# CBSE Physics | Solved Numerical on Force and Laws of Motion

## Solved Numerical on Force and Laws of Motion

### Numerical- Solving Tips

Step 1 : Read the question carefully and visualise the situation (think as if the process is happening in front of you).

Step 2 : Draw a relevant figure and write the parameters given.

Step 3 : Make sure all values are in the same set of units.

Step 4 : Plan the concept/ formulae which are to be used.

Step 5 : Avoid doing calculations in each step. Try to do all the calculations at final step to save time.

### Question 1. A motorcycle of mass of 2000 kg is moving over a horizontal road, with uniform velocity. If this motorcycle has to be stopped with a negative acceleration of 1.5 ms–2, then what is the force of friction between the types of the motorcycle and the road?

Solution:

Here, m = 2000 kg; a = – 1.5 ms–2; F = ?

Using, F = ma, we get

F = 2000 kg (– 1.5) ms –2 = – 3000 kg ms– 2

= – 3000 N

Negative sign shows that force of friction (F) acts in a direction opposite to the motion of the motorcycle.

### Question 2. 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 ms–1 to 7 ms–1. 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?

Solution:

Given u = 3 ms–1, v = 7 ms–1, t = 2s, m = 5 kg

Applying

Also F = ma

u = 3 ms–1, t = 5 s, a = 2 ms– 2, v = ?

Applying, v = u + at

v = 3 + 2 5 = 13 ms–1

### Question 3. A bus of mass 5000 kg starts from rest and rolls down a hill. If it travels a distance of 200 m in 10s. Calculate (i) acceleration of the bus and (ii) the force acting on the bus.

Solution:

Here, mass m = 5000 kg; u = 0; s = 200 m; t = 10 s;

a = ?; F = ?

Using we get

or a = 4 ms – 2.

Using , F = ma, we get

F = 5000 kg 4 ms– 2 = 20,000 kg ms– 2

= 20,000 N

### Question 4. The velocity-time graph of a ball m moving on the surface of floor is shown in figure below. Calculate the force acting on the ball, if mass of the ball is 20 g.

Solution:

The velocity-time graph shows that the velocity of the ball at t = 0 is zero. That is initial velocity of ball, u = 0

Velocity of ball at t = 4 s is 20 ms– 1

That is, final velocity, v = 20 ms – 1

time, t = 4 s

a = ? ; F = ?

Step 1. Acceleration of the ball a =

Or a =

Or a = 5 ms– 2

Step 2. Also, mass of ball, m = 20 g = kg =kg

Force acting on the ball, F = ma

or F =

= 0.1 N

### 5. The velocity-time graph of a ball moving on the surface of a floor is shown in fig below. Find the force acting on the ball if the mass of the ball is 50 g.

Solution:

The velocity-time graph shows that velocity of the ball at t = 0 is 30 ms–1

That is, initial velocity of the ball, u = 30 ms–1

The velocity of the ball at t = 6s is zero.

That is, final velocity of the bal, = 0

time, t = 6s ; F = ?

Step 1. Acceleration of the ball, a =

=

= – 5 ms–2

Negative sign shows that the ball is retarded or decelerated.

Step 2. Also, mass of ball, m = 50 g

Force acting on the ball, F = ma

= – 0.25 N [ 1 kg ms – 2 = 1 N]

### Question 6. An iron sphere of 1 kg is moving a velocity of 20 ms– 1 on a cemented floor. It comes to rest after traveling a distance of 50 m. Find the force of friction between the sphere and the floor.

Solution.

Here, m = 1 kg.; u = 20 ms – 1; ; s = 50 m; F = ?

Step 1. Using = 2as, we get

a =

= – 4 ms– 2

Step 2. Using F = ma, we get

F = 1 kg (– 4 ms– 2)

= – 4 kg ms– 2 = – 4N

Negative sign shows that the force is retarding force.

Force of friction F = 4N.

### Question 7. A wooden block of mass m1 kg accelerates at 10 ms–2 when a force of 5 Newton acts on it. Another block of mass m2 kg accelerates at 20 ms–2 when the same force acts on it. Find the acceleration, if both the blocks are tied together and same force acts on this combination.

