CBSE Class 9 - Ch9 - Force and Laws of Motion

Newtons' Cradle
Find out which Law is demonstrated here?
credits:Wikipedia

Force and Laws of Motion

NCERT Solution and Q & A

Q1: Define inertia.

Answer: Inertia is the resistance of a body to change in its state of rest or motion. It is an inherent property of the body. The property of inertia is due to the mass of the body. Greater the mass, higher will be the inertia.

There are three kinds of intertia:

1. Intertia of rest
2. Interia of motion
3. Interia of direction.

(Some physicists consider inertia of direction is part of inertia of motion).

Q2(NCERT): Which of the following has more inertia: 
(a) a rubber ball and a stone of the same size? 
(b) A bicycle and a train? 
(c) A five-rupee coin and a one-rupee coin?

Answer: The property of inertia is due to the mass of the body. Greater the mass, higher will be the inertia.
(a) The stone will have more inertia because it has more mass than the ball.

(b) Mass of a train is more than the mass of a bicycle, thus the train will have more inertia.
(c) Mass of a five rupee coin is more than that of a one-rupee coin. Hence, the five rupee coin has greater inertia than the one-rupee coin.

Q3: State Newton's First Law of motion.

Answer: Newton's First Law of motion states 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 in the state of the motion with the same velocity and same direction unless an external force is applied on it.

Q4: State why Newton's first law of motion is called the law of inertia.

Answer: Inertia is a tendency of the object to resist change in its state. Newton's first law of motion also states similar i.e. the object will remain in its present state unless an external force is applied. That's why Newton's first law is called Law of inertia.

Q5: Justify Newton's first law gives notion (or definition) about force.

Answer: According to Newton's first law of motion, an object tends to continue it is the present state unlean and external force is applied to change its state. For example, when a ball is rolled over a glass surface, it stops after covering a distance. It is because the frictional forces between the ball and the glass are being applied here. If there is a frictionless surface in a vacuum chamber, the ball will continue to run and cover a longer distance. It implies the external force (frictional force in this case) is almost absent and the ball continues to move to its present state.
Thus Newton's first law provides a notion of force.

Q6(NCERT): In the following example, try to identify the number of times the velocity of the ball changes:
“A football player kicks a football to another player of his team who kicks the football towards the
goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of
his own team”.
Also, identify the agent supplying the force in each case.

Answer:  We need to apply first law of motion and identify what force is applied which changes the velocity (state i.e. rest, direction, motion) of the ball. A force when applied can produce acceleration in the body.

Event	Agent	force and Action
A football player kicks a football	player	push force, the ball moves from rest.
to another player of his team who kicks the football towards the goal.	another player	force changes the direction of the ball.
The goalkeeper of the opposite team collects the football	goalkeeper	force to stop the ball (ball velocity is zero now)
and kicks it towards a player of his own team	goalkeeper	force moves the ball from rest to motion in opposite direction.

Q7: What a force can do?

Answer: A force is a push or pull which produces acceleration in the body on which it acts. A force acting on a body can cause:

1. it can alter the speed (or velocity) of the moving object.
2. it can change the direction of motion of a body.
3. it can change the shape of an object

Q8(NCERT): Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.

Answer: Some leaves of a tree get detached when we shake its branches vigorously is due to inertia of rest. When the tree moves to and fro, leaves tend to remain in the rest state and resist the change. Due to this, some leaves get detached and fall down.

Q9: Study the following scenarios and identify what type of inertia tends to resist the change.
(a) When a bus starts suddenly, the bus passengers standing in the bus tend to fall backwards.
(b) When a passenger jumps out of moving bus, he falls down.
(c) A cyclist on a levelled road does not come to rest immediately even he stops pedalling. 
(d) When the playing card is flicked with the finger the coin placed over it falls in the tumbler.
(e) An athlete often jumps before taking a long jump. 
(f) When a car enters a curved path, the car passengers tend to tilt outwards.

Answer:
(a) Inertia of rest
(b) Inertia of motion.
(c) Inertia of motion.
(d) Inertia of rest.
(e) Inertia of motion.
(f) Inertia of direction.

Q10(NCERT): Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?

Answer: Inertia tends to resist the change in state. When a moving bus applies brakes to stop, our body is in the inertia of motion state. It tends to oppose the change of state of rest and we fall forward.

Similarly, when the bus is at rest and starts, our body is in the inertia of rest state. It opposes the forward motion of the bus and hence we fall backwards.

Q11: Is force required to keep a moving object in motion (yes/no)?

Answer: No.

Q12: Define momentum.

Answer: Momentum is defined as the product of its mass and velocity.
i.e. Momentum (P) = mass(m) × velocity(v).

Momentum is a measure of the quantity of motion of a body. It is a vector quantity. It has both direction and magnitude. and its direction is same as the direction of velocity. The SI unit of momentum is kg-m/s.


Q13: A body of mass 25 kg has a momentum of 125 kg m/s. What is the velocity of the body?

Answer:  Since P = m × v
⇒ v = P / m = 125 / 25 = 5m/s


Q14(CBSE 2010): Why is it easier to stop a tennis ball in comparison to a cricket ball moving at the same speed?

