# Force and Laws of Motion Class 9 Numerical

Important Force and Laws of Motion Class 9 extra Numerical for practice.

Contents

Here are a few Important Force and Laws of Motion Class 9 Numerical for practice. These Force and Laws of Motion Class 9 Numerical will help you make better concepts.

Note: Before looking for the solution, first try yourself, and when it becomes very difficult or if you want to check your solution, only then open the answer and explanation.

I have also provided some questions from your books themselves but with changes in the data. So try solving those questions and put your answer in the comment box. I will surely react to your answer.

## Force and Laws of Motion Class 9 Numerical

Before you solve the questions, let’s revise the important formulas that you read in the chapter Force and Laws of Motion.

### Class 9 Important Force and Laws of Motion Formula

• Force = mass × acceleration
• Force = rate of change in momentum = (Final momentum – Initial Momentum)/time
• Momentum (p) = mass × velocity
• Chane in Momentum = mass × (final velocity – Initial Velocity) = m×(v-u)
• Impulse = Force × time
• Impulse = chane in momentum = mass×(v-u)
• Conservation of Momentum = m1u1+m2v2=m1v1+m2v2

Question 1.

A constant force acts on an object of mass 5 kg for a duration of 2 s. It increases the object’s velocity from 3 m/s to 7 m/s. Find the magnitude of the applied force. Now, if the force was applied for a duration of 5 s, what would be the final velocity of the object?

F = m(v-u)/t

Substitution of values in this relation gives F = 10 N

The final velocity can be calculated by

v= u+at = 13 m/s.

Question 2.

Which would require a greater force – accelerating a 2 kg mass at 5 m/s² or a 4 kg mass at 2 m/s²?

Use

F = ma

and solve it

Question 3.

A motorcar is moving with a velocity of 108 km/h and it takes 4 s to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar if its mass along with the passengers is 1000 kg.

Initial velocity- 108 km/h = 30 m/s, Final Velocity = 0 m/s

Use the Equation of motion and find a = -30/4 = -7.5 m/s2

Now, F = ma

Solve and get F = -7500 N (-ve sign means, the force by brakes applied is opposite of the direction of motion)

Question 4.

A body of mass 1 kg undergoes a change of velocity of 4 m/s in 4 s what is the force acting on it?

Δv= 4m/s, t= 4 s, m=1 kg

Acceleration is given by, a= Δv/t

a= 1 m/s²

Now the force is given by

F= ma

F= 1 N

Question 5.

A driver accelerates his car first at the rate of 4 m/s² and then at the rate of 8 m/s². Calculate the ratio of the forces exerted by the engines.

F1=ma1 and F2= ma2

So, Ratio of force exerted is given by = F1/F2= ma1/ma2 = a1/a2=1:2

Question 6.

A cricket ball of mass 0.20 kg is moving with a velocity of 1.2 m/s. Find the Impulse on the ball and the average force applied by the player if he is able to stop the ball in 0.10 s?

Impulse= Change in momentum

I = Δp = mΔv = -0.24 kgm/s

Also, Impulse = F×t

Put the value of t and get the Force

Video Solution of Question 6.

Do Subscribe Also for more Videos. Do you still face problems? Comment there in the video for a reply.

Question 7.

Two objects of masses of 100 gm and 200 gm are moving in along the same line and direction with velocities of 2 m/s and 1 m/s respectively. They collide and after the collision, the first object moves at a velocity of 1.67 m/s. Determine the velocity of the second object.

By the law of conservation of momentum,

m1u1+m2u2 = m1v2+m2v2

On substituting the values

0.1×2+0.2×1=0.1×1.67+0.2v2

Or, v2=1.165 m/s

It will move in the same direction after the collision

Question 8.

An object of mass 1 kg traveling 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.

Initial Momentum

$p_{1}=m_{1}v_{1}+m_{2}v_{2}$

p1=10 kgm/s = 10 kgm/s

Now after the collision, they stick together to move with velocity V

Final Momentum

p2= (m1+m2)V = 6V

Now As per the law of conservation of Momentum, the external force is absent

Initial Momentum = Final Momentum

p1 = p2

10 = 6V

V=1.67 m/s

Now

p2 =6V= 10 kg-m/s

Question 9.

A man weighing 60 kg runs along the rails with a velocity of 18 km/h and jumps into a car of mass 1 quintal (100 kg) standing on the rails. Calculate the velocity with which the car will start traveling along the rails.

