Physics Numerical: Force and Laws of Motion Class 9 Numerical

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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.

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1. Force and Laws of Motion Class 9 Numerical
1.1. Class 9 Force and Laws of Motion Important Formulae
1.2. Force and Laws of motion Class 9 Numerical Solved Pdf

Force and Laws of Motion Class 9 Numerical

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

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

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

Class 9 Force and Laws of Motion Important Formulae

  • 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

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?

Click for Answer/Explanation

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.

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?

Click for Answer/Explanation

Use

F = ma

and solve it

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.

Click for Answer/Explanation

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)

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?

Click for Answer/Explanation

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

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.

Click for Answer/Explanation

F1=ma1 and F2= ma2

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

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?

Click for Answer/Explanation

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 Q6, Do Subscribe Also for more Videos. Do you still face problems? Comment there in the video for a reply.

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.

Click for Answer/Explanation

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

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.

Click for Answer/Explanation

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

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.

Click for Answer/Explanation

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

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.

Click for Answer/Explanation

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.

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?

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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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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.

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 Q18, Do Subscribe Also for more Videos. Do you still face problems? Comment there in the video for a reply.

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?

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.

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

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

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

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

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.

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?

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)

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.
Click for Answer/Explanation
  • 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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9 thoughts on “Physics Numerical: Force and Laws of Motion Class 9 Numerical”

  2. 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

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