Answer: Tennis ball is lighter than (less mass) than a cricket ball. A tennis ball moving with same speed has less momentum ( mass × velocity) than a cricket ball. ∴ It is easier to stop tennis ball having less momentum.


Q15(CBSE 2010): Which is having a higher value of momentum? A bullet of mass 10 g moving with a velocity of 400 m/s or a cricket ball of mass 400 g thrown with the speed of 90 km/hr.

Answer:  Momentum (P) = mass(m) × velocity(v).
mass of bullet = 10g = 10  × 10^-3kg = 10^-2kg
velocity of bullet = 400 m/s
momentum of bullet =  10^-2kg × 400 m/s = 4 kg m/s

mass of cricket ball = 400g = 400  × 10^-3kg = 0.400 kg
velocity of ball = 90 km/hr = 90 × 1000m / 3600s = 25m/s

momentum of ball = 0.400 × 25 = 10kg m/s
∴ The cricket ball has higher momentum.

Following questions asked in CBSE examination (2010) are similar to Q14 and Q15.

Q: Two similar vehicles are moving with the same velocity on the roads such that one of them is loaded and the other one is empty. Which of the two vehicles will require a larger force to stop it? Give reasons?

Q: Which one has greater inertia: a stone of mass 1 kg or a stone of mass 5 kg?

Q16: The term 'mass' is analogous to physical quantity 
(a) Weight
(b) Intertia
(c) Force
(d) Acceleration

Answer: (b) Intertia


Q17: What are balanced and unbalanced forces?

Answer: If a set of forces acting on a body does not change the state of rest or of motion of an object such forces are called balanced forces. In this case, object is said to be in an equilibrium state. The vector sum of all forces is zero i.e. ∑F_x = 0.

If a set of forces acting on body results in the change in state i.e. either changes the speed or direction, such forces are called unbalanced forces.  In this case, the object is in non-equilibrium state and the resultant force is non-zero. i.e. ∑F_x ≠ 0.

Q18: The object shown below moves with a constant velocity. Two forces are acting on the object. Considering negligible friction, the resultant force will be:

(a)  17N leftwards
(b)  10N leftwards
(c)  3N leftwards
(d)  7N rightwards

Answer: (c)  3N leftwards.


Q19: If the set of forces acting on an object are balanced, the object
(a) must be at rest.
(b) must be moving
(c) must not be accelerating.
(d) none of these.

Answer: (c) must not be accelerating. If the forces are balanced, the object will remain in its present state i.e. if it is moving, it will keep on moving with uniform speed. If it is at rest, it will remain at rest.

Q20(NCERT): An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.

Answer:  According to Newton's first law, in order to change the state of motion of a body, an external unbalanced force must be applied.
If an object is moving at a uniform speed and the object experiences a net zero external unbalanced force. It implies the object is in an equilibrium state and it will remain in its present state i.e. it will keep on moving with the same speed and direction.

Q21(NCERT): When a carpet is beaten with a stick, dust comes out of it. Explain.

Answer: Due to the inertia of rest. When a carpet is beaten with a stick, dust particles tend to remain in the inertia of rest and resist the change. Hence, the dust particles come out of the carpet.

Q22(NCERT): Why is it advised to tie any luggage kept on the roof of a bus with a rope?

Answer: While the bus is moving, luggage tends to remain in the inertia of motion state. When the bus stops, the luggage tends to resist the change and due to the inertia of motion it moves forward and may fall off. That's why it is advised to tie any luggage kept on the roof of a bus with a rope.


Q23: State Newton's second law of motion.

Answer: The second law of motion states that the rate of change of momentum of an object is
proportional to the applied unbalanced force in the direction of the force.

Mathematically, the law tells that force is the product of mass and acceleration.
i.e. F = ma
where m = mass of the object, a = acceleration and
F is the net force (i.e. F = ∑F_x = F₁ + F₂  + ... + F_n)


Q24: Which of the following graph represents unbalanced force.

Answer: An unbalanced force results in acceleration. Figure (b) represents this condition.
In fig. (a), v-t graph represents uniform motion ⇒ acceleration (a) = 0.
Fig. b, the v-t graph shows the velocity is not uniform, it is decreasing ⇒ acceleration is non-zero.
Fig. c is a Distance-time graph which shows uniform motion ( a straight line).


Q25: How Newton's Second law of motion is different from the First law?

Answer: Newton’s first law of motion deals with the behaviour of objects on which all existing forces are balanced. It gives us a notion of external force required to change the inertial state of the object.

While the second law of motion deals with the behaviour of objects on which all existing forces are not balanced. It gives us a measure of the force.

Q26: Name the categories of forces based on interaction.

Answer: Force exists only as a result of an interaction of two objects. Based on it, forces are categorised as:

• contact forces  and 
• non-contact-forces

Q27: What are contact forces? Give examples

Answer: Contact forces are forces in which the two interacting objects are physically in contact with each other. Examples of contact forces are:

1. Tension force
2. Normal force
3. Air resistance
4. Buoyant force
5. Friction force

Q28: What are non-contact forces? Give examples.

