# 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 9th Physics | Force and Laws of Motion: Part 1

## Force and Laws of Motion: Part 1

### Force

Definition :

A pull or push acting on a body which changes or tends to change the speed of the body or direction of motion of the body or the shape and size of the body is known as force.

A force influences the shaped and motion of an object. A single force will change its velocity (i.e. accelerate it) and possibly its shape.

Two equal and opposite force may change its shape or size. It is a vector quantity, having both magnitude and direction, and is measured in Newtons.

The main types of force are gravitational, magnetic, electric and nuclear.

The S.I. unit of force is Newton and denoted by N.

The CGS unit of force is Dyne and denoted by D.

### Effect of Force:

A force when applied on an object may show the following effects:

(i) A force may cause motion in an object at rest

(ii) A force can bring the moving object to rest

(iii) A force can change the speed of a moving body

(iv) A force may change the size and shape of the object.

(v) A force may change the direction of the motion of a moving object.

Note:

A force acting on an object does not necessarily show the above mentioned effects. For example, when we push a wall (ie force is applied on the wall), the wall does not move.

Similarly, if we press a steel ball, it is not deformed. This is because, in these cases, the force applied by us is not sufficient to cause the changes

### Balanced and Unbalanced Force

When the forces acting on a body cancel out the effects of each other in such a way that the resultant force is zero.

Then these forces are known as balanced force. Balanced force may change the shape or the size of an object.

Example: –

Let us consider a rigid almirah kept at rest on a horizontal surface. Let two forces F1 and F2 act simultaneously on the almirah from opposite directions.

If the two forces are equal the effects produced by one force get cancelled by the effect produced by other. The net force or the resultant force is then zero. The almirah continues to remain at rest.

### Unbalanced force: –

When two forces of unequal magnitudes, act in opposite directions on an object simultaneously then the object moves in the directions of the larger force.

These forces acting on the object are known as unbalanced forces.

Example: –

When a boy drags a box on the floor, then an unbalanced force is acting on the box. The force of pull is applied by the boy acts on the box in the direction of pull and the force of friction between the lower surface of the box and the surface of the floor acts in the opposite direction.

The box moves in the direction of unbalanced force (= force of pull- force of friction) i.e. in the direction of pulling force.

Note: –

When balanced forces act on a body the acceleration of the body is zero. In otherwards, if a number of forces acting on a body produce zero acceleration, then the forces are balanced.

And if a number of forces acting on a body produce an acceleration, then the forces are unbalanced.

### Newton’s Law of Motion

Aristotle, a great Greek philosopher believed that, to maintain the motion of a body, a continuous force was needed.

However this concept of motion was totally rejected by a great Italian physicist Galileo Galileo in 17th century. He gave the famous principle of inertia, which will be discussed a bit latter.

The idea of Galileo helped Newton to state the first law of motion which is also known as law of inertia.

### 1. Newton’s first law of motion: –

According to this law, a body continues to be in its state of rest or uniform motion in a straight line unless compelled by some external force (i.e. unbalanced force) to change that state.

Newton’s first law of motion gives the definition of force-

### Inertia:-

The tendency of a body to oppose any change in its state of rest or uniform motion or direction is called inertia of the body.

The principle of inertia was given by Galileo Galilei in 17th century. He verified this fact by performing the following experiment.

Consider a double inclined plane ABCD as shown in fig. 2.1 (a). The surface of these planes are very smooth so that there is no force of friction.

A glass ball released from height h was observed. It was found that it rises up to the some height h on the plane CD.

Now, the inclination of eth plane CD was decreased as shown in fig. 2.1 (b). Again the ball released from height h on the plan AB was observed.

It was found that again the ball reached to the some height h on the plane CD after traveling more distance. Now the inclination of the plane CD was reduced to zero so that the plane CD was made horizontal as shown in fig. 2.1 (c).

The ball released from height h on the plane AB was found to travel forever.

Inertia is the inherent property of all the objects.

### Types of inertia: –

(i) Inertia of rest

(ii) Inertia of motion

(iii) Inertia of direction

### (i) Inertia of rest : –

The tendency of a body to oppose any change in its state of rest is known as inertia of rest.

For example: – When a bus suddenly starts moving forward, the passengers in the bus fall backward. This is because, the lower part of the bodies of the passengers being in contact with the floor of the bus, come in motion along with the bus.

On the other hand, the upper part of their bodies remains at rest due to inertia of rest. Hence the passengers fall backward.

### (ii) Inertia of motion: –

The tendency of a body to oppose any change in its state of uniform motion is known as inertia of motion.

Example: –

The passengers fall forward when a fast moving bus stops suddenly.

This happens because, the lower part of the bodies of the passengers come to rest as soon as the bus stops. But the upper part of their bodies continue to move forward due to the inertia of motion.

### (iii) Inertia of direction: –

The tendency of a body to oppose any change in its direction of motion is known as inertia of direction.

Fore Example: – When a bus suddenly takes a turn, the passengers sitting casually experience a jerk in the outward direction.

This happens because the passengers tend to remain in their original direction of motion due to inertia of direction.