# 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

- During its upward motion
- During its downward motion
- 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:

- T
- `T-(mv^2)/t`
- `T+(mv^2)/t`
- 0

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

**Answer:** (a) T

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; 10^{5} 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