Answer: Non-contact forces are forces in which the two interacting objects are not in physical contact which each other, but are able to exert a push or pull. These are also called Action-at-a-distance forces. Examples are:

1. Gravitational force
2. Magnetic Forces
3. Electric Force
4. Nuclear Force

Q29 (CBSE 2010): Using the second law of motion, derive the relation between force and acceleration.

Answer: Let an object of mass, m is moving along a straight line with an initial velocity, u.
Let the final velocity = v m/s
Acceleration = a ms^-2
Time taken = t s
and Force applied = F.

Initial momentum (before the force applied) = P₁ = mu
Final momentum (after the force applied) = P₂ = mv

Change in momentum ∝ P₂ - P₁
and Rate of Change in momentum ∝ (P₂ - P₁)/t
⇒                ∝ (mv - mu)/t      ∝ m(v - u)/t

According to Newton's second law, Rate of Change in momentum = F
⇒ F ∝ m(v - u)/t
⇒ F ∝ ma                                (∵ v = u + at)
⇒ F = kma                (where k  is a constant of proportionality).

The SI units for m is 1 kg,  a is ms^-2 , thus k = 1
⇒ F = ma


SI unit of force is Newton (N) = kg-ms^-2

Q30: Define 1N (Newton).

Answer: One Newton is the force acts on a body of mass 1 kg and it produces an acceleration of
1 ms^-2.
⇒  1N  = 1kg × 1 ms^-2

Q31: What is the unit of force in c.g.s system?

Answer:  dyne (g cm s^-2)


Q32: What's the relation between 1N and 1 dyne?

Answer: 1 N = 1kg × 1 ms^-2
and 1 dyne = 1g × 1 cms^-2
1 N = 1kg × 1 ms^-2 = 1000g × 100 cms^-2
⇒ 1N =  10⁵ dyne


Q33: Is Newton's second law of motion is consistent with the first law? How?

Answer:  Yes Newton's second law is consistent with first law. The first law of motion can be
mathematically stated from the mathematical expression for the second law of motion.

Since F = ma = m(v - u)/t
or Ft = mv - mu.

⇒ F = 0 when v = u for any given time t.
⇒ The object will continue moving with uniform velocity, u throughout the time, t.

Similarly when object is at rest, u = v = 0. ⇒ F = 0. No external force is being applied, the object will remain at rest.


Q34(NCERT): A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because

(a) the batsman did not hit the ball hard enough.
(b) velocity is proportional to the force exerted on the ball.
(c) there is a force on the ball opposing the motion.
(d) there is no unbalanced force on the ball so the ball would want to come to rest.

Answer: (c) there is a force on the ball opposing the motion.
Frictional force exerted by the ground opposes the motion due to which the ball comes to rest.
(Note: Frictional force always acts in the direction opposite to the direction of motion.)


Q35(NCERT): A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400m  in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 metric tonnes. (Hint: 1 metric tonne = 1000 kg.)

Answer: Given, Truck starts from rest, initial velocity (u) = 0 m/s
Distance covered (S) = 400m
Time taken (t) = 20s
mass = 7 metric tonne = 7 × 1000 = 7000 kg
final velocity(v) = ?
acceleration (a) = ?
Force (F) = ?

Using equation of motion, S = ut + ½at², acceleration (a) is
i.e. 400 m = 0 + ½(a)(20)²
⇒ 400 × 2 = a × 400
⇒ a = 2 m/s²                                 ...(answer)

Using equation v = u + at
final velocity (v) = 0 + 2 × 2 = 4 m/s                                ...(answer)

Since Force (F) = mass(m) × acceleration (a)
⇒ F = 7000 kg × 2 = 14000 N                                          ...(answer)


Q36(CBSE 2010/NCERT A3): A hammer of mass 500 g, moving at 50 m/s, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?

Answer: Given, mass of hammer(m) = 500g = 0.5 kg
   initial velocity of hammer(u) = 50 m/s
   final velocity of hammer (v) = 0 m/s
   Duration (t) = 0.01s

   Using second law of motion, F = ma = m(v - u)/t
 ⇒ F = 0.5(0 - 50)/0.01 = (0.5 × -50)/0.01 = -2500N    ... (answer)
The -ve sign shows the force of 2500N acts in the opposite direction of motion.


Q37(NCERT): What is the momentum of an object of mass m, moving with a velocity v?
(a) (mv)²
(b) mv²
(c) ½ mv²
(d) mv

Answer: (d) mv

Q38(NCERT): A stone of 1 kg is thrown with a velocity of 20 ms^-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice? (Calculate time is taken and acceleration also).

Answer: mass of stone (m) = 1kg
 initial velocity of stone (u) = 20 m/s
 final velocity of stone (v) = 0 m/s
 Distance covered (S) = 50m
 time taken (t) = ?
 acceleration (a) = ?
 The friction force between stone and ice (F) = ?

