Force and Laws of Motion- Important Questions

Very Short Answer Questions [1 Marks Question]

Ques. What is Newton’s first law of motion known as?

1. Law of Inertia
2. Law of Momentum
3. Law of Action & Reaction
4. None of these

Ans. a. Law of Inertia

Ques. What is the reason for bicycles to slow down when we stop pedalling?

Ans. The reason is that the frictional forces act opposite to the direction of motion.

Ques. What is the physical quantity that changes the shape of a ball?

Ans. Force.

Ques. What is the force acting, if the initial velocity is zero?

1. Retarding
2. Acceleration
3. Both
4. None

Ans: a. Acceleration

Ques. What is the C.G.S unit of force?

Ans. Dyne is the C.G.S unit of force.

Ques. What is 1 newton force?

Ans. In a body mass of 1 kg, 1 newton is the magnitude of force which produces an acceleration of 1 m/s².

Ques. Why do people in the bus are pushed backwards when the bus starts suddenly?

1. Inertia due to rest
2. Inertia due to motion
3. Inertia due to direction
4. Inertia

Ans. a. Inertia due to Rest

Ques. What is an impact force?

Ans. Impact force can be defined as the force produced by the impact of a fast moving object on another.

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Short Answer Questions [2 Marks Question]

Ques. Why does dust come out when a carpet is beaten with a stick?

Ans. Newton’s First Law of Motion, the law of Inertia explains the reason for dust coming out of the carpet when beaten with a stick. In the beginning, the dust particles as well as the carpet are in a state of rest. If the carpet is beaten with a stick, it makes the carpet move. On the other hand, the dust particles will resist the change in motion due to inertia of rest. Hence, the carpet’s forward motion will exert a backward force on the dust particles and makes them move in the opposite direction. This is the reason for the dust to come out of the carpet.

Ques. What are the factors that friction depends on?

Ans. The force of friction depends on the nature of the surfaces in contact. The force of friction is directly proportional to the body’s weight sliding over the surface.

Ques. When do you consider the forces acting on a body as balanced? Explain with an example. What are the expected changes that the balanced forces bring about in an object? (CBSE 2010, 2015)

Ans. Balanced forces are two forces of equal magnitude acting in opposite directions on a body. If you push a wall but the wall does not move at all, then balanced forces are said to be acting on the wall. There is no change in the state of the object when balanced forces are acting on them.

Ques. An automobile vehicle has a mass of 1500kg. If the vehicle is to be stopped with a negative acceleration of 1.7ms⁻², what must be the force between the vehicle and the road?

Ans: We know that,

Mass of vehicle: m=1500kg

Negative acceleration: a=−1.7ms⁻²

To find that,

Force of friction between road and vehicle.

We already know that - F=ma

= F= (1500) (−1.7)

= F= −2550N

Therefore, if the vehicle is to be stopped with a negative acceleration of 1.7ms⁻², the force between the vehicle and road is −2550N.

Ques. What is Newton's second law of motion?

Ans: According to Newton’s Second law of motion, when an unbalanced external force acts on an object, its velocity changes, i.e. the object gets accelerated. In other words, the time rate of change of the momentum of a body is equal in both magnitude as well as direction to the force applied on it.

To explain mathematically,

– F= ma, in which ‘F’ is the force, ‘m’ is the mass of the object, and ‘a’ is the acceleration.

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Long Answer Questions [3 Marks Question]

Ques. The leaves may get detached from a tree if we vigorously shake its branch- Explain

Ans. If we vigorously shake the tree, some of the leaves may get detached from the tree because the branches come into motion while the leaves tend to continue in their state of rest. This is because of the inertia of rest of the leaves. With the change in direction rapidly, the force of shaking will act on the leaves. This results in the leaves detaching and falling off from the tree.

Ques. A bullet of 10 g strikes a sand bag at a speed of 10³ ms^-1 and then gets embedded after travelling 5 cm. Answer the below given questions.

(i) Find the resistive force exerted by the sand on the bullet.

(ii) Find the time taken by the bullet to come to rest. [NCERT Exemplar]

Ans. (i) m = 10 g = 10/100 kg, u = 10³ m / s, v = 0, s = 5/100 m

v² – u² = 2as

0 – (10³)² = 2.a.5/100

a = −1000×1000/2×5 × 100

= -10⁷ ms^-2

F = m . a = -105 N

(ii) v = u + at

0 = 10³ – 10⁷t

10⁷t = 10³

t = 103/107 = 10-4 s

Ques. A bullet of mass 50g is fired from a rifle of mass 4kg, with an initial velocity of 35ms⁻¹. How do you calculate the initial recoil velocity of the rifle?

Ans: Given that.

Mass of rifle: m₁=4kg

Mass of bullet: m₂=50g=0.05kg

Initial velocity of rifle: u₁=0ms⁻¹ (it is stationary during firing)

Initial velocity of bullet: u₂=0ms⁻¹(it starts from rest, inside the barrel of the rifle)

Fired velocity of bullet: v₂=35ms⁻¹

To find that: Recoil velocity of rifle: v₁

For an isolated system, when we apply the law of conservation of momentum, the total initial momentum for an event is equal to total initial momentum,

m₁v₁+m2v₂=m₁u₁+m₂u₂

Here,

m₁u₁+m₂u₂ is the total sum of momentum of rifle and bullet before firing

m₁v₁+m₂v₂ is the total sum of momentum of rifle and bullet after firing.

