Exercises – Force and Laws of Motion (Class 9)

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Question 1

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.

Answer:

Yes, it is possible.

Condition:

If the object is already moving and all external forces are balanced (net force = 0), then the object will continue to move with a constant (non-zero) velocity in a straight line.

Example:

A car moving on a straight road at constant speed.

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Question 2

When a carpet is beaten with a stick, dust comes out of it. Explain.

Answer:

When the carpet is beaten, the carpet moves suddenly, but the dust particles tend to remain at rest due to inertia of rest.
As a result, the dust gets detached and falls off.

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Question 3

Why is it advised to tie any luggage kept on the roof of a bus with a rope?

Answer:

When the bus starts, stops, or turns suddenly, the luggage tends to maintain its state of motion due to inertia and may fall off.
Tying the luggage prevents it from falling by providing the necessary force.

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Question 4

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.

Correct Answer:

(c) there is a force on the ball opposing the motion.

Explanation:

Friction between the ball and the ground opposes motion and gradually reduces the speed to zero.

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Question 5

A truck starts from rest and rolls down a hill with a constant acceleration. It travels 400 m in 20 s. Find its acceleration and the force acting on it if its mass is 7 tonnes.

Given:

• Initial velocity, $u=0$

• Distance, $s=400\text{m}$

• Time, $t=20\text{s}$

• Mass, $m=7\text{ tonnes}=7000\text{kg}$

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Step 1: Find acceleration

Using:

$$s=ut+\frac{1}{2}a{t}^2$$
$$400=0+\frac{1}{2}a(20{)}^2$$
$$400=200a$$
$$a=2{\text{m s}}^{-2}$$

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Step 2: Find force

Using:

$$F=ma$$
$$F=7000\times 2=14000\text{N}$$

Answer:

• Acceleration = 2 m s⁻²

• Force acting = 14,000 N

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Question 6

A stone of 1 kg is thrown with a velocity of 20 m s⁻¹ on ice and comes to rest after 50 m. Find the force of friction.

Given:

• Mass, $m=1\text{kg}$

• Initial velocity, $u=20{\text{m s}}^{-1}$

• Final velocity, $v=0$

• Distance, $s=50\text{m}$

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Step 1: Find acceleration

Using:

$$v^2=u^2+2as$$
$$0=400+100a$$
$$a=-4{\text{m s}}^{-2}$$

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Step 2: Find friction force

$$F=ma=1\times (-4)=-4\text{N}$$

Answer:

Force of friction = 4 N (opposite to motion)

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Question 7

An 8000 kg engine pulls 5 wagons of 2000 kg each. Engine force = 40000 N, friction = 5000 N. Find:
(a) net accelerating force
(b) acceleration

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Step 1: Total mass

$$m=8000+(5\times 2000)=18000\text{kg}$$

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Step 2: Net force

$$F_{\text{net}}=40000-5000=35000\text{N}$$

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Step 3: Acceleration

$$a=\frac{F}{m}=\frac{35000}{18000}$$

$$a\approx 1.94{\text{m s}}^{-2}$$

Answer:

• Net accelerating force = 35,000 N

• Acceleration = 1.94 m s⁻²

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Question 8

An automobile of mass 1500 kg is stopped with a negative acceleration of 1.7 m s⁻². Find the force between the vehicle and the road.

Given:

• Mass, $m=1500\text{kg}$

• Acceleration, $a=-1.7{\text{m s}}^{-2}$

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Calculation:

$$F=ma=1500\times (-1.7)=-2550\text{N}$$

Answer:

Force = 2550 N (opposite to direction of motion)

Question 9

What is the momentum of an object of mass m, moving with velocity v?

Correct answer:
(d) mv

Explanation:
Momentum $p=m\times v$

Question 10

Using a horizontal force of 200 N, a wooden cabinet is moved across the floor at constant velocity. What is the friction force acting on the cabinet?

Answer:

When an object moves with constant velocity, net force = 0.

$$\text{Applied force}=\text{Friction force}$$

$$\text{Friction force}=200\text{N}$$

Friction force = 200 N (opposite to motion)

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Question 11

A student says that when we push a parked truck, the equal and opposite forces cancel each other, so the truck does not move. Comment on this logic.

Answer:

The student’s logic is incorrect.

• According to Newton’s third law, action and reaction forces act on different objects, not on the same object.

• The force applied by the person acts on the truck, while the reaction force acts on the person.

• The truck does not move because its large mass (high inertia) and friction with the road require a much larger force to produce noticeable acceleration.

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Question 12

A hockey ball of mass 200 g moving at 10 m s⁻¹ returns with velocity 5 m s⁻¹. Find the magnitude of change in momentum.

