Chapter - 9 Force and Laws of Motion

Inertia of an object tends to resist any change in its state of rest or state of motion. When a carpet is beaten with a stick, then the carpet comes to motion. But, the dust particles try to resist their state of rest. According to Newton's first law of motion, the dust particles stay in a state of rest, while the carpet moves. Hence, the dust particles come out of the carpet.

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.

1. 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.
2. Mass of a train is more than the mass of a bicycle. Hence, inertia of the train is greater than that of the bicycle.
3. 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 the ball changes four times as follows.

As a football player kicks the football, its speed changes from zero to a certain value. As a result, the velocity of the ball gets changed. In this case, the player applied a force to change the velocity of the ball.

Another player kicks the ball towards the goal post. As a result, the direction of the ball gets changed. Therefore, its velocity also changes. In this case, the player applied a force to change the velocity of the ball.

The goalkeeper collects the ball. In other words, the ball comes to rest. Thus, its speed reduces to zero from a certain value. The velocity of the ball has changed. In this case, the goalkeeper applied an opposite force to stop/change the velocity of the ball.

The goalkeeper kicks the ball towards his team players. Hence, the speed of the ball increases from zero to a certain value. Hence, its velocity changes once again. In this case, the goalkeeper applied a force to change the velocity of the ball.

Some leaves of a tree get detached when we shake its branches vigorously. This is because when the branches of a tree are shaken, it moves to and fro, but its leaves tend to remain at rest. This is because the inertia of the leaves tend to resist the to and fro motion. Due to this reason, the leaves fall down from the tree when shaken vigorously.

Due to the inertia of the passenger.

Everybody tries to maintain its state of motion or state of rest. If a body is at rest, then it tries to remain at rest. If a body is moving, then it tries to remain in motion. In a moving bus, a passenger moves with the bus. As the driver applies brakes, the bus comes to rest. But, the passenger tries to maintain his state 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. This is because when the bus accelerates, the inertia 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.

Due to the backward reaction of the water being ejected 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 = 35 m/s
Initially, the rifle is at rest.
Thus, its initial velocity, v = 0
Total initial momentum of the rifle and bullet system = (m1 + m2)v = 0
Total momentum of the rifle and bullet system after firing:
= m1v1 + m2v2 = 4(v1) + 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.

Mass of one of the objects, m1 = 100 g = 0.1 kg
Mass of the other object, m2 = 200 g = 0.2 kg
Velocity of m1 before collision, v1 = 2 m/s
Velocity of m2 before collision, v2 = 1 m/s
Velocity of m1 after collision, v3 = 1.67 m/s
Velocity of m2 after collision = v4
According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
m1v1 + m2v2 = m1v3 + m2v4
⇒ 0.1×2 + 0.2×1 = 0.1×1.67 + 0.2×v4
⇒ 0.4 = 0.67 + 0.2×v4
⇒ v4 = 1.165 m/s
Hence, the velocity of the second object becomes 1.165 m/s after the collision.

Yes. Even when an object experiences a net zero external unbalanced force, it is possible that the object is travelling with a non-zero velocity. This is possible only when the object has been moving with a constant velocity in a particular direction. Then, there is no net unbalanced force applied on the body. The object will keep moving with a non-zero velocity. To change the state of motion, a net non-zero external unbalanced force must be applied on the object.

Inertia of an object tends to resist any change in its state of rest or state of motion. When a carpet is beaten with a stick, then the carpet comes to motion. But, the dust particles try to resist their state of rest. According to Newton's first law of motion, the dust particles stay in a state of rest, while the carpet moves. Hence, the dust particles come out of the carpet.

When the bus accelerates and moves forward, it acquires a state of motion. However, the luggage kept on the roof, owing to its inertia, tends to remain in its state of rest. Hence, with the forward movement of the bus, the luggage tends to remain at its original position and ultimately falls from the roof of the bus. To avoid this, it is advised to tie any luggage kept on the roof of a bus with a rope.

(c)

A batsman hits a cricket ball, which then rolls on a level ground. After covering a short distance, the ball comes to rest because there is frictional force on the ball opposing its motion. Frictional force always acts in the direction opposite to the direction of motion. Hence, this force is responsible for stopping the cricket ball.

Initial velocity of the truck, u = 0 m/s
Time taken, t = 20 s
Distance covered by the stone, s = 400 m
According to the second equation of motion:
s = ut + $\frac{1}{2}a{t}^2$
We have
400 = 0 × 20 + $\frac{1}{2}\times a\times {\left(20\right)}^2$
⇒ 400 = 200a
⇒ a = 2 m/s²

Now, Force = mass × acceleration

⇒ F = 7000 × 2 = 14000 N

Initial velocity of the stone, u = 20 m/s
Final velocity of the stone, v = 0 m/s
Distance covered by the stone, s = 50 m
According to the third equation of motion:
v² = u² + 2as
Where,
Acceleration, a
(0)² = (20)² + 2 × a × 50
a = – 4 m/s2
The negative sign indicates that acceleration is acting against the motion of the stone.
Mass of the stone, m = 1 kg
From Newton's second law of motion:
Force,
F = Mass x Acceleration
F= ma
F= 1 × (– 4) = – 4 N
Hence, the force of friction between the stone and the ice is – 4 N.