Solution.

For First Block: m = m1 kg.; a = 10 m/s2

Using F = ma, we get

m1 =

Or m1 = 0.5 kg

For Second Block:

m = m2 kg; a = 20 ms– 2; F = 5 N

Using, F = ma, we get

m2 =

When both the blocks are tied together

Mass of the combination, m = (m1 + m2) = ( 0.5 kg + 0.25 kg) = 0.75 kg

F = 5 N

Using F = ma, we get

a = 6.67 ms – 2

### Question 8. A car of mass 1000 kg moving with a velocity of 36 km/h hits a wall and comes to rest in 5 s. Find the force exerted by the car on the wall.

Solution.

Here, m = 1000 kg

Initially velocity, u = 36 km/h =

Final velocity, = 0

Time, t = 5 s.

F = ?

Using, F = ma, we get

F = 2000 N

Thus force exerted by the car on the wall = 2000N

### Question 9. A ball of mass 100 g moving with velocity of 10 ms – 1 is stopped by a boy in 0.2s. Calculate the force applied by the boy to stop the ball.

Solution.

Here, mass of ball, m = 100g =kg = 0.1 kg

Initial velocity of ball, u = 10 ms–1

Final velocity of ball, = 0

Time, t = 0.2 s.

F = ?

Step 1. Using, a = , we get

a = ms–2

step 2. Using, F = ma, we get

F = 0.1 kg ( – 50 ms– 2) = – 5 kg ms– 2 = – 5 N

Negative sign shows that force applied by the boy is the opposing force i.e. a force in a direction opposite to the direction of motion of the ball.

### Question 10. A car of mass 1000 kg and bus of mass 8000 kg are moving with same velocity of 36 kmh– 1. Find the forces to stop both the car and the bus in 5 s.

Solution.

Here, initial velocity, u = 36 km h–1 = ms–1= 10 ms–1

Final velocity, = 0

Time t = 5 s.

Step 1. a = , we get

a =

Mass of car, m = 1000 kg

Force required to stop the car,

F = ma = 1000 kg (– 2 ms – 2) = – 2000 N

Setp 2. Mass of the bus, m = 8000 kg

Force required to stop the bus,

F = ma = 8000 kg (– 2 ms– 2) = – 16000 N

Negative sign with force shows that the force is applied in a direction opposite to the direction of the motion of the car and bus. Such a force is known as retarding force.

### Question 11. A body of mass 100 g is at rest on a smooth surface. A force of 0.2 newton acts on it for 5 seconds. Calculate the distance traveled by the body.

Solution.

Here, m = 100g =

F = 0.2 N

t = 5 s

u = 0

S = ?

Step 1. F = ma, we get

a = = 2 ms– 2

Step 2. To find distance traveled, using

S = , we get

S = (2 ms–2) (25 s)2 = 625 m

### Question 12. A gun fires a bullet of mass 50g. The bullet moving with a velocity of 100 ms– 1 strikes a wooden plank and comes to rest in 0.05 s. Calculate (i) the force exerted by the wooden plank on the bullet and (ii) the distance of penetration of the bullet in the wooden plank.

Solution.

Here, m = 50 g = kg

u = 100 ms – 1

t = 0.05 sec

= 0 (finally bullet comes to rest)

F = ? ; S = ?

(i) Using, F = ma, we get

F =

=

= – 100 kg ms – 2 = –100 N

Thus, force exerted by the wooden plank on the bullet = 100 N.

(ii) Using, s =

or s = , we get

s = 100 ms–1 0.05 s + ( 0 – 100 ms–1) 0.05 s

= 5 m – 2.5 m = 2.5 m.

### Question 13. A mechanic strikes a nail with a hammer of mass 500 moving with a velocity of 20 ms – 1.The hammer comes to rest in 0.02 s after striking the nail. Calculate the force exerted by the nail on the hammer.

Solution.

Here, m = 500 g = kg; u = 20 ms– 1

= 0; t = 0.02s; F = ?

Using, F = ma

Since a = , So

F =

=kg – 500N

Therefore, force exerted by nail on the hammer = 500 N

### Question 14. A bullet of mass 100 g is fired from a gun of mass 20 kg with a velocity of 100 ms– 1. Calculate the velocity of recoil of the gun.

