Class 11 Physics

# Laws of Motion

## NCERT Solution

### Part 1

Question 1: Give the magnitude and direction of the net force acting on

(a) A drop of rain falling down with constant speed

Answer: Since the speed is constant hence acceleration is zero, and as a result the net force is zero.

(b) A cork of mass 10 g floating on water

Answer: The cork is floating on water because buoyant force is balancing the downward force due to gravity. So, net force is zero.

(c) A kite skillfully held stationary in the sky

Answer: Kite is at rest, so net force is zero.

(d) A car moving with a constant velocity of 30 km/h on a rough road

Answer: Constant velocity implies zero acceleration. Hence, net force is zero.

(e) A high speed electron in space far from all material objects, and free of electric and magnetic fields

Answer: Electron is free of electric and magnetic fields so net force is zero.

Question 2: A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble

1. During its upward motion
2. During its downward motion
3. At the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle with 45° with the horizontal direction?

Ignore air resistance

Answer: Whether the pebble is moving upward or downward, the only force acting on it is due to gravitational acceleration which always acts towards down.

F = m × a

= 0.05 × 10 = 0.5 N

When the pebble is thrown at 45° angle there are two components of force, i.e. horizontal and vertical components. But in all these cases the net effect of horizontal component will be zero.

Question 3: Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg

(a) Just after it is dropped from the window of a stationary train

Answer: F = m × a

=0.1 × 10 = 1 N

(b) Just after it is dropped from the window of a train running at a constant velocity of 36 km/h

Answer: Constant velocity of the train means zero acceleration. So, in this case also the net force acting on stone = 1 N

(c) Just after it is dropped from the window of a train accelerating with 1 m s-2

Answer: The moment stone is dropped from the train its acceleration in horizontal direction becomes zero. So net force acting on stone = 1 N

(d) Lying on the floor of a train which is accelerating with 1 m s-2, the stone being at rest relative to the train.

Answer: Here, F = m × a

= 0.1 × 1 = 0.1 N (horizontal direction)

Question 4: One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is:

1. T
2. T-(mv^2)/t
3. T+(mv^2)/t
4. 0

T is the tension in the string. (choose correct alternative)

When a particle is moving on a circular path, centripetal force is provided by the tension on the string

F = T =(mv^2)/l

Question 5: A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m s-1. How long does the body take to stop?

Answer: F = m × a

Or, 50 = 20 × a

Or, a = 50 ÷ 20 = 2.5 m s-2

Now, we have u = 15 m/s and v = 0

Time taken to stop can be calculated as follows:

v=u+at

Or, 0=15-2.5t

Or, 2.5t=15

Or, t=(15)/(2.5)=6 s

Question 6: A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s-1 to 3.5 m s-1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

Answer: In the first step we need to calculate acceleration

Given, u = 2 m/s, v = 3.5 m/s and t = 25 s

v=u+at

Or, 3.5=2+25a

Or, 25a=3.5-2=1.5

Or, a=(1.5)/(25)=0.06 m s-2

Now, F = ma

= 3 × 0.06 = 0.18 N

As the speed is increasing so direction of force is in direction of motion.

Question 7: A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of acceleration of the body.

Answer: The two perpendicular forces can be shown by following diagram The resultant force can be calculated using Pythagoras theorem

h^2=p^2+b^2

=8^2+6^2=100

Or, h=sqrt(100)=10 N

Acceleration can be calculated as follows:

F = m × a

Or, 10 = 5 × a

Or, a = 10 ÷ 5 = 2 m s-2

Acceleration is in the direction as shown by resultant force R in the diagram

Question 8: The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.

Answer: Given, u = 36 km/h =36xx5/(18)=10 m/s, v = 0, t = 4 s, m = 400 + 65 = 465 kg

Acceleration can be calculated as follows:

v=u+at

Or, 0=10+4a

Or, 4a = -10

Or, a=-(10)/4=-2.5 m s-2

F = m × a

= 465 × -2.5 = -1162.5 N

The force is in opposite direction to motion

Question 9: A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s-2. Calculate the initial thrust (force) of the blast.

Answer: Given, m = 20,000 kg, a = 5 m s-2 and g = 10 m s-2

As per Newton’s second law of motion, the net thrust acting on the rocket is given as follows:

F-mg=ma

Or, F=ma+mg

Or, F=20000(5+10)

= 20000 × 15 = 300000 N = 3 7times; 105 N

Question 10: A body of mass 0.40 kg moving initially with a constant speed of 10 m s-1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time be x = 0, and predict its position at t = -5 s, 25s, 100s.

Answer: Given, m = 0.40 kg, u = 10 m/s, N = - 8 N

Let us first calculate acceleration

F=ma

Or, -8=0.4a

Or, a=-8/(0.4)=-20 m s-2

Position at – 5 s

s=ut+1/2at^2

=10xx(-5)-1/2xx0xx(-5)^2

= - 50 m

Position at 25 s

s=10xx25-1/2xx20xx25^2

=250-6250=-6000 m

Position at 100 s

As the force is being applied for 30 s so we need to find the position at 30 s

s=10xx30-1/2xx20xx30^2

=300-9000=-8700 m

Now, we need to find the velocity after 30 s

v=u+at

=10-20xx30=10-600=-590 m/s

Acceleration will be zero after 30 s

So, distance covered between 30 and 100 s

= - 590 × 70 = -41300

Total distance = - 8700 – 41300 = - 50000 m = - 50 km