Free NCERT Solutions For Class 11 Physics Chapter 5 : Laws of Motion Free PDF Download

Question 1: Give the magnitude and direction of the net force acting on
(a) a drop of rain falling down with 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 a rough road,
(e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields.

Solution 1:
(a) Zero net force
The rain drop is falling with a constant speed. Hence, it acceleration is zero. As per Newton’s second law of motion, the net force acting on the rain drop is zero.
(b) Zero net force
The weight of the cork is acting downward. It is balanced by the buoyant force exerted by the water in the upward direction. Hence, no net force is acting on the floating cork.
(c) Zero net force
The kite is stationary in the sky, i.e., it is not moving at all. Hence, as per Newton’s first law of motion, no net force is acting on the kite.
(d) Zero net force
The car is moving on a rough road with a constant velocity. Hence, its acceleration is zero. As per Newton’s second law of motion, no net force is acting on the car.
(e) Zero net force
The high speed electron is free from the influence of all fields. Hence, no nett force is acting on the electron.

Question 2:
A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction an d magnitude of
the 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 was thrown at an angle of 45° with the horizontal direction? Ignore air resistance.

Ans:
0.5 N, in vertically downward direction, in all cases Acceleration due to gravity, irrespective of the direction of motion of an object, always acts downward. The gravitational force is the only force that acts on the pebble in all three cases. Its magnitude is given by Newton’s second law of motion as:

Ans:
Answer: (i)
When a particle connected to a string revolves in a circular path around a centre, the centripetal force is provided by the tension produced in the string. Hence, in the given case, the net force on the particle is the tension T, i.e.,
Where F is the net force acting on the particle

Question 16: Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string
when the masses are released.

Question 22: If, in Exercise 21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the
stone after the string breaks:
(a) the stone moves radially out wards,
(b) the stone flies off tangentially from the instant the string breaks,
(c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of
the particle?

Ans:
Answer: (b)
When the string breaks, the stone will move in the direction of the velocity at that instant. According to the first law of motion, the direction of velocity vector is tangential to the path of the stone at that instant. Hence, the stone will fly off tangentially from the instant the string breaks.

Question 23:
Explain why
(a) a horse cannot pull a cart and run in empty space,
(b) passengers are thrown forward from their seats when a speeding bus stops suddenly,
(c) it is easier to pull a lawn mower than to push it,
(d) a cricketer moves his hands backwards while holding a catch.

Solution 23:
(a) In order to pull a cart, a horse pushes the ground backward with some force. The ground in turn exerts an equal and opposite reaction force upon the feet of the horse. This reaction force causes the horse to move forward. An empty space is devoid of any such reaction force. Therefore, a horse cannot pull a cart and run in empty space.
(b) When a speeding bus stops suddenly, the lower portion of a passenger’s body, which is in contact with the seat, suddenly comes to rest. However, the upper portion tends to remain in motion (as per the first law of motion). As a result, the passenger’s upper body is thrown forward in the direction in which the bus was moving.

(c) While pulling a lawn mower, a force at an angle

is applied on it, as shown in the
following figure.
The vertical component of this applied force acts upward. This reduces the effective
weight of the mower.
On the other hand, while pushing a lawn mower, a force at an angle

is applied on it, as
shown in the following figure.
In this case, the vertical component of the applied force acts in the direction of the weight of
the mower. This increases the effective weight of the mower.
Since the effective weight of the lawn mower is lesser in the first case, pulling the lawn
mower is easier than pushing it.
According to Newton’s second law of motion, we have the equation of motion:

Additional Excersice has been done in pdf