Using equation of motion, v² - u² = 2aS , we have
(0)² - (20)² = 2 × a × 50m
-400 = 100a
⇒ a = -4 m/s²

The negative sign indicates that it is retardation acting in the opposite side of the motion.

Since  Force (F) = mass(m) × acceleration (a)
F = 1kg  × (-4m/s²)
⇒ F = -4 kg m/s² = -4 N
The -ve sign shows the force acts in the opposite direction of motion.

using equation v = u + at
i.e. 0 = 20 - 4t
⇒ -4t = -20
⇒ t = 5s


Q39: An object of mass 5 kg is moving with a velocity of 4 m/s. A constant force of 20 N acts on the object. What will be the velocity after 3 s.

Answer: Give, mass (m) = 5 kg
Initial velocity (u) = 4 m/s
Force (F) = 20 N
Time taken (t) = 3 s
acceleration (a) = ?
final velocity (v) = ?

Since F = ma  ⇒ a = F/m
∴ a = 20/5 = 4 m/s²

Using equation v = u + at
v = 4 + 4×3 = 4 + 12 = 16 m/s


Q40(CBSE 2010): While catching a fast moving ball, fielder gradually pulls his hand backwards. Give reasons.

Answer: By doing so, the fielder increases the time during which the high velocity of the moving ball
decreases to zero.  The acceleration of the ball is decreased and therefore the impact of catching the fast moving ball is also reduced.

We can prove it mathematically,

Let u be the initial velocity of the ball before it reaches fielder's hands. Let m be the mass of the ball.
Initial momentum of the ball = mu.

When the ball reaches field hands, it stops. ∴ final velocity (v) = 0.
And the final momentum = mv = 0

The ball will exert a force (F) on fielder's hands = Change in momentum/time (t₁)

In this case F₁ = (0 - mu)/ t₁ =  -mu/t₁

If fielder increases the time (say t₂)  i.e. t₂ > t₁ , the force will reduce
i.e.  F₂ = - mu/ t₂  and F₂ < F₁


Q41(CBSE 2010): A man pushes a box of mass 50 kg with a force of 80 N. What will be the acceleration of the box due to this force? What would be the acceleration if the mass were doubled?

Answer: Given,
   mass of the box (m) = 50kg
   Force (F) acting on box = 80N
   acceleration (a) = ?

   Since  Force (F) = mass(m) × acceleration (a)
  ⇒ a = F/m = 80/50 = 1.6 m/s²

   If mass becomes m = 2 × 50 = 100 kg
   acceleration (a) = 80 / 100 = 0.8 m/s²
Thus, the acceleration will be halved if the mass were doubled.


Q42: An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 ms^-2.

Answer: Given, mass of vehicle (m) = 1500 kg
negative acceleration (a) = -1.7 m/s²

According to Newton's 2nd law of motion,
Force (F) = mass(m) × acceleration (a)

F = 1500  × -1.7 m/s² =  -2550N

∴ The negative sign indicates a force of 2550N is to be applied in the opposite direction of motion.


Q43: State Newton's third law of motion.

Answer: It states that to every action there is always an equal and opposite reaction.
When two objects A and B act on each other, the force exerted by A on B (F_AB) is equal to the force exerted by object B on A (F_BA) in magnitude but are in opposite directions.

i.e. F_AB  =   -F_BA
or  F_AB  + F_BA = 0

One of the force (say F_AB) can be said action force then another one is reaction force (i.e. F_BA). Action-reaction forces always act on different bodies and their line of action is the same.

Q44: When two objects act on each other, does action-reaction force pair cancel out each other?

Answer: No. The action-reaction forces act on different bodies.


Q45: While driving on a highway, an insect strikes on the car windshield and splatters. Which experiences greater impact (force): an insect or the windshield?

Answer:  It is an example of the third law of motion. Both insect and the car windshield experiences the same amount of force but in opposite direction. But the force is too great for the insect that its body splatters while the windshield is able to withstand the impact.


Q46(NCERT): If action is always equal to the reaction, explain how a horse can pull a cart.

Answer:  According to Newton's third law of motion, action force is equal to the reaction but acts on two different bodies and in opposite directions. When a horse pushes the ground, the ground reacts and exerts a force on the horse in the forward direction. The force is able to overcome friction force of the cart and it moves.
(Note: A detailed explanation of house cart problem can be read here)


Q47: Identify the action and reaction forces in the following cases:
(a) A television (TV) lying on a table
(b) Pushing a wall with your hand
(c) Firing a bullet from a gun.
(d) Walking of a person on a ground.

Answer:
(a) A TV lying on a table: TV is interacting with the table surface. Applying Newton's third law of motion, action force acts on the table due to the TV (F_TV-Table) and a reaction force acts on TV by the table i.e. (F_Table-TV). Both these forces are equal in magnitude but in opposite directions.
⇒ F_Table-TV = - F_TV-Table

(b) Pushing a wall with your hand: When you push a wall, you exert a force (action) on the wall. Similarly, the wall exerts a (reaction) force on your hands in contact with the wall. According to Newton's third law of motion, both these forces form the action-reaction pair and act on different bodies. Both are equal in magnitude but opposite in direction.