Substituting the values in –

m₁v₁+m₂v₂=m₁u1+m₂u₂

= (4×v₁) + (0.05×35) = (4×0) + (0.05×0)

= (4×v₁) + (17.5) = 0

= (4×v₁) = −17.5

= v₁= −4.375ms⁻¹

In this, the negative sign indicates the backward direction in which the rifle moves when it recoils. Therefore, the recoil velocity of the rifle is 4.375ms⁻¹.

Ques. A body of mass 300 g kept at rest breaks into two parts due to internal forces. One of the parts of mass 200 g moves at a speed of 12 m/s towards the east. Calculate the velocity of the other part?

Ans. It is known that at the beginning, the body was at rest.

Thus, the linear momentum of the body is p = mu = 0.

Due to internal forces, the body breaks.

As the external force acting on it is zero, its linear momentum will remain constant, i.e. again, zero.

p₁ = m₁v₁ = (200 g) × (12 m/s), towards the east.

The linear momentum of the other part must have the same magnitude and should be opposite in direction. Therefore it moves towards the west.

If its speed is v₂, its linear momentum is

p₂ = m₂v₂ = (100 g) × v₂.

Then, m₁v₁ = m₂v₂.

Hence, (200 g) × (12 m/s) = (100 g) × v₂ or, v₂ = 24m/s.

The velocity of the other part is 24 m/s towards the west.

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Very Long Answer Questions [5 Marks Question]

Ques. A body of mass 10kg starts from rest and then rolls down an inclined plane. It rolls down 10m in 2s. Calculate,

1. The acceleration attained by the body
2. The velocity of the body at 2s
3. The force acting on the body

Ans. a. Given that,

Mass of body: m=10 kg

Initial velocity of body: u= 0ms⁻¹ (as it is starting to roll)

Distance travelled on inclined plane: s=10m

Time duration of rolling: t=2s

To find acceleration attained by the body.

We know that – s= ut+1/2at²

Thus, 10= (0×2) + 1/2(a×22)

=10= 1/2(a×4)

=10= (a×2)

=a=5ms⁻²

2. Given that,

Mass of body: m=10 kg

Initial velocity of body: u=0ms⁻¹ (as it is starting to roll)

Distance travelled on inclined plane: s=10m

Time duration of rolling: t=2s

To find velocity of body at t=2s

It is known that – v=u+at

Thus, v=0 + (5×2)

=v=10ms⁻¹

100. Given that,

Mass of body: m=10 kg

Initial velocity of body: u=0ms⁻¹ (as it is starting to roll)

Distance travelled on inclined plane: s=10m

Time duration of rolling: t=2s

To find force acting on the body,

= F=ma

= F= (10×5)

= F=50N

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

Ans. Given that,

Mass of object 1: m₁=100g=0.1kg

Mass of object 2: m₂=200g=0.2kg

Initial velocity of object 1: u₁=2ms⁻¹

Initial velocity of object 2:u₂=1ms⁻¹

Final velocity of object 1:v₁=1.67ms⁻¹

To find that,

Final velocity of object 2:v₂

According to the law of conservation of momentum, for an isolated system, i.e. the total initial momentum for an event is equivalent to total initial momentum,

m₁v₁+m₂v₂=m₁u₁+m₂u₂

Here,

m₁u₁+ m₂u₂ is the total sum of momentum of objects before collision

m₁v₁+ m₂v₂ is the total sum of momentum of objects after collision

Substituting the values in –

m₁v₁+m₂v₂=m₁u₁+m₂u₂

= (0.1×1.67)+(0.2×v₂)=(0.1×2)+(0.2×1

= (0.167)+(0.2×v₂)=(0.2)+(0.2)

= (0.167)+(0.2×v₂)=0.4

= 0.2×v₂=0.4−0.167

= 0.2×v₂=0.233

= v₂=1.165ms⁻¹, the velocity of the second object.

Ques. Assume that a stone is dropped from a 100 m high tower. How long does it take to fall?

1. The first 50 m
2. The second 50m

Ans: a. Given that,

Initial velocity of stone: u=0ms⁻¹ (as it is starts from rest, before being dropped)

Height of the tower: s=100m

Distance travelled in case A: s₁=50m (first half distance)

Distance travelled in case B: s₂=50m (next half distance)

To find: Time duration of travel: t during the first 50m.

It is known that – s=ut+1/2at2

As we know, the stone is dropped from a height; we can consider its acceleration to be equal to the acceleration due to gravity.

Acceleration of stone: Acceleration due to gravity a=g=10ms−2

s1=ut+1/2at²

=50= (0×t)+1/2(10×t²)

=50=5×t²

=t2=10

=t=\(?10\) =3.16s

=t1=3.16s

2. Ans: Given that,

Initial velocity of stone: u=0m^s−1 (as it is begins from rest, before being dropped)

Height of the tower: s=100m

Distance travelled in case A: s₁=50m (first half distance)

Distance travelled in case B: s₂=50m (next half distance)

To find: Time duration of travel: t during the second 50m.

It is known that – s=ut+1/2at²

Hence, the stone is dropped from a height, we can consider its acceleration to be equal to the acceleration due to gravity.

Acceleration of stone: Acceleration due to gravity a=g=10ms⁻²

Time duration for the next 50m can be found by subtracting time for the first half distance from the time for the total distance of travel.

s=ut+1/2at₂

=100= (0×t) + 1/2(10×t²)

=100=5×t²

=t2=20

=t= \(?20\) =4.47s⇒t=20=4.47s

=t=4.47s

Thus the time for the second half is – t²=t−t1

=t²= 4.47−3.16

=t²=1.31s

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