CBSE - Force and Laws of Motion

Inertia is the measure of the mass of the body. The greater is the mass of the body; the greater is its inertia and vice-versa.

(a) Mass of a stone is more than the mass of a rubber ball for the same size. Hence, inertia of the stone is greater than that of a rubber ball.

(b) Mass of a train is more than the mass of a bicycle. Hence, inertia of the train is greater than that of the bicycle.

(c) Mass of a five rupee coin is more than that of a one-rupee coin. Hence, inertia of the five rupee coin is greater than that of the one-rupee coin.

The velocity of football changes four times.
First, when a football player kicks to another player, second when that player kicks the football to the goalkeeper. Third when the goalkeeper stops the football. Fourth when the goalkeeper kicks the football towards a player of his own team.

Agent supplying the force:
→ First case – First player
→ Second case – Second player
→ Third case – Goalkeeper
→ Fourth case – Goalkeeper

Some leaves of a tree get detached when we shake its branches vigorously because branches comes in motion while the leaves tend to remain at rest due to inertia of rest.

In a moving bus, a passenger moves with the bus due to inertia of motion. As the driver applies brakes, the bus comes to rest. But, the passenger tries to maintain to inertia of motion. As a result, a forward force is exerted on him.
Similarly, the passenger tends to fall backwards when the bus accelerates from rest because when the bus accelerates, the inertia of rest of the passenger tends to oppose the forward motion of the bus. Hence, the passenger tends to fall backwards when the bus accelerates forward.

A horse pushes the ground in the backward direction. According to Newton's third law of motion, a reaction force is exerted by the Earth on the horse in the forward direction. As a result, the cart moves forward.

When a fireman holds a hose, which is ejecting large amounts of water at a high velocity, then a reaction force is exerted on him by the ejecting water in the backward direction. This is because of Newton's third law of motion. As a result of the backward force, the stability of the fireman decreases. Hence, it is difficult for him to remain stable while holding the hose.

Mass of the rifle, m₁= 4 kg
Mass of the bullet, m₂= 50g= 0.05 kg
Recoil velocity of the rifle= v₁
Bullet is fired with an initial velocity, v₂= 35m/s
Initially, the rifle is at rest.
Thus, its initial velocity, v= 0
Total initial momentum of the rifle and bullet system= (m₁+m2₂)v= 0
Total momentum of the rifle and bullet system after firing:
= m₁v₁ + m₂v₂= 0.05 × 35= 4v₁ + 1.75

According to the law of conservation of momentum:
Total momentum after the firing = Total momentum before the firing  4v₁ + 1.75= 0
v₁= -1.75 / 4= -0.4375 m/s
The negative sign indicates that the rifle recoils backwards with a velocity of 0.4375 m/s.

Yes, an object may travel with a non-zero velocity even when the net external force on it is zero. A rain drop falls down with a constant velocity. The weight of the drop is balanced by the up thrust and the velocity of air. The net force on the drop is zero.

When the carpet is beaten, it is suddenly set into motion. The dust particles tend to remain at rest due to inertia of rest, therefore the dust comes out of it.

When a bust starts suddenly, the lower part of the luggage kept on the roof being in contact with the bus begins to move forward with the speed of bus, but the upper part tends to remain at rest due to inertia of rest. Therefore, the upper part is left behind and hence luggage falls backward. So, it is advised to tie any luggage kept on the roof of a bus with a rope.

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

The ball slows down and comes to rest due to opposing forces of air resistance and frictional force on the ball opposing its motion. Therefore the choice (c) there is a force on the ball opposing the motion is correct.

Initial velocity, u = 0 
Distance travelled, s = 400 m
Time taken, t = 20 s

We know, s = ut + ½ at²

Or, 400 = 0 + ½ a (20)²
Or, a = 2 ms^–2
Now, m = 7 MT = 7000 kg, a = 2 ms^–2
Or, F = ma = 7000 × 2 = 14000 N Ans.

Initial velocity of the stone, u= 20 m/s
Final velocity of the stone, v= 0
Distance covered by the stone, s= 50 m
Since, v² - u²
2 = 2as,
Or, 0 - 202 = 2a × 50,
Or, a = – 4 ms^-2
Force of friction, F = ma = – 4N

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 exerted by the engine, F = 40000 N
Frictional force offered by the track, F_f = 5000 N 
Net accelerating force, Fa = F − F_f = 40000 − 5000 = 35000 N
Hence, the net accelerating force is 35000 N.