(c) Firing a bullet from a gun: When a gun fires a bullet, a force is exerted (action) on a bullet and the gun experiences an equal recoil (reaction) force in opposite direction.

(d) Walking of a person on a ground: When a person pushes the ground (action) backwards by his foot, the ground also exerts an equal force (reaction) on his foot in the forward direction.

Q48: A TV set is lying on a table. The TV set experiences a gravitational force (pull)  downwards (F_W) by the Earth. It also experiences a force on it due to the table in contact i.e. (F_Table-TV) in the upward direction. Do these forces form an action-reaction pair? Do these forces follow Newton's Third Law of motion?

Answer: No because these two forces are acting on the same body, these are not due to Newton's third law of motion. These two forces do not form the action-reaction pair.

In fact, Newton's first law of motion does apply. Since the TV set is at rest because these forces are balanced. i.e. F_W = - F_Table-TV

A similar question (as Q48) was asked in CBSE 2010 examination, hope you may answer it!


Q 49(CBSE Exam/NCERT): According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move. 

Answer: Two cases arise here:
Case I: No external force is applied. In this case, the truck weight (gravitational force) acting downwards, is cancelled by the contact force applied on the truck by the ground.
i.e.
F_Weight = - F_ground

Case II: If a person or group of persons push truck, the static friction acting on the truck horizontally, cancels this push-force. Static friction is a self-adjusting force. More push is applied, more static friction will oppose it (up to a limit) and forces are balanced. In order to move the truck, the push-force must overcome static friction.

i.e.  F_Weight = - F_ground
and  F_Push = - F_Friction

Q50 (NCERT): Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.

Answer: A large amount of water coming out from the nozzle of the hose at high velocity will have large momentum. Due to a large momentum, water comes out exerting a large force. According to Newton's third law of motion, the hose will also experience an equal reaction force but in opposite direction. Because of this reaction force, it is difficult for a fireman to hold a hose.

Q51 (CBSE 2010): It is difficult to balance our body when we accidentally slip on a peel of a banana. Explain why?

Answer: A frictional force always acts parallel to the surface and is directed to oppose sliding. Banana skin reduces friction (or frictional force) and thus brings a body in the unbalanced state and we tend to fall.


Q52(NCERT): Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?

Answer: A friction force (here it is kinetic or sliding friction) always acts parallel to surface but in the direction opposing the motion. Since the wooden cabinet is moving with uniform speed and a horizontal force of 200N acts on it. According to Newton's first law of motion, forces are balanced. Hence, a frictional force of 200 N is exerted on the cabinet.


Q53: State Law of conservation of momentum.

Answer: According to the law of conservation of momentum, in the absence of external unbalanced force the total momentum of a system of objects remains unchanged or conserved by a collision.

i.e. Total Initial Momentum Before Reaction = Total Initial Momentum After Reaction

This law holds good for any number of objects.


Q54: Law of conservation of momentum is applicable to an isolated system. What do you mean by an isolated system?

Answer: Isolated system refers to the absence of external unbalanced force or net force is zero.


Q55(NCERT EXEMPLAR): Suppose a ball of mass m is thrown vertically upward with an initial speed v, its speed decreases continuously till it becomes zero. Thereafter, the ball begins to fall downward and attains the speed v again before striking the ground. It implies that the magnitude of initial and final momentums of the ball are same. Yet, it is not an example of conservation of momentum. Explain why?

Answer: The reason is that there exists external unbalanced force i.e. gravitation pull of the earth due to which ball falls down and is accelerating. Therefore it is not an example of conservation of momentum.

Q56: What is a collision? What are the types of collision?

Answer: The interaction between two or more bodies causing the exchange of momentum is called
collision. It is categorised as of two types:

• elastic collision (both momentum and kinetic energy are conserved)
• inelastic collision (the only momentum is conserved)

Q57(NCERT): From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m/s. Calculate the initial recoil velocity of the rifle.

Answer: Initially both rifle and bullet are at rest. Given,
mass of rifle (m₁) = 4 kg
initial velocity of rifle (u₁) = 0 m/s
final velocity of rifle (v₁) = ?

mass of bullet (m₂) =  50g = 50 × 10^-3 kg = 0.05 kg
initial velocity of bullet (u₂) = 0 m/s
final velocity of bullet (v₂) =  35 m/s


Applying the law of conservation of momentum,
    total  initial momentum = total final momentum

i.e. (m₁u₁) + (m₂u₂) = (m₁v₁) + (m₂v₂)
⇒   0 + 0 = (4v₁) + (0.05 × 35)
⇒   - 4v₁ = 0.05 × 35
⇒  v₁ = - 0.4375 m/s                           ... (answer)
The negative sign indicates that rifle recoils in opposite direction to bullet i.e. backwards


Q58: Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 ms^-1 and 1 ms^-1, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 ms^-1. Determine the velocity of the second object.

