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 for better concepts.

You can find the solution at the bottom of the article, and also the solved pdf.

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.

## Force and Laws of Motion Class 9 Numerical

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

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

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?

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.

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

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

Now, F = ma

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

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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 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?

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 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 = 0.12 kgm/s

Also, Impulse = F×t

Put value of t and get Force

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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

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 law of conservation of Momentum as 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.

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

Let v be the velocity with car start travelling

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 on to 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

### Force and Laws of motion Class 9 Numerical Solved Pdf

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