NCERT Solution for Class 11 Physics Chapter 4: Laws of Motion

(a) As the drop of rain is falling with constant speed, in accordance with the first law of motion, the net force on the drop of rain is zero.

(b) As the cork is floating on water, its weight is being balanced by the upthrust (equal to.weight of water displaced). Hence net force on the cork is zero.

(c) Net force on a kite skilfully held stationary in the sky is zero because it is at rest.

(d) Since the car is moving with a constant velocity, the net force on the car is zero.

(e) Since an electron is far away from all material agencies producing electromagnetic and gravitational forces, the net force on the electron is zero.

(a) When the pebble is moving upward, the acceleration g is acting downward, so the force is acting downward is equal to F = mg = 0.05 kg x 10 ms-2 = 0.5 N.

(b) In this case also F = mg = 0.05 x 10 = 0.5 N. (downwards).

(c) The pebble is not at rest at the highest point but has a horizontal component of velocity. The direction and magnitude of the net force on the pebble will not alter even if it is thrown at 45° because no other acceleration except ‘g’ is acting on the pebble.

(a) Mass of stone = 0.1 kg

Net force, F = mg = 0.1 x 10 = 1.0 N. (vertically downwards).

(b) When the train is running at a constant velocity, its acceleration is zero. No force acts on the stone due to this motion. Therefore, the force on the stone is the same (1.0 N.).

(c) The stone will experience an additional force F’ (along horizontal) i.e.,F = ma = 0.1 x l = 0.1 N

As the stone is dropped, the force F’ no longer acts and the net force acting on the stone F = mg = 0.1 x 10 = 1.0 N. (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 x 1 = 0.1 N.

It acts along the direction of motion of the train.

(i) T
The net force T on the particle is directed towards the centre. It provides the centripetal force required by the particle to move along a circle.

Here m = 20 kg, F = – 50 N (retardation force)
As F = ma

Here mass of three-wheeler mi = 400 kg, mass of driver = m2 = 65 kg, initial speed of auto,
u = 36 km/h = 36 x — m/s 10 ms-1, final speed, v – 0 and t = 4s.

Here, m = 20000 kg = 2 x 104 kg
Initial acceleration = 5 ms-2
Clearly, the thrust should be such that it overcomes the force of gravity besides giving it an upward acceleration of 5 ms-2.
Thus the force should produce a net acceleration of 9.8 + 5.0 = 14.8 ms-2.
Since, thrust = force = mass x acceleration
F = 2 x 104 x 14.8 = 2.96 x 105 N.

u = 0, a = 2 ms-2, t 10 s
Using equation, v = u + at, we get
v = 0 + 2 x 10 = 20 ms-1
(a) Let us first consider horizontal motion. The only force acting on the stone is force of gravity which acts vertically downwards.
Its horizontal component is zero. Moreover, air resistance is to be neglected. So, horizontal motion is uniform motion.
.-. vx = v = 20 ms-1
Let us now consider vertical motion which is controlled by force of gravity.
u=0, a = g = 10 ms-2, t = (11 — 10) s = 1 s

(b) The moment the stone is dropped from the car, horizontal force on the stone is zero. The only acceleration of the stone is that due to gravity. This gives a vertically downward acceleration of 10 ms-2. This is also the net acceleration of the stone.

 Let the bob be oscillating as shown in the figure.
(a) When the bob is at its extreme position (say B), then its velocity is zero. Hence on cutting the string the bob will fall vertically downward under the force of its weight F = mg.
(b) When the bob is at its mean position (say A), it has a horizontal velocity of v = 1 ms-1 and on cutting the string it will experience an acceleration a = g = 10 ms-2 in vertical downward direction. Consequently, the bob will behave like a projectile and will fall on ground after describing a parabolic path.

Here, m = 70 kg, g = 10 m/s2
The weighing machine in each case measures the reaction R i.e., the apparent weight.
(a) When the lift moves upwards with a uniform speed, its acceleration is zero.
R = mg = 70 x 10 = 700 N
(b) When the lift moves downwards with a = 5 ms-2
R = m (g – a) = 70 (10 – 5) = 350 N
(c) When the lift moves upwards with a = 5 ms-2
R = m (g + a) = 70 (10 + 5) = 1050 N
(d) If the lift were to come down freely under gravity, downward acc. a = g
:. R = m(g -a) = m(g-g) = Zero.

Let m1, m2 be the masses of products and v1,v2 be their respective velocities. Therefore, total linear momentum after disintegration = m1v1 +m2 v2. Before disintegration, the nucleus is at rest.
Therefore, its linear momentum before disintegration is zero.
According to the principle of conservation of linear momentum,

Negative sign shows that v1 and v2 are in opposite directions.

Initial momentum of each ball before collision
= 0.05 x 6 kg ms-1 = 0.3 kg ms-1
Final momentum of each ball after collision
= – 0.05 x 6 kg ms-1 = – 0.3 kg ms-1 Impulse imparted to each ball due to the other
= final momentum – initial momentum = 0.3 kg m s-1 – 0.3 kg ms-1
= – 0.6 kg ms-1 = 0.6 kg ms-1 (in magnitude)
The two impulses are opposite in direction.

 m = 0.02 kg, M = 100 kg, v = 80 ms-1, V = ?

Negative sign indicates that the gun moves in a direction opposite to the direction of motion of the bullet.

(b) The velocity is tangential at each point of circular motion. At the time the string breaks, the particle continues to move in the tangential direction according to Newton’s first law of motion.