Answer:

mass of the first object (m₁) = 100g = 0.1 kg
velocity of the first object (u₁) before collision = 2 m/s
velocity of the first object (v₁) after collision = 1.67 m/s

mass of the second object (m₂) = 200g = 0.2 kg
velocity of the second object (u₂) before collision = 1 m/s
velocity of the second object (v₂) after collision = ?  m/s

Since no unbalanced external force is acting on the system, total momentum is conserved.
i.e.  total  initial momentum = total final momentum

⇒ (m₁u₁) + (m₂u₂) = (m₁v₁) + (m₂v₂)
⇒ (0.1 × 2) + (0.2 × 1) = (0.1 ×1.67) + (0.2 × v₂)
⇒  0.2 + 0.2 = 0.167 + 0.2v₂
⇒ 0.2v₂ = 0.4 - 0.167 = 0.233
⇒ v₂  = 0.233 / 0.2 = 1.165  m/s


Q59(NCERT): Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 m/s before the collision during which they stick together. What will be the velocity of the combined object after the collision?

Answer: Given
mass of the first object (m₁) = 1.5 kg
velocity of the first object (u₁) before collision = 2.5 m/s

mass of the second object (m₂) = 1.5kg
velocity of the second object (u₂) before collision = -2.5 m/s
The negative sign indicates the second object moves in the direction opposite to the first one.

mass of the combined objects after collision (m₃) =  m₁+ m₂= 3.0 kg
velocity of the combined objects after collision (v₃) =?

Being an isolated system, total momentum is conserved i.e.
(m₁u₁) + (m₂u₂) =  (m₃v₃)
⇒ (1.5 × 2.5) + (1.5 × -2.5) = 3.0 × v₃
⇒  0 = 3.0 × v₃
⇒ v₃ = 0 m/s

Q60: What is impulse?

Answer: The effect of a force applied for a short duration is called impulse. It is the product of force (F) and the time duration (t) for which the force is applied.
i.e. Impulse = F × t

Impulse is a vector quantity and its SI unit is N-s.

Since F = ma = m(v-u)/t
⇒ F × t = mv - mu
⇒ Impulse is equal to change in momentum.


Q61(NCERT): A hockey ball of mass 200 g travelling at 10 ms^-1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms^-1. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

Answer: Given, mass of ball = 200g = 0.2 kg
initial velocity of ball (u) = 10 m/s
final velocity of ball (v) = -5 m/s
(When ball struck by Hockey stick, it returns to its original position, means the direction of ball reverses. Hence -ve sign of final velocity).

∴ initial momentum P_i = mu = 0.2 × 10 = 2 kg m/s
and final momentum (P_f) = mv = 0.2 × - 5 = -1 kg m/s

Change in momentum (ΔP) = P_f - P_i = -1 - 2 = -3 kg m/s
Momentum is a vector quantity. The -ve sign indicates the direction of momentum i.e. same as final velocity.



Q62(NCERT/CBSE 2010): Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. 

(a) Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). 

(b) Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result, the insect died. 

(c) Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. 
Comment on these suggestions.

Answer: Let us consider the following scenario,

mass of the car (m_c) = 1500 kg
speed of car before collision (u_c) = 90 km/hr = 25 m/s

mass of the insect (m_i) = 10g = 0.01 kg
speed of insect before collision (u_i) = -5 m/s
(-ve sign indicates its direction is opposite to car's motion)

final speed of (car + insect) after collision (v) = ? m/s

Since both car and insect are moving with uniform speed, and there is not external force applied. Hence, total momentum is conserved, i.e.

(m_cu_c) + (m_iu_i) =  (m_c+m_i)v

⇒ (1500 × 25) + (0.01 × -5) = (1500 + 0.01) × v
⇒ 37500 - 0.05 = 1500.01 × v
⇒ 37499.95 = 1500.01 × v
⇒ v = 37499.95 / 1500.01= 24.9998 m/s

So the velocity of insect changes from -5 m/s to 24.9998 m/s   (big change)
The velocity of car changes from 25 m/s to 24.9998 m/s (almost a negligible change)

Change in momentum of the car = (m_cv) - (m_cu_c) = m_c(v - u_c)
   = 1500 × (24.9998 - 25) = - 0.3 N-s

Change in momentum of the insect = (m_iv) - (m_iu_i) = m_i(v - u_i)
   = 0.01 × ( 24.9998 + 5) = 0.3 N-s

⇒ The momentum gained by the insect is equal to the momentum lost by the car.

(a) Kiran is incorrect in terms of change in momentum. However, s/he is correct, the change in velocity in the insect is higher than that of the car.

(b) Akhtar observation is correct that speed of the car is higher than that of insect. However, he is incorrect that the insect experiences the larger force. According to Newton's third law of motion, both experienced the same amount of force.

(c) Rahul is correct that both experienced same force. If he says the amount of change in momentum in both object is same, then he is correct.


Q63: A bullet of mass 10 g travelling horizontally with a velocity of 150 m/s strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also, calculate the magnitude of the force exerted by the wooden block on the bullet.