Here m= 60 kg, u1= 18 km/hr = 5 m/s, M=100 kg, u2=0

Let v be the velocity with which the car starts traveling

Now

mu1+Mu2 = (M+m)v

60×5 = 160v

v = 1.875 m/s

Question 10

A boy of mass 50 kg running 5 m/s jumps onto a 20 kg trolley traveling in the same direction at 1.5 m/s. Find their common velocity.

Here m = 50 kg, u1 = 5m/s, M = 20 kg, u2 =  1.5m/s

Now

mu1 + Mu2 = (M+m)v

50×5+20×1.5=70v

or v = 4 m/s

Now try solving these few questions by yourself. Comment your Answer in the comment box below.

Question 11.

If a constant force acts on an object of mass 15 kg for a duration of 3 s. It increases the object’s velocity from 10 m/s to 15 m/s. How much force is applied? Now, if the force was applied for a duration of 4s, what would be the final velocity of the object?

For Solution, watch the video below. Also, SUBSCRIBE the Channel for more questions

Question 12.

A bullet of mass 40 g is horizontally fired with a velocity of 120 m/s from a gun of mass 1.5 kg. Find the recoil velocity of the gun

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Question 13

If a horizontal force of 800 N drags a box across a floor at a constant velocity, then find the force of friction that will be exerted on the box.

Question 14

A block of mass 3 kg has a velocity of u m/s. When a force of 18 N acts on the block, it reduces the velocity from u m/s to u/2 m/s after the block has covered a distance of 9 m. Find u.

Video Solution of Question 18.

Do Subscribe Also for more Videos. Do you still face problems? Comment there in the video for a reply.

Question 15

If a force of 50 N is applied on a heavy box of 50 kg and it doesn’t move. What is the frictional force exerted on the box?

Question 16

A football of mass 700 gm moving with a velocity of 10 m/s is brought to rest by a player in 0.02 s. Find the impulse of the force and the average force applied by the player.

Question 17

Which would require a greater force- accelerating a 50 g mass at 10 m/s2 or 75 g mass at 2 m/s2?

Question 18

What is the acceleration produced by a force of 50 N when applied on a body of mass 5 kg?

Question 19

Find the force needed to accelerate a body of 8 kg by 7 m/s2?

Question 20

Find the ratio of SI to cgs units of linear momentum.

Question 21

A bus starting from rest is rolling down a hill with constant acceleration. If it travels a distance of 500 m in 25 seconds. Find the acceleration of the bus and the force acting on it. Consider the mass of the bus 5 metric tonnes.

Question 22

A stretching force of 100 N is applied at one end of a spring balance and an equal stretching force is applied at the other end at the same time. What would be the reading on the spring balance?

Question 23

A machine gun has a mass of 30 kg. The gun fires bullets of 35 g at the rate of 400 bullets per minute with a speed of 400 m/s. What force must be applied to the gun so that the gun is kept in position? (Ans: F= 93.3 N)

Question 24.

A car with a mass of 1200 kg is traveling at a constant velocity of 20 m/s on a straight road. Suddenly, the driver applies the brakes, and the car comes to a complete stop in 5 seconds. Calculate-

• The initial momentum of the car
• The deceleration (negative acceleration) of the car.
• The force applied by the brakes to stop the car.
• Momentum(p)=Mass(m)×Velocity(v) = 24000 kg m/s
• To calculate the deceleration

use the following equation from the first equation of motion- v = u + at ⇒ a = (v-u)/t = – 4 m/s2

• Force(F)=Mass(m)×Acceleration(a) = -4800 N

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### 10 thoughts on “Force and Laws of Motion Class 9 Numerical”

1. Some more solved numerinals examples plz plz plz plz very urgent iwant tomorrow

2. In Q6 0.20×1.2=0.24 therefore impulse will be 0.24 kg.m/s and f will be -2.4 N

3. The site is very good every que has ans and which que we can do easily theres the formula to do it by ourselves I would recommend to use this site for such numericals

4. In q 11 force is 25N
And the final velocity is 16.66mpersecond

• No , in Qno 11 the acceleration is 1.67m/s and the final velocity is 16.68m/s.

• Don’t be haste in solving questions. Do watch the video and subscribe the channel. 5/3 = 1.67. So, there is nothing wrong.

• What will be the force in Q.6. I am having trouble pls reply

• -2.4N

5. Acceleration equal to 1.6ms^-1
Velocity =16.4m/s

6. 12-3.2m/s