 (a) A horse by itself cannot move in space due to law of inertia and so cannot pull a cart in space.
(b) The passengers in a speeding bus have inertia of motion. When the bus is suddenly stopped the passengers are thrown forward due to this inertia of motion.
(c) In the case of pull, the effective weight is reduced due to the vertical component of the pull. In the case of push, the vertical component increases the effective weight.
(d) The ball comes with large momentum after being hit by the batsman. When the player takes a catch it causes a large impulse on his palms which may hurt the cricketer. When he moves his hands backward the time of contact of ball and hand is increased so the force is reduced.

This graph can be of a ball rebounding between two walls situated at position 0 cm and 2 cm. The ball is rebounding from one wall to another, time and again every 2 s with uniform velocity.

Here acceleration of conveyor belt a = 1 ms-2, μs= 0.2 and mass of man m = 65 kg. t As the man is in an accelerating frame, he experiences a pseudo force Fs = ma as shown
in fig. (a). Hence to maintain his equilibrium, he exerts a force F = – Fs = ma = 65 x 1 = 65 N in forward direction i.e., direction of motion of belt.
.’. Net force acting on man = 65 N (forward)
As shown in fig. (b), the man can continue to be stationary with respect to belt, if force of friction

The net force at the lowest point is (mg – T1) and the net force at the highest point is (mg + T2). Therefore, alternative (a) is correct.
Since mg and T1 are in mutually opposite directions at lowest point and mg and T2 are in same direction at the highest point.

Here, mass of helicopter, m1= 1000 kg
Mass of the crew and passengers, m2 = 300 kg upward acceleration, a = 15 ms-2 and g = 10 ms-2
(a)Force on the floor of helicopter by the crew and passengers = apparent weight of crew and passengers
= m2 (g + a) = 300 (10 + 15) N = 7500 N
(b)Action of rotor of helicopter on surrounding air is obviously vertically downwards, because helicopter rises on account of reaction to this force. Thus, force of action
F = (m1+ m2) (g + a) = (1000 + 300) (10 + 15) = 1300 x 25 = 32500 N
(c)Force on the helicopter due to surrounding air is the reaction. As action and reaction are equal and opposite, therefore, force of reaction, F = 32500 N, vertically upwards.

In one second, the distance travelled is equal to the velocity v.
Volume of water hitting the wall per second, V = av where a is the cross-sectional area of the tube and v is the speed of water coming out of the tube.
V = 10-2 m2 x 15 ms-1 = 15 x 10-2 m3 s-1
Mass of water hitting the wall per second
= 15 x 10-2 x 103 kg s-1 = 150 kg s-1 [v density of water = 1000 kg m-3] Initial momentum of water hitting the wall per second
= 150 kg s-1 x 15 ms-1 = 2250 kg ms-2 or 2250 N Final momentum per second = 0 Force exerted by the wall = 0 – 2250 N = – 2250 N Force exerted on the wall = – (- 2250) N = 2250 N.

(a) The force on 7th coin is due to weight of the three coins lying above it. Therefore,
F = (3 m) kgf = (3 mg) N
where g is acceleration due to gravity. This force acts vertically downwards.
(b) The eighth coin is already under the weight of two coins above it and it has its own weight too. Hence force on 7th coin due to 8th coin is sum of the two forces i.e.
F = 2 m + m = (3 m) kg f = (3 mg) N The force acts vertically downwards.
(c) The sixth coin is under the weight of four coins above it.
Reaction, R = – F = – 4 m (kg) = – (4 mgf) N Minus sign indicates that the reaction acts vertically upwards, opposite to the weight.

In 1st case, man applies an upward force of 25 kg wt.r (same as the weight of the block). According to Newton’s third law of motion, there will be a downward reaction on the floor.
The action on the floor by the man.
= 50 kg wt. + 25 kg wt. = 75 kg  wt = 75 kg x 10 m/s2 = 750 N.
In case II, the man applies a downward force of 25 kg wt. According to Newton’s third law, the reaction is in the upward direction.
In this case, action on the floor by the man
= 50 kg wt – 25 kg wt. = 25 kg wt. = 25 kg x 10 m/s2 = 250 N.
Therefore, the man should adopt the second method.

(a) When the monkey climbs up with an acceleration a, then T – mg = ma
where T represents the tension (figure a).

(c) When the monkey climbs up with uniform speed, then
T mg = 40 kg x 10 ms-2 = 400 N The rope will hot break.
(d) When the monkey is falling freely, it would be a state of weightlessness. So, tension will be zero and the rope will not break.

(i) When the wall exists and blocks A and B are pushing the wall, there can’t be any motion i.e., blocks are at rest. Hence,
(a) reaction of the partition = – (force applied on A) = 200 N towards left.
(b) action-reaction forces between A and B are 200 N each. A presses B towards right with an action force 200 N and B exerts a reaction force on A towards left having magnitude 200 N.
(ii) When the wall is removed, motion can take place such that net pushing force provides the acceleration to the block system. Hence, taking kinetic friction into account, we have

(a) Force experienced by block, F = ma = 15 x 0.5 = 7.5 N Force of friction,Ff= p mg = 0.18 x 15 x 10 = 27 N. i.e., force experienced by block will be less than the friction.So the block will not move. It will remain stationary w.r.t. trolley for a stationary observer on ground.
(b) The observer moving with trolley has an accelerated motion i.e., he forms non-inertial frame in which Newton’s laws of motion are not applicable. The box will be at rest relative to the observer.

If the coin is to revolve with the record, then the force of friction must be enough to provide the necessary centripetal force.

When the motorcyclist is at the highest point of the death-well, the normal reaction R on the motorcyclist by the ceiling of the chamber acts downwards. His weight mg also acts downwards. These two forces are balanced by the outward centrifugal force acting on him.

Here v is the speed of the motorcyclist and m is the mass of the motorcyclist (including the mass of the motor cycle). Because of the balancing of the forces, the motorcyclist does not fall down.
The minimum speed required to perform a vertical loop is given by equation (1) when R = 0.