# NCERT Solution , Laws of Motion

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

(a) a drop of rain falling down a constant speed.

(b) a cork of mass 10 g floating on water,

(c) a kite skillfully held stationary in the sky,

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

(e) a high speed electron in space far from all gravitational objects, and free of electric and magnetic fields.

Solution : (a) As the raindrop is falling with a constant speed, its acceleration a = 0, hence net force F = ma = 0

(b) As the cork is floating on water, its weight is being balanced by the up thrust (= weight of water displaced), hence net force on the cork = 0

(c) As the kite is held stationary, net force on the kite is zero, in accordance with Newton’s 1st law

(d) As no fields (gravitational, electrical or magnetic) act on the electron, net force = 0

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

(a) during its upward motion,

(b) during its downward motion,

(c) at the highest point, where it is momentarily at rest.

Do your answers change if the pebble were thrown at an angle, say, 45° with the horizontal direction ?

Solution:  If a body is thrown vertically upwards or downwards, the acceleration acting on the body is uniform i.e. acceleration due to gravity = g, in the downward direction. Therefore, the net force on the pebble in all the three cases is vertically downwards.

As m = 0.05 kg and a = 9.8 ms-2

in all three cases, F = ma = 0.05 × 9.8 = 0.49 N, vertically downwards.

If the pebble were thrown at an angle of 45° with the horizontal direction, it will have horizontal and vertical components of velocity. These components do not affect the force on the pebble. Hence, our answers do not alter any case. However, in case (c), the pebble will not be at rest. It will have horizontal component of velocity at the highest point.

Question.3. Give the magnitude and direction of the net force acting on

(a) a stone of mass 0.1 kg just after it is dropped from the window of a stationary train,

(b) the same stone as above just it is dropped from the window of a train running at a constant velocity of 36 km/hr.

(c) The same stone as above just after it is dropped from the window of train acceleration with 1 ms-2

(d) the same stone as above just after it is dropped from the window of a train accelerating with 1 ms-2 the stone being at rest relative to the train. Neglect the resistance of air throughout and take g = 9.8 ms-2.

Solution.  (a) Given, m = 0.1 kg, a = g = 9.8 ms-2.

Net force , F = ma

= 0.1 × 9.8 = 0.98 N , which acts vertically downwards.

(b) When the train is running at a constant velocity, its acceleration = 0, i.e. no force acts on the stone due to this motion

force on the same F = weight of stone

= mg = 0.1 × 9.8

= 0.98 N, which acts vertically downwards.

(c)  When the train is running with a acceleration, 1 ms-2, additional force, F’ = ma = 0.1 × 1 = 0.1 N acts on the stone in the horizontal direction. But once the stone is dropped from the train, F’ becomes zero and the net force on the stone, F = m g = 0.1 × 9.8 = 0.98 N, acting vertically downwards.

(d) As the stone is lying on the floor of the train, its acceleration is same as that of the train.

force acting on the stone, F = ma = 0.1 × 1 = 0.1 N. This force is along the horizontal direction of motion of the train. Note that weight of the stone in this case is being balanced by the normal reaction.

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

(i) T (ii) $T – \frac{m v^2}{r}$  (iii) $T + \frac{m v^2}{r}$ (iv) 0

where T is the tension in the string . Choose correct alternative.

Solution: The net force on the particle directed towards the centre is T . This provides the necessary centripetal force to the particle moving in the circle .

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 ms-1. How long does the body take to stop?

Solution:Here, F = -50 N, m = 20 kg, u = 15 ms-1 and v = 0

Now, F = ma

a = F/m = -50/20 = -2.5 ms-2

From relation, v = u + at

0 = 15 – 2.5 t

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 to 3.5 m/s in 25 s. The direction of motion of the body remains unchanged. What is the magnitude and direction of the force ?

Solution: Here, m = 3.0 kg, u = 2.0 ms-1 and v = 3.5 ms-1, t = 25 s;

Now, F = ma

$F = \frac{m(v – u)}{t}$

$F = \frac{3(3.5 – 2.0)}{25}$

F = 0.18 N

The force is along the direction of motion.

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

Solution:  Here m = 5 kg , F1  = 8 N and F2  = 6 N

Resultant force, $F = \sqrt{(8^2 + 6^2)}$

F = 10 N

$tan\theta = \frac{6}{8} = 0.75$

which is the direction of the resultant force, and hence the direction of the acceleration.

Hence ,   a = F/m = 10/5 = 2 ms-2

Question.8. The driver of a three wheeler moving with speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4s  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 65kg.