(b) Acceleration of the train = a
The engine exerts a force of 40000 N on all the five wagons. 
Net accelerating force on the wagons, Fa = 35000 N
Mass of the wagons, m = Mass of a wagon x Number of wagons
Mass of a wagon = 2000 kg
Number of wagons = 5
∴ m = 2000 × 5 = 10000 kg
Total mass, M = m  = 10000 kg
From Newton’s second law of motion:
Fa= Ma
a=Fam   = 35000 10000    = 3.5 ms^-2
Hence, the acceleration of the wagons and the train is 3.5 m/s².

(c) Mass of all the wagons except wagon 1 is 4 × 2000 = 8000 kg
Acceleration of the wagons = 3.5 m/s²
Thus, force exerted on all the wagons except wagon 1 
= 8000 × 3.5 = 28000 N
Therefore, the force exerted by wagon 1 on the remaining four wagons is 28000 ​ N.
Hence, the force exerted by wagon 1 on wagon 2 is 28000 ​ N.

Mass of the automobile vehicle, m= 1500 kg 
Final velocity, v= 0 (finally the automobile stops)
Acceleration of the automobile, a= −1.7 ms⁻²
From Newton’s second law of motion:
Force = Mass x Acceleration = 1500 x (−1.7) = −2550 N
Hence, the force between the automobile and the road is −2550 N, in the direction opposite to the motion of the automobile.

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
Mass of the object = m
Velocity = v
Momentum = Mass x Velocity 
Momentum = mv

The cabinet will move with constant velocity only when the net force on it is zero.
Therefore, force of friction on the cabinet = 200 N, in a direction opposite to the direction of motion of the cabinet. 

Mass of one of the objects, m₁ = 1.5 kg
Mass of the other object, m₂ = 1.5 kg 
Velocity of m1 before collision, u₁ = 2.5 m/s
Velocity of m2, moving in opposite direction before collision, u₂ = −2.5 m/s
Let v be the velocity of the combined object after collision. By the law of conservation of momentum,
Total momentum after collision = Total momentum before collision,
Or, (m₁ + m₂) v = m₁u₁ + m₂u₂
Or, (1.5 + 1.5) v = 1.5 × 2.5 +1.5 × (–2.5) [negative sign as moving in opposite direction]
Or, v = 0 ms^–1

The logic is that Action and Reaction always act on different bodies, so they can not cancel each other. When we push a massive truck, the force of friction between its tyres and the road is very large and so the truck does not move.

Mass of the hockey ball, m = 200 g = 0.2 kg
Hockey ball travels with velocity, v₁ = 10 m/s
Initial momentum = mv₁
Hockey ball travels in the opposite direction with velocity, v₂ = −5 m/s
Final momentum = mv₂
Change in momentum = mv₁ − mv₂ = 0.2 [10 − (−5)] = 0.2 (15) = 3 kg m s⁻¹
Hence, the change in momentum of the hockey ball is 3 kg m s⁻¹.

Initial velocity, u= 150 m/s
Final velocity, v= 0 (since the bullet finally comes to rest)
Time taken to come to rest, t= 0.03 s
According to the first equation of motion, v= u + at
Acceleration of the bullet, a
0 = 150 + (a × 0.03 s)a = -150 / 0.03 = -5000 m/s²

(Negative sign indicates that the velocity of the bullet is decreasing.)

According to the third equation of motion:
v²= u2²2as
0 = (150)²+ 2 (-5000)
= 22500 / 10000
= 2.25 m

Hence, the distance of penetration of the bullet into the block is 2.25 m.
From Newton’s second law of motion:
Force, F = Mass × Acceleration 
Mass of the bullet, m = 10 g = 0.01 kg 
Acceleration of the bullet, a = 5000 m/s²
F = ma = 0.01 × 5000 = 50 N
Hence, the magnitude of force exerted by the wooden block on the bullet is 50 N.

Mass of the object, m₁ = 1 kg
Velocity of the object before collision, v₁ = 10 m/s 
Mass of the stationary wooden block, m₂ = 5 kg
Velocity of the wooden block before collision, v₂ = 0 m/s
∴ Total momentum before collision = m₁ v₁+ m₂ v₂
= 1 (10) + 5 (0) = 10 kg m s⁻¹ 

It is given that after collision, the object and the wooden block stick together.
Total mass of the combined system = m₁ + m₂
Velocity of the combined object = v
According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
m₁v₁ + m₂ v₂ = (m₁ + m₂) v
1 (10) + 5 (0) = (1 + 5) v
v = 10 / 6
= 5 / 3

The total momentum after collision is also 10 kg m/s.
Total momentum just before the impact = 10 kg m s−1
Total momentum just after the impact = (m1 + m2) v = 6 × 5 / 3 = 10 kg ms-1
Hence, velocity of the combined object after collision = 5 / 3 ms-1