Solution.

Mass of bullet m = 100 g =

Velocity of bullet, = 100 ms – 1

Mass of gun, M = 20 kg

Let recoil velocity of gun = V

Step 1.Before firing, the system (gun + bullet) is at rest therefore, initial momentum of the system = 0

Final momentum of the system = momentum of bullet + momentum of gun

= + MV = V = 10 + 20 V

Step 2. Apply law of conservation of momentum

Final momentum = Initial momentum

i.e. 10 + 20 V = 0

20 V = – 10

Or V = – 0.5 ms – 1

Negative sign shows that the direction of recoil velocity of the gun is opposite to the direction of the velocity of the bullet.

### Question 15. An iron sphere of mass 10 kg is dropped from a height of 80 cm. If the downward acceleration of the ball is 10 ms– 2, calculate the momentum transferred to the ground by the ball.

Solution.

Here, Initial velocity of sphere, u = 0

Distance travels, S = 80 cm = 0.8 m

Acceleration of sphere, a = 10 ms – 2

Step 1. Final velocity of sphere when it just reaches the ground can be calculated using

= 2as

= ms– 2 0.8 m

= 16 m2 s– 2

or =

= 4 ms–1

Momentum of the sphere just before it touches the ground =

= 10 kg4 ms– 1 = 40 kg ms–1

Step 2. On reaching the ground, the iron sphere comes to rest, so its final momentum = 0

According to the law of conservation of momentum,

Momentum transferred to the ground = momentum of the sphere just before it comes to rest

= 40 kg ms– 1

### Question 16. Two small glass spheres of masses 10 g and 20 g are moving in a straight line in the same direction with velocities of 3 ms – 1 and 2 ms–1 respectively. They collide with each other and after collision, glass sphere of mass 10g moves with a velocity of 2.5 ms– 1. Find the velocity of the second ball after collision.

Solution.

Here, m1 = 10 g = kg = 10– 2 kg

m2 = 20 g = 2 10– 2kg

u1 = 3 ms–1; u2 = 2 ms– 1

= 2.5 ms –1; = ?

Momentum of first sphere before collision = m1u1

= 10– 2 kg 3 ms–1

= 3 10–2 kg ms–1

Momentum of second sphere before collision = m2u2

= 2 10– 2 kg 2 ms–1

= 4 10– s2 kg ms–1

Total momentum of both the sphere before collision = m1u1 + m2u2

= 3 10– 2 kg ms–1 + 4 10– 2 kg ms–1

= 710–2 kg ms –1

Now, momentum of first sphere after collision =

= 10–2 kg 2.5 ms –1

= 2.510–2 kg ms–1

Momentum of second sphere after collision =

= 210– 2 kg ms–1

Total momentum of both the spheres after collision

=

= 2.5 10– 2 kg ms– 1 + 210– 2 kg m–1

Now, according to the law of conservation of momentum.

Total momentum after collision = Total momentum before collision

2.5 10– 2 + 2 10– 2 = 7 10– 2

Or 210– 2 = 710– 2 – 2.510– 2

= 4.510– 2

=

= 2.25 ms–1

### Question 17. Two bodies each of mass 0.5 kg are moving in a straight line but opposite in direction with the same velocity of 2 ms– 1. They collide with each other and stick to each other after collision. What is the common velocity of these bodies after collision?

Solution.

Let one body is moving towards left side and second body is moving to the right side. So velocity of the body to the left side is taken as positive and velocity of body moving to the right side is taken as negative.

Momentum of body moving to left side before collision = mv

= 0.5kg 2 ms–1 = 1 kg ms–1

Momentum of body moving to right side before collision = – mv

= – 0.5 kg 2 ms–1

= – 1 kg ms–1

Total momentum of both the bodies before collision

= 1 kg ms–1 – 1 kg ms–1 = 0

Let V = common velocity of both the bodies after striking to each other

Total momentum of both the bodies after collision

= (m1 + m2) V

= (0.5 + 0.5) V = V kg ms– 1

According to the law of conservation of momentum

Total momentum after collision = Total momentum before collision

i.e., V = 0

Thus, the combination of two bodies comes to rest after collision.