(a) Force exerted by the engine, F = 40000 N
Frictional force offered by the track, Ff = 5000 N
Net accelerating force, Fa = F - Ff = 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 × Number of wagons
Mass of a wagon = 2000 kg
Number of wagons = 5
∴ m = 2000 × 5 = 10000 kg
Total mass (including the mass of engine), M = m + 8000 = 18000 kg
F_a = Ma
⇒ a = $\frac{F_a}{M}$ = $\frac{35000}{18000}$ = 1.944 m/s²

(c) Mass of 4 wagons (excluding wagon 1) m = 2000 × 4 = 8000 kg
Acceleration of the train = 1.944 m/s²

∴ The force of wagon 1 on wagon 2 = ma = 8000 × 1.944 = 15552 N

Mass of the automobile vehicle, m = 1500 kg
Final velocity, v = 0 m/s
Acceleration of the automobile, a = –1.7 m/s²
From Newton's second law of motion:
Force = Mass × Acceleration = 1500 × (–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.

(d) mv
Mass of the object = m
Velocity = v
Momentum = Mass × Velocity
Momentum = mv

A force of 200 N is applied in the forward direction. Thus, from Newton's third law of motion, an equal amount of force will act in the opposite direction. This opposite force is the fictional force exerted on the cabinet. Hence, a frictional force of 200 N is exerted on 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, v₁ = 2.5 m/s
Velocity of m₂, moving in opposite direction before collision, v₂ = -2.5 m/s
(Negative sign arises because mass m₂ is moving in an opposite direction)
After collision, the two objects stick together.
Total mass of the combined object = 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
1.5(2.5) + 1.5 (-2.5) = (1.5 + 1.5) v
3.75 - 3.75 = 3 v
v = 0
Hence, the velocity of the combined object after collision is 0 m/s.

The truck has a large mass. Therefore, the static friction between the truck and the road is also very high. To move the car, one has to apply a force more than the static friction. Therefore, when someone pushes the truck and the truck does not move, then it can be said that the applied force in one direction is cancelled out by the frictional force of equal amount acting in the opposite direction. Therefore, the student is right in justifying that the two opposite and equal cancel each other.

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
Hence, the change in momentum of the hockey ball is 3 kg m/s

Now, it is given that the bullet is travelling with a velocity of 150 m/s.
Thus, when the bullet enters the block, its velocity = 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= $\frac{-150}{0.03}$ = 5000 m/s²

(Negative sign indicates that the velocity of the bullet is decreasing.)
According to the third equation of motion:
v²= u² + 2as
0 = (150)² + 2 ( - 5000) s

s = $\frac{-{\left(150\right)}^2}{-2\left(5000\right)}=\frac{22500}{10000}$

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 x 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

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
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
m₁ v₁+ m₂ v₂ = (m₁+ m₂) v
1 (10) + 5 (0) = (1 + 5) v

v = $\frac{10}{6}=\frac{5}{3}m/s$

The total momentum after collision = 10 and velocity of combined object is 5/3 m/s.

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
Final momentum = mv= 100 × 8 = 800 kg m/s

Force exerted on the object, F = $\frac{mv - mu}{t}$

= $\frac{m(v-u)}{t} = \frac{800 - 500}{6} = \frac{300}{6} = 50$

Initial momentum of the object is 500 kg m/s.
Final momentum of the object is 800 kg m/s.

Force exerted on the object is 50 N.

According to the law of conservation of momentum:

Momentum of the car and insect system before collision = Momentum of the car and insect system after collision

Hence, the change in momentum of the car and insect system is zero.

The insect gets stuck on the windscreen. This means that the direction of the insect is reversed. As a result, the velocity of the insect changes to a great amount. On the other hand, the car continues moving with a constant velocity. Hence, Kiran's suggestion that the insect suffers a greater change in momentum as compared to the car is correct. The momentum of the insect after collision becomes very high because the car is moving at a high speed. Therefore, the momentum gained by the insect is equal to the momentum lost by the car.

Akhtar made a correct conclusion because the mass of the car is very large as compared to the mass of the insect.

Rahul gave a correct explanation as both the car and the insect experienced equal forces caused by the Newton's action-reaction law. But, he made an incorrect statement as the system suffers a change in momentum because the momentum before the collision is equal to the momentum after the collision.

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 kg m/s
= 40 kg m/s