Initial velocity of the object, u = 5 m/s
Final velocity of the object, v = 8 m/s
Mass of the object, m = 100 kg
Time take by the object to accelerate, t = 6 s
Initial momentum = mu = 100 × 5 = 500 kg m s⁻¹
Final momentum = mv = 100 × 8 = 800 kg m s⁻¹
Force exerted on the object, F = mv - mu / t
= m (v-u) / t
= 800 - 500
=  300 / 6
= 50 N

Initial momentum of the object is 500 kg m s⁻¹.
Final momentum of the object is 800 kg m s.⁻¹
Force exerted on the object is 50 N.

Akhtar, Kiran and Rahul were riding in a motorocar 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. 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). 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. 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

The suggestion made by Kiran that the insect suffered a greater change in momentum as compared to the change in momentum of the motor car is wrong. 
The suggestion made by Akhtar that the motor car exerted a larger force on the insect because of large velocity of motor car is also wrong. The explanation put forward by Rahul is correct. On collision of insect with motor car, both experience the same force as action and reaction are always equal and opposite. Further, changes in their momenta are also the same. Only the signs of changes in momenta are opposite, i.e., change in momenta of the two occur in opposite directions, though magnitude of change in momentum of each is the same.

Mass of the dumbbell, m = 10 kg
Distance covered by the dumbbell, s = 80 cm = 0.8 m
Acceleration in the downward direction, a = 10 m/s²
Initial velocity of the dumbbell, u = 0
Final velocity of the dumbbell (when it was about to hit the floor) = v
According to the third equation of motion:
v² = u² + 2as
v² = 0 + 2 (10) 0.8
v = 4 m/s 
Hence, the momentum with which the dumbbell hits the floor is 
= mv = 10 × 4 = 40 kg m s⁻¹

Mass of the motor car = 1200 kg
Only two persons manage to push the car. Hence, the acceleration acquired by the car is given by the third person alone.
Acceleration produced by the car, when it is pushed by the third person, 
a = 0.2 m/s²
Let the force applied by the third person be F.
From Newton’s second law of motion:
Force = Mass × Acceleration 
F = 1200 × 0.2 = 240 N
Thus, the third person applies a force of magnitude 240 N.
Hence, each person applies a force of 240 N to push the motor car. 

Mass of the hammer, m = 500 g = 0.5 kg 
Initial velocity of the hammer, u = 50 m/s
Time taken by the nail to the stop the hammer, t = 0.01 s
Velocity of the hammer, v = 0 (since the hammer finally comes to rest)
From Newton’s second law of motion: 
Force, f = m(v-u) / t
= 0.5(0-50) / 0.01
= -2500 N

The hammer strikes the nail with a force of −2500 N. Hence, from Newton’s third law of motion, the force of the nail on the hammer is equal and opposite, i.e., +2500 N.

Mass of the motor car, m = 1200 kg 
Initial velocity of the motor car, u = 90 km/h = 25 m/s 
Final velocity of the motor car, v = 18 km/h = 5 m/s 
Time taken, t = 4 s 
According to the first equation of motion:
v = u + at
5 = 25 + a (4)
a = − 5 m/s²
Negative sign indicates that its a retarding motion i.e. velocity is decreasing.
Change in momentum = mv − mu = m (v−u)
= 1200 (5 − 25) = − 24000 kg m s⁻¹
Force = Mass × Acceleration
= 1200 × − 5 = − 6000 N
Acceleration of the motor car = − 5 m/s²
Change in momentum of the motor car = − 24000 kg m s⁻¹
Hence, the force required to decrease the velocity is 6000 N.
(Negative sign indicates retardation, decrease in momentum and retarding force)

The following is the distance-time table of an object in motion:

(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) There is an unequal change of distance in an equal interval of time.
Thus, the given object is having a non - uniform motion. Since the velocity of the object increases with time, the acceleration is increasing.
(b) The object is in accelerated condition. According to Newton's second law of motion, the force acting on an object is directly proportional to the acceleration produced in the object. So, we can say unbalanced force is acting on the object.

1. Which of the following statement is not correct for an object moving along a straight path in an accelerated motion?
(a) Its speed keeps changing
(b) Its velocity always changes
(c) It always goes away from the earth
(d) A force is always acting on it
Ans. (c) It always goes away from the earth
Explanation: To move away from the earth, an object needs the acceleration which is more than acceleration due to gravity. Only moving on a straight path is not enough for an object to escape the gravitation of the earth.