Solution: Here, u = 36 km/h = 10 m/s , t = 4s, m = 400 + 65 = 465 kg

Now, Retarding force, $F = ma = \frac{m(v – u)}{t }$

F = 465(0 – 10)/4 = -1162.5 N

Question.9. A rocket with lift off mass 20000 kg is blasted upwards with net initial acceleration of 5 ms-2. Calculate the initial thrust of the blast.

Solution: Here, m = 20000 kg, a = 5 ms-2

The thrust should be such that it overcomes the force of gravity besides giving it an upwards acceleration of 5 ms-2. Thus the force should produce a net acceleration of 9.8 + 5.0 = 14.8 ms-2.

As thrust = force = mass × acceleration.

Force, F = 20000 × 14.8

= 2.96 × 105 N

Question.10. A body of mass 0.40 kg moving initially with a constant speed of 10 m/s 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, and the position of the particle at that time to be x = 0, predict its position at t = -5 s , 25 s , 100 s ?

Solution: Here, m = 0.4 kg, u = 10 ms-1 due north

F = -8 N, (negative sign shows the force directed opposite).

Therefore, a = F/m = -8/0.4

= -20 ms-2 (0 ≤ t ≤ 30 s)

When t = -5 s ,

x = ut = 10(-5) = -50 m

When t = 25 s ,

$x = ut + \frac{1}{2} at^2$

$x = 10 \times 25 + \frac{1}{2} (-20)(25)^2$

= – 6000 m

upto t = 30 s, motion is under acceleration,

i.e. $x_1 = ut + \frac{1}{2} at^2$

$x_1 = 10 \times 30 + \frac{1}{2} (-20) (30)^2$

= -8700 m

At t = 30 s , v = u + at

= 10 – 20 × 30 = -590 m/s

During t = 30 to 100 s,

x2 = vt = -590 × 70

x2 = – 41300 m

Total distance, x1 + x2 = -(8700 + 41300)m = -50 km

Question.11. A truck starts from rest and accelerates uniformly with 2 ms-2. At t = 10s, a stone is dropped by a person standing on the top of the truck (6m high from ground). What are the (a) velocity and (b) acceleration of the stone at t = 11s ? Neglect air resistance, and take g =9.8ms-2.

Solution: Given, u = 0 , t = 10 s , a = 2 ms-2

The velocity of the truck when the stone is dropped,

v = u + at = 0 + 2 ×10 = 20 ms-1.

(a) When the stone is dropped, horizontal velocity, vx = v = 20 ms-1; vx remains constant.

In vertical direction, a = g = 9.8 ms-2, u = 0, t = 1 1– 10 = 1s.

vy = v = u + at = 0 + 9.8 × 1 = 9.8 ms-1

Resultant velocity = √(202 + 9.82)

If θ = angle which vx makes with resultant velocity,

θ = tan-1(vy/vx) = tan-1(9.8/20) = tan-1(0.49) = 29°

(b) When the stone is dropped from the car, horizontal force on the stone is zero. Only acceleration of the stone = acceleration due to gravity = 9.8 ms-2 in the vertically downward direction. The path of the stone would be parabolic.

Question.12 . A bob of mass 0.1 kg hung from the ceiling of room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 m/s.
What is the trajectory of the bob, if the string is cut when the bob is

(a) at one of its extreme positions

(b) at its mean position ?

Solution: (a)At each extreme position, velocity of the bob is zero. If the extreme position, it is only under the action of ‘g’. Hence the bob will fall vertically downwards.

(b)At each mean position, velocity of the bob is 1 ms-1 along the tangent to the arc, which is in the horizontal direction. If the string is cut at mean position, the bob will behave as horizontal projectile. Hence, it will follow a parabolic path.

Question.13. A man of mass 70 kg stands on a weighing machine in lift, which is moving

(a) upwards with a uniform speed of 10 ms-1,

(b) downwards with a uniform acceleration of 5 ms-2,

(c) upwards with a uniform acceleration of 5 ms-2.

What would be the reading on the scale in each case ? What would be the reading if the lift mechanism failed and it came down freely under the gravity? Take g = 9.8 ms-2.

Solution: Given m = 70 kg, a = 9.8 ms-2

The weighing machine gives the reading of the reaction force, R, which is apparent weight.

(a) When lift is moving upwards with a uniform speed, acceleration = 0,

R = mg = 70 × 9.8 = 686 N

(b) When lift is moving downwards with acceleration = 5 ms-2,

m g – R = m a

R = m(g – a)

R = 70 × (9.8 – 5) = 336 N

(c)When lift is moving upwards with acceleration = 5 ms-2,

R – m g = m a

R = m(g + a)

R = 70 × (9.8 + 5) = 1036 N

When lift is coming down freely under gravity, acceleration, a = g

m g – R = m a

R = m g – ma

R = m(g – g) = 0

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