Answer:  Given,
  mass of the bullet (m) = 10g = 0.01 kg
  initial velocity of the bullet (u) = 150 m/s
  final velocity of the bullet (v)  = 0 m/s
  time taken (t) = 0.03s
  Distance (S) = ? m
  Force (F) = ? N
  acceleration (a) = ? m/s²

  Using equation v = u + at
  0 = 150 + a × 0.03
⇒ a = -150/0.03 = -5000 m/s²

  The -ve sign of acceleration indicates that it is retardation.
  Using equation of motion  v² - u² = 2aS, we have
  (0)² - (150)² =  2 × (-5000) × (S)
   - 22500 = -10000 × (S)
⇒ S = 22500/10000 = 2.25 m

    Using Newton's second law of motion i.e. F = ma
    F = 0.01  × (-5000) = -50N
 The -ve sign of force indicates that the force opposes the motion.


Q64(NCERT): A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate:
(a) the net accelerating force;
(b) the acceleration of the train; and
(c) the force of wagon 1 on wagon 2.

Answer:
(a)Force generated by engine = 40000N
    Friction force by track = -5000N
∴ Net force (F_net) acting on wagons = 40000 - 50000 = 35000N

(b) acceleration of wagons = F /(total mass of wagons)
Note here we are assuming net force acting on wagons by the engine and the mass of the engine is not considered in the system.
Total mass of 5 wagons = 5 × 2000kg  = 10000kg
acceleration (a) =  35000N / 10000kg = 3.5 m/s²
⇒ all wagons are moving with the same acceleration i.e. 3.5 m/s²

(c) Consider the free body diagram of  the first wagon

Two forces are acting on the first wagon and it is moving with acceleration 'a'
F_net is the force applied by the engine.
T₁ is the action-reaction force (Newton's third law) on wagon-1 due to wagon-2 . Same force T₁ is also acting on wagon-2 by the first wagon. Since there is acceleration,

⇒ F_net - T₁ = m₁a                        (where  m₁ is the mass of wagon-1)
⇒ T₁ = F_net - m₁a = 35000 - (2000 × 3.5 ) = 35000 - 7000 = 28000 N

Note: If you consider engine part of the system, total mass = mass of engine + wagons = 18000N. Acceleration (a) = 35000/18000 = 1.944 m/s². Then engine and wagons will move with the same acceleration. In some books, two accelerations are computed. In some books, two accelerations are computed. One including engine plus wagons. And another for wagons separately, which I think is incorrect.

Q65(NCERT): An object of mass 1 kg travelling in a straight line with a velocity of 10 m/s collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the
same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.

Answer: Given, mass of object (m₁) = 1 kg
initial velocity of object before collision (u₁) = 10 m/s

mass of wooden block (m₂) = 5kg
velocity of wooden block before collision (u₂) = 0 m/s
final velocity of object+wooden black (v)  after collision = ? m/s

Since momentum is conserved,
total initial momentum = total final momentum
i.e.  (m₁u₁) + (m₂u₂) = (m₁+ m₂)v
⇒ 1 × 10 + 0 = (1 + 5)v
⇒ v = 10/6 m/s  = 5/3 m/s

Initial momentum = (m₁u₁) + (m₂u₂) = 1 × 10 + 0 = 10 kg m/s
Final momentum = (m₁+ m₂)v = (1 + 5) × (5/3) = 6 × (5/3) = 10 kg m/s

Q66(NCERT): The following is the distance-time table of an object in motion:

Time in seconds	Distance in meters
0	0
1	1
2	8
3	27
4	64
5	125
6	216
7	343

(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?

(b) What do you infer about the forces acting on the object?

Answer:  (a) Since the distance travelled in equal intervals of time is not equal rather it is increasing. It means the object is moving with a non-uniform velocity. It is accelerating.
By looking at the table we find the distance (S) ∝ t³          ...(I)
Let us check if acceleration is constant. We know that for fixed 'a', equation S = ut + ½at² is valid.
In this case, u = 0, (since at t = 0, S  = 0), ⇒ S = ½at²   or  (S) ∝ t²                  ... (II)
But from the table, we get S ∝ t³, we conclude, that 'a' is increasing.

(b) Since the object is in accelerated condition, According to Newton's Second Law, F ∝ a.
We can say, unbalanced force is acting on the object.

Q67(NCERT): Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2m s^-2. With what force does each person push the motorcar?
(Assume that all persons push the motorcar with the same muscular effort.)

Answer: Let each person applies force = F N
Initially, car is moving with uniform speed. When two persons apply force, there is no change in state, the car moves with same uniform speed.  It means the force of the two persons is cancelled by the kinetic friction of the car.

⇒  2F + F_{kinetic-friction} = 0                                        ... (I)

When the third person applies there occurs a change in state, due to which acceleration (a) occurs.
⇒ 3F + F_{kinetic-friction} = ma
where m is the mass of the car = 1200 kg and a is acceleration = 0.2 ms^-2.

Using Eq. I, we get
   F = ma = 1200  × 0.2 = 240N
∴ Each person applies 240N force.


Q68(NCERT): A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also, calculate the magnitude of the force required.