Given:

• Mass, $m=200\text{g}=0.2\text{kg}$
  

• Initial velocity, $u=+10{\text{m s}}^{-1}$

• Final velocity, $v=-5{\text{m s}}^{-1}$ (opposite direction)

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Calculation:

Initial momentum:

$$p_1=mu=0.2\times 10=2{\text{kg m s}}^{-1}$$

Final momentum:

$$p_2=mv=0.2\times (-5)=-1{\text{kg m s}}^{-1}$$

Change in momentum:

$$\mathrm{\Delta }p=p_2-p_1=-1-2=-3$$

Magnitude of change in momentum = 3 kg m s⁻¹

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Question 13

A bullet of mass 10 g moving at 150 m s⁻¹ stops in 0.03 s. Find the distance penetrated and the force exerted.

Given:

• Mass, $m=10\text{g}=0.01\text{kg}$

• Initial velocity, $u=150{\text{m s}}^{-1}$

• Final velocity, $v=0$

• Time, $t=0.03\text{s}$

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Step 1: Acceleration

$$a=\frac{v-u}{t}=\frac{0-150}{0.03}=-5000{\text{m s}}^{-2}$$

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Step 2: Distance penetrated

$$s=ut+\frac{1}{2}a{t}^2$$

$$s=150(0.03)+\frac{1}{2}(-5000)(0.03{)}^2$$
$$s=4.5-2.25=2.25\text{m}$$

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Step 3: Force exerted

$$F=ma=0.01\times (-5000)=-50\text{N}$$

Answers:

• Distance of penetration = 2.25 m

• Force exerted = 50 N (opposite to motion)

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Question 14

A 1 kg object moving at 10 m s⁻¹ sticks to a stationary 5 kg block. Find momentum before and after collision and final velocity.

Given:

• $m_1=1\text{kg},u_1=10{\text{m s}}^{-1}$

• $m_2=5\text{kg},u_2=0$

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Momentum before collision

$$p=(1\times 10)+(5\times 0)=10{\text{kg m s}}^{-1}$$

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Momentum after collision

Momentum is conserved:

$$p=10{\text{kg m s}}^{-1}$$

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Final velocity

$$v=\frac{p}{m_1+m_2}=\frac{10}{6}=1.67{\text{m s}}^{-1}$$

Answers:

• Momentum before impact = 10 kg m s⁻¹

• Momentum after impact = 10 kg m s⁻¹

• Final velocity = 1.67 m s⁻¹

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Question 15

A 100 kg object accelerates from 5 m s⁻¹ to 8 m s⁻¹ in 6 s. Find initial momentum, final momentum and force.

Given:

• $m=100\text{kg}$

• $u=5{\text{m s}}^{-1}$

• $v=8{\text{m s}}^{-1}$

• $t=6\text{s}$

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Initial momentum

$$p_1=mu=100\times 5=500{\text{kg m s}}^{-1}$$

Final momentum

$$p_2=mv=100\times 8=800{\text{kg m s}}^{-1}$$

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Force

$$a=\frac{v-u}{t}=\frac{3}{6}=0.5{\text{m s}}^{-2}$$
$$F=ma=100\times 0.5=50\text{N}$$

Answers:

• Initial momentum = 500 kg m s⁻¹

• Final momentum = 800 kg m s⁻¹

• Force = 50 N

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Question 16

An insect hits a moving car’s windscreen. Comment on the views of Akhtar, Kiran and Rahul.

Correct Explanation:

Rahul is correct.

Reason:

• According to Newton’s third law, the insect and the car exert equal and opposite forces on each other.

• The change in momentum of both is equal in magnitude, but:

  • The insect has a very small mass, so it undergoes large acceleration and gets crushed.

  • The car has a very large mass, so its acceleration is negligible.

❌ Why others are wrong:

• Kiran: Change in momentum is equal, not greater for insect.

• Akhtar: Force on both bodies is equal, not larger on insect.

Question 17

How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm?
(Take downward acceleration = 10 m s⁻²)

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Given:

• Mass of dumb-bell,

  $$m=10\text{ kg}$$

• Height,

  $$h=80\text{ cm}=0.8\text{ m}$$

• Acceleration due to gravity,

  $$g=10{\text{ m s}}^{-2}$$

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Step 1: Find velocity just before hitting the floor

Using the equation of motion:

$$v^2=u^2+2gh$$

Since the dumb-bell is dropped from rest:

$$u=0$$
$$v^2=2\times 10\times 0.8$$
$$v^2=16$$
$$v=4{\text{ m s}}^{-1}$$

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Step 2: Calculate momentum

Momentum transferred to the floor equals the momentum of the dumb-bell just before impact.

$$p=mv$$
$$p=10\times 4$$
$$p=40{\text{ kg m s}}^{-1}$$

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Final Answer

Momentum transferred to the floor = 40 kg m s⁻¹