# CBSE Physics | Force and Laws of Motion: Law of Conservation of Momentum

## Force and Laws of Motion: Part 3

### Law of Conservation of momentum: –

According to this law, the total momentum of an object remains constant if no net external force is applied on the object.

### Illustration of the law of conservation of momentum: –

Let a moving ball collides with a stationary ball lying on the ground, observe what happens after collision. The moving ball will slow down i.e. its velocity decreases after colliding with the stationary ball.

On the other hand, the stationary ball beings to move i.e. its velocity increase after collision. We know, momentum of a body = mass of the body velocity of the body

Therefore, the momentum of moving ball decrease after collision and the momentum of the stationary ball increase after collision.

Thus, we find that when two balls collide with each other, then moving ball loses momentum and the stationary ball gains momentum. The loss of momentum of one ball is equal to the gain of momentum of other ball.

However, the total momentum of these colliding balls before and after the collision remains the same. This is the law of conservation of momentum.

### Proof of the law of conservation of momentum: –

Consider a system consisting of two bodies A and B of masses m1 and m2 respectively. suppose these bodies are moving with velocities u1and u2 as shown in fig below Let u1 > u2

These bodies collide with each other for a small interval of time t. At the time of collision FAB is the force on A due to B and FBA is the force on B due to A. According to Newton’s third law of motion FAB = – FBA.

Due to these forces, the momentum of the bodies changes. Let V1 and V2 is the velocities of body

A and body B respectively after collision.

Now, initial momentum of body A = m1u1

Initial momentum of body B = m2u2

Final momentum of body A = m1v1

Final momentum of body B = m2 v2

Total momentum of the system before collision = m1 u1 + m2 u2

Total momentum of the system after collision = m1v1 + m2 v2

Change in momentum of the body A = final momentum of body A – Initial momentum of body A

= m1v1 – m1u1

and change in momentum of the body B = final momentum of body B – initial momentum of body B

= m2 v2 – m2 u2

According to Newton’s second law of motion

and

According to Newton’s third law of motion FAB = –F BA

or

or

or

or

Hence, total momentum of the system before collision = Total momentum of the system after collision.

This is the law of conservation of momentum.

### Application of conservation of Momentum

#### (i) Rocket Propulsion (Movement of A rocket in the upward direction)

The movement of a rocket in the upward direction can also be explained with the help of the law of conservation of momentum.

The momentum of rocket before it is fired is zero. When the rocket is fired, gases are produced in the combustion chamber of the rocket due to the burning of fuel. These gases come out of the rear of the rocket with high speed. The direction of the momentum of the gases coming out of the rocket is in the downward direction. To conserve the momentum of the system (rocket + gases), the rocket moves upward with a momentum equal to the momentum of gases. The rockets continue to move upward as long as the gases are ejected out of the rocket.

#### (ii) Inflated balloon lying on the surface of a floor moves forward when pierced with a pin: –

The momentum of the inflated balloon before it is pierced wit a pin is zero. When it is pierced with a pin, air in it comes out with in the backward direction. To conserve the momentum, the balloon moves in the forward direction.

#### (iii) Recoil of gun:

The recoil of a gun can also be explained with the help of the law of conservation of momentum. Before firing, the gun and the bullet are at rest, therefore, momentum of the system before firing is zero.

When the bullet is fired, it leaves the gun in the forward direction with certain momentum. Since no external force acts on the system, so the momentum of the system must be zero after firing. This is possible only if the gun moves backward with a momentum equal to the momentum of the bullet. That is why gun recoils or moves backward.

#### Recoil velocity of a gun

Let, mass of the bullet = m

Velocity of the bullet after firing = v

Mass of the gun = M.

Velocity of the gun after firing = V

Since the system is at rest before firing, so moment of the system (gun + bullet) before firing = 0

According to the law of conservation of momentum,

Momentum of system after firing = Moment of system before firing

i.e. MV + mv = 0

or, MV = – mv

or,

Negative sign shows that the direction of the velocity of the gun after firing is opposite to the direction of the velocity of the bullet.