there is an equal and opposite reaction.
Step 1: Pulling hands back reduces rate of change of momentum (2nd law), not 3rd law.
Step 2: Reason is true (3rd law) but doesn't explain the Assertion.
Answer: (b)

(v) Assertion (A) : Momentum of a body is the product of mass and acceleration.
Reason (R) : The rate of change of momentum of a body is equal to the product of mass and acceleration.
Step 1: Momentum is mass × velocity, NOT acceleration (Assertion is false).
Step 2: Rate of change of momentum is mass × acceleration (Reason is true).
Answer: (c)

(vi) Assertion (A) : When we stop pedalling a bicycle, it slows down.
Reason (R) : Force of friction always acts in the direction of motion.
Step 1: Bicycle slows down due to opposing friction (Assertion is true).
Step 2: Friction always acts OPPOSITE to motion, not in the direction (Reason is false).
Answer: (d)

(vii) Assertion (A) : On moon, a man feels heavier than on earth.
Reason (R) : This is due to the lower value of 'g' on moon.
Step 1: Man feels LIGHTER on moon, so Assertion is false.
Step 2: True that 'g' is lower on moon.
Answer: (c)

(viii) Assertion (A) : If the distance between bodies of masses M₁ and M₂ is increased by a factor of 4, the gravitational force reduces by 1/16.
Reason (R) : The gravitational force is inversely proportional to the square of distance between two bodies.
Step 1: Distance × 4 means Force / 4² = Force / 16.
Step 2: Reason correctly states the inverse square law explaining this.
Answer: (a)

(ix) Assertion (A) : Weight is the force with which the earth attracts a body.
Reason (R) : It is a measure of the quantity of matter contained in the body, at rest.
Step 1: Weight is gravitational force (Assertion is true).
Step 2: Mass (not weight) is the measure of matter (Reason is false).
Answer: (d)

(x) Assertion (A) : A swimmer pushes the water backward to move forward in a swimming pool.
Reason (R) : Water exerts a greater but opposite force on the swimmer, helping him to move forward.
Step 1: Swimmer pushes water backward (action). Assertion is true.
Step 2: Water exerts EQUAL (not greater) opposite force. Reason is false.
Answer: (d)

(F) CASE STUDY BASED QUESTIONS :

1. In a supermarket, Avantika could easily push an empty trolley but found it difficult to push a trolley full of groceries. The loaded trolley accelerated very slowly even when the same force was applied. To keep the trolley moving, she had to apply a greater force.
(a) Why did the loaded trolley accelerate less?
(b) According to you, which law of motion explains this observation?
(c) Why did Avantika have to apply a greater force to the loaded trolley?
(d) Correlate this event with one other similar example from daily life.
Step 1: (a) Loaded trolley has more mass, hence more inertia.
Step 2: (b) Newton's second law of motion (acceleration is inversely proportional to mass).
Step 3: (c) Greater mass requires a greater force to produce the same change in velocity (F = ma).
Step 4: (d) It is more difficult to push a heavy stationary car than a stationary bicycle.
Answer: (a) Because the loaded trolley has more mass and hence more inertia. (b) Newton's second law of motion. (c) Greater mass requires a greater force to produce the same change in velocity. (d) It is more difficult to push a heavy stationary car than a stationary bicycle.

2. During the annual sports day, a game of tug of war was held. The two teams pulled the rope in opposite directions. Although the rope did not move initially for two minutes, both teams felt a strong pulling force.
(a) Why did both teams feel a pull?
(b) Which law of motion comes into play in the above situation?
(c) Name the forces acting in this situation.
(d) State two other examples of this force.
Step 1: (a) Both teams were exerting equal and opposite tension forces on the rope.
Step 2: (b) Newton's third law of motion (action and reaction).
Step 3: (c) Tension force in the rope and frictional force between the ground and feet.
Step 4: (d) Pushing against a rigid wall without it moving, or two spring balances pulled in opposite directions showing the same reading.
Answer: (a) Because both teams were exerting equal and opposite tension forces on the rope. (b) Newton's third law of motion. (c) Tension force in the rope and frictional force. (d) Pushing against a rigid wall without it moving, or two spring balances pulled in opposite directions.