Answer:  Given,
    mass of  the car (m) = 1200 kg
    initial velocity of car (u) = 90 km/h = 90 × 1000m/3600s = 25 m/s
    final velocity of car (v) = 18 km/h = 5 m/s
    time taken (t) = 4s
    acceleration (a) = ? m s^-2.
    Force applied (F) = ? N

    Using equation of motion v = u + at
⇒ 5 = 25 + 4a
⇒ 4a = -25+ 5 = -20
⇒ a = -5 m s^-2
   The -ve sign indicates it is retardation.

Change in momentum = mv - mu = m(v - u) = 1200 × (5 - 25) = 1200 × -20 = -24000 kg m s^-1.

According to 2nd law of motion, Force(F)  = mass (m) × acceleration (a)
⇒ F = 1200 × -5 = -6000 N
-ve sign indicates a force of 6000N acts on car opposing its motion.


Q69: A spring scale reads 20 N as it pulls a 4.0 kg object across a table. What is the magnitude of the force exerted by the object on the spring scale?
(a) 40 N
(b) 20 N
(c) 4 N
(d) 5 N

Answer: (b) 20 N (Hint: Newton's 3rd Law Action-reaction force).


Q70: A boy throws a ball upwards with a speed of 10m/s. The mass of the ball is 400g. Calculate
(i) Weight of the ball.
(ii) Initial momentum of the ball.
(iii) How high the ball will go?
(iv) The momentum of the ball at its highest point.
(Take acceleration towards gravity (g) = 10 m/s²)

Answer: Given,
 mass of the ball (m) = 400g = 0.4 kg
 initial speed of the ball (u) = 10 m/s
 When ball will reach at its highest point, its velocity will become zero and it will start to fall.
 Velocity at the highest point (v) = 0 m/s

  If we take upward direction positive(+ve), the downward direction will be taken as (-ve).
∴ acceleration due to gravity (g) = -10 m/s²

(a) Weight of the ball = mass × acceleration due to gravity(g) = 0.4 × -10 = -4N
-ve sign indicates weight (force) acts downwards.

(b) Initial momentum = mass(m) × initial velocity(u) = 0.4 × 10 = 4 kg m/s²

(c)  Consider equation of motion,  v² - u² = 2aS,
The height (h) will be,
(0)² - (10)² = 2 (-10)h
⇒ -100 = -20h
⇒ h = 100/20 = 5m

(d) Momentum at the highest point = mass (m) × final velocity(v) = 10 × 0 = 0 kg m/s²

Q71(HOTS): Are there any limitations to Newton's Laws of motion?

Answer: Newton's laws of motion hold true for common practical applications. However, it has certain limitations:

1. Second Law (F = ma) assumes mass is constant. When an object travels at the speed of the light, its mass is also affected. Newton's law is not applicable in this case. 
2. Newton's Laws fail to explain the motion of electrons around the nucleus.
3. These laws also fail to explain about black holes and bending of gravity due to starlight.
4. It also fails to explain why a leaf falls in river stream, its path cannot be determined (chaos theory) 

Later these challenges were explained by Einstein's theory of relativity and quantum physics which became the basis of modern physics.

Q72: Two objects having their masses in ratio 3:5 are acted upon by two forces each on one object. The forces are in the ration of 5:3. Find the ratio in their accelerations.

Answer: Since masses are in ratio 3:5.
Let the mass of the objects be 3x and 5x.
Let  F₁ and F₂are the two forces with a₁ and a₂ accelerations.
∴ F₁ = m₁a₁ = 3xa₁
and F₂ = m₂a₂ = 5xa₂
Since F₁:F₂ = 5:3, we have
3xa₁:5xa₂ :: 5:3
⇒ a₁:a₂ = 5×5:3×3
⇒ a₁:a₂ = 25:9

Q73: The speed-time graph of a car is shown below. The car weighs 1000kg. 
(i) Find the distance covered by the in fits two seconds.
(ii) Find how much force is applied by the car brakes in the fifth second so that the car comes to an halt by sixth second.

Answer: Given mass of the car(m) = 1000kg
(i) As shown in the graph, Distance covered in first 2 secs = Area of Δ inscribed in the first 2s.
  =  ½ × base × height =  ½ × 2 × 15 = 15m

(ii) Force (F) = mass(m) × acceleration (a)
Time taken by force to stop the car = 6 - 5 = 1 second
Initial Velocity (i.e. velocity at B) = 15 m/s
Final velocity (at point C) = 0 m/s
Using equation v = u + at
a = (v - u)/t = (0 - 15)/1 = -15 m/s² (-ve sign indicates it is retardation)
Force applied by brakes = 1000 × -15 = -15000N = -15KN


Q74: Does Newton's third law apply to a system where bodies do not actually touch each other?  

Answer:  Yes, whenever the bodies are in actual contact or not (e.g., attraction or repulsion between two magnets), Newton's third law is applicable. 


Q75: An imaginary co-ordinate system which is either at rest or in uniform and where Newton's laws are valid is called an _______ frame of reference. (Fill in the blanks)


Answer: Inertial

See Force and Laws of Motion (Worksheet)

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