2. According to the third law of motion, action and reaction
(a) always act on the same body
(b) always act on different bodies in opposite directions
(c) have same magnitude and directions
(d) act on either body at normal to each other
Ans. (b) always act on different bodies in opposite directions
Explanation: Action and reaction act on different bodies but in opposite directions. They have the same magnitude.

3. A goalkeeper in a game of football pulls his hands backwards after holding the ball shot at the goal. This enables the goal keeper to
(a) exert larger force on the ball
(b) reduce the force exerted by the ball on hands
(c) increase the rate of change of momentum
(d) decrease the rate of change of momentum
Ans. (b) reduce the force exerted by the ball on hands
Explanation: Pulling the hand backwards allows enough time to reduce the momentum of the ball. This helps in reducing the force exerted by the ball on hands.

4. The inertia of an object tends to cause the object
(a) to increase its speed
(b) to decrease its speed
(c) to resist any change in its state of motion
(d) to decelerate due to friction
Ans. (c) to resist any change in its state of motion
Explanation: Inertia is the property because of which an object resists any change in its state of motion.

5. A passenger in a moving train tosses a coin which falls behind him. It means that motion of the train is 
(a) accelerated
(b) uniform
(c) retarded
(d) along circular tracks
Ans. (a) accelerated
Explanation: Had the motions of the train been uniform, the coin would have fallen in his hand. Had the motion been retarded, the coin would have fallen ahead of him. So,  motion is accelerated.

6. An object of mass 2 kg is sliding with a constant velocity of 4 ms–1 on a frictionless horizontal table. The force required to keep the object moving with the same
velocity is
(a) 32 N
(b) 0 N
(c) 2 N
(d) 8 N
Ans. (b) 0 N
Explanation: Since no friction is opposing the motion, hence no force is required to keep the object in uniform motion.

7. Rocket works on the principle of conservation of
(a) mass
(b) energy
(c) momentum
(d) velocity
Ans. (c) momentum

8. A water tanker filled up to 2/3 of its height is moving with a uniform speed. On sudden application of the brake, the water in the tank would
(a) move backward
(b) move forward
(c) be unaffected
(d) rise upwards
Ans. (b) move forward
Explanation: On sudden application of brake, the tanker would come to rest but water would remain in motion. Due to this, the water in the tank would move forward. 
 

Steel- As the mass is a measure of inertia, the ball of same shape and size, having more mass than other balls will have highest inertia. Since steel has greatest density and greatest mass, therefore, it has highest inertia.

Yes. the balls will start rolling in the direction in which the train was moving. Due to the application of the brakes, the train comes to rest but due to inertia the balls try to remain in motion, therefore, they begin to roll. Since the masses of the balls are not the same, therefore, the inertial forces are not same on both the balls. Thus, the balls will move with different speeds.

According to law of conservation of momentum; the momentum of bullet will be equal to the momentum of rifle. In case of light rifle; velocity will be more than the velocity of heavier rifle so that momentum (product of mass and velocity) for both shall be equal. Due to this, the lighter rifle will hurt the shoulder more. 

When a cart is moving on the road, it has to encounter friction. To maintain a constant speed, some force need to be applied continuously to overcome the friction. Hence, the horse needs to continuously apply a force in order move the cart with a constant speed. 

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?

Ans. Law of conservation of momentum is applicable to isolated system (no external force is applied). In this case, the change in velocity is due to the gravitational force of earth.

Separation between them will increase. Initially the momentum of both of them are zero as they are at rest. In order to conserve the momentum, the one who throws the ball would move backward. The second will experience a net force after catching the ball and herefore will move backwards that is in the direction of the force.

The working of the rotation of sprinkler is based on third law of motion. As the water comes out of the nozzle of the sprinkler, an equal and opposite reaction force comes into play. So, the sprinkler starts rotating. 

Acceleration can be given as follow:

Acceleration can be given as follow: 

When mass is doubled and force is halved;

So, acceleration becomes one-fourth.

What is momentum? Write its SI unit. Interpret force in terms of momentum. Represent the following graphically.

(a) momentum versus velocity when mass is fixed.
(b) momentum versus mass when velocity is constant

Ans. Momentum = mass � velocity
SI unit of momentum is kg ms-1
Force = Rate of change in momentum

Derive the unit of force using the second law of motion. A force of 5 N produces an acceleration of 8 ms^-2 on a mass m1 and an acceleration of 24 ms ² on a mass m2. What acceleration would the same force provide if both the masses are tied together?

F = m a = kg ms^-2 
This unit is also called newton. Its symbol is N.