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NCERT Solutions Class 11 Physics Chapter 5 Laws of Motion – Here are all the NCERT solutions for Class 11 Physics Chapter 5. This solution contains questions, answers, images, explanations of the complete chapter 5 titled Of Laws of Motion taught in Class 11. If you are a student of Class 11 who is using NCERT Textbook to study Physics, then you must come across chapter 5 Laws of Motion After you have studied the lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 11 Physics Chapter 5 Laws of Motion in one place.

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NCERT Solutions Class 11 Physics Chapter 5 Laws of Motion

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Class 11
Subject Physics
Book Physics Part I
Chapter Number 5
Chapter Name

Laws of Motion

NCERT Solutions Class 11 Physics chapter 5 Laws of Motion

Class 11, Physics chapter 5, Laws of Motion solutions are given below in PDF format. You can view them online or download PDF file for future use.

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Question & Answer

Q.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.

Ans : (a) Zero net force 
The raindrop is falling with a constant speed. Hence, it acceleration is zero. 
Newton's second law of motion, the net force acting on the rain drop is zero. 
(b) Zero net force 
As per 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 car. 
(e) Zero net force 
The high speed electron is free from the influence of all fields. Hence, no net force is 
acting on the electron. 
Q.2: A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and 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: 
\(\begin{array}{l}{F=m \times a} \\ {\text { Where, }} \\ {F=\text { Net force }} \\ {m=\text { Mass of the pebble }=0.05 \mathrm{kg}} \\ {a=\mathrm{g}=10 \mathrm{m} / \mathrm{s}^{2}} \\ {\therefore F=0.05 \times 10=0.5 \mathrm{N}}\end{array}\)
The net force on the pebble in all three cases is 0.5 N and this force acts in the 
downward direction. 
If the pebble is thrown at an of 450 with the horizontal, it will have both the 
horizontal and vertical components Of velocity. At the highest point, only the vertical 
component of velocity becomes zero. However, the pebble will have the horizontal 
component of velocity throughout its motion. This component of velocity produces no 
effect on the net force acting on the pebble. 
Q.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,
(b) just after it is dropped from the window of a train running at a constant velocity of 36 km/h, 
(c) just after it is dropped from the window of a train accelerating with 1 m \(\mathbf{s}^{-2}\) , 
(d) lying on the floor of a train which is accelerating with 1 m \(\mathbf{s}^{-2}\), the stone being at rest relative to the train. Neglect air resistance throughout.

Ans : (a) l N; vertically downward 
Mass of the stone, m 0.1 kg 
Acceleration Of the stone, a = g = 10 \(\mathrm{m} / \mathrm{s}^{2}\) 
As per Newton's second law of motion, the net force acting on the stone, 
F ma = mg 
=0.1 x 10=1 N 
(b) l N; vertically downward 
The train is moving with-I a constant velocity. Hence, its acceleration is zero in the 
direction of its motion, i.e., in tie horizontal direction. Hence, no force is acting on ene 
stone in the horizontal direction. 
The net force acting on the stone is because of acceleration due to gravity and it always 
acts vertically downward. The magnitude of this force is 1 N.
(c) l N; vertically downward 
It is given that the train is accelerating at the rate Of 1 \(\mathrm{m} / \mathrm{s}^{2}\) 
Therefore, the net force acting on the stone, F' = ma — -0.1 x 1-0.1 N 
This force is acting in the horizontal direction. Now, when the stone is dropped, the 
horizontal force F,' stops acting on the stone. This is because of the fact that the force 
acting on a body at an instant depends on the situation at that instant and not on earlier 
Situations.
Therefore, the net force acting on the stone is given only by acceleration due to gravity. 
\(F=m g=1 \mathrm{N}\)
This force acts vertically downward.  
(d) O.1 N, in the direction of motion of the train 
The weight of the stone is balanced by the normal reaction of the floor. The only acceleration is provided by the horizontal motion of the train. 
Acceleration of the train, a = 0.1 \(\mathrm{m} / \mathrm{s}^{2}\)  
The net force acting on the stone will be in the direction of motion of the train. Its 
magnitude is given by: 
F = ma 
-0.1 x 1-0.1 N 
Q.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 :
(i) \(T\)
(ii) \(T-\frac{m w^{2}}{l}\)
(iii) \(T+\frac{m w^{2}}{l}\)
(iv) 0

Ans : (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 
\(F=T=\frac{m v^{2}}{l}\)
Where F is the net force acting on the particle. 
Q.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 \(\mathbf{s}^{-1}\). How long does the body take to stop ?

Ans : Retarding force, F = —50 N 
Mass of the body, m = 20 kg 
Initial velocity of the body, u = 15 m/s 
Final velocity Of the body, V = O 
using Newton's second law of motion, the acceleration (a) produced in the body can be 
calculated as: 
\(\begin{array}{l}{F=m a} \\ {-50=20 \times a} \\ {\therefore a=\frac{-50}{20}=-2.5 \mathrm{m} / \mathrm{s}^{2}}\end{array}\)
using the first equation Of motion, the time (t) taken by the body to come to rest can be 
calculated as: 
\(\begin{array}{l}{v=u+a t} \\ {\therefore t=\frac{-u}{a}=\frac{-15}{-2.5}=6 \mathrm{s}}\end{array}\)

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