Laws of Motion is one of the most weighing topics in JEE Main Physics Syllabus. The topic deals with the study of objects or say various bodies and the force acting on them, given by Sir Isaac Newton. JEE Main Laws of Motion is scoring in nature and is also easy to understand. Check JEE Main physics syllabus

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The topic of Laws of Motion in JEE Main is a crucial portion covering a weightage of 6.6 % of the entire paper with almost 2 questions on the topic. In order to secure a meritorious position in the JEE Main exam, the candidates are required to prepare the section on Laws of Motion with the help of the previous year question papers. The thorough understanding of the topic might help the candidates to qualify for a better position. Check JEE Main Sample Papers

The applications of Laws of Motion are spread through various fields of sports like playing basketball, swimming, shooting, etc. Further, the topic also has its application in physics and statistics. Laws of Motion generally involves various applications and studies through statistics and figures which are ultimately easily and fluently understood by the students.

For reference, a key of previous year questions and solutions are given below for the topic of Laws of Motion which are more frequently asked in the exam every year. Solving these questions would help the students to prepare well for the upcoming JEE Main examination. Beginners are suggested to use the authentic study materials for the same as the NCERTs of Physics and textbooks of class 10 – 12. After that, they could definitely reach out to the below-given set of questions for a more thorough revision. Check Important Books for JEE Main Preparation

JEE Main Test Series Laws of Motion: Previous Year Questions and Solution

Majorly all the questions in this section carry a weightage of 6.6 % in JEE Main Question Paper and are of year 2016, 2017, 2018 and 2019.

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Q.1 A block of mass 2kg rests on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block is___________.

Solution: 

Q. 2 During peddling of a bicycle, the force of friction exerted by the ground on the two wheels is such that it acts

1. in the backward direction on the front wheel and in the forward direction on the rear wheel

2. in the forward direction on the front wheel and in the backward direction on the rear wheel

3. in the backward direction on both, the front and the rear wheels

4. in the forward direction on both, the front and the rear wheels.

Solution: (a) due to peddling, the point of contact of the real wheel has a tendency to move backwards. So frictional force opposes the back ward tendency i,e., frictional force acts in the forward direction. But the back wheel accelerates the front wheel in the forward direction. To oppose this frictional force acts in the backward direction on the front wheel.

Q. 3 A car is moving in a circular horizontal track of radius 10m with a constant speed of lOm/s. A plumb bob is suspended from the roof of the car by a light rigid rod. The angle made by the rod with the track is (g = 10 m/s2) 

1. Zero

2. 30°

3. 45°

4. 60°

Solution:

Q. 4 In the arrangement, shown in the figure, the ends P and Q of an unstretchable string move downwards with uniform speed u. Pulleys A and B are fixed. Mass M moves upwards with a speed

Solution. 

Q. 5 A small block is shot into each of the four tracks as shown below. Each of the tracks rises to the same height. The speed with which the block enters the track is the same in all cases. At the highest point of the track, the normal reaction is maximum in

Solution.

Q. 6 An insect crawls up a hemispherical surface very slowly (see fig.) The coefficient of friction between the insect and the surface is 1/3.If the line joining the center of the hemispherical surface to the insect makes an angle alpha with the vertical the maximum possible value of alpha is given by

Solution.

Q. 7 A block P of mass m is placed on a horizontal frictionless plane. A second block of same mass m is placed on it and is connected to a spring of spring constant k. The two blocks are pulled by distance A. Block Q oscillates without slipping. What is the maximum value of frictional force between the two blocks

Solution.

1. Block Q oscillates but does not slip in the block P. It means that the acceleration is same for both blocks P and block Q. There is the force of friction between the two blocks while the horizontal plane is frictionless. The spring is connected to upper block. The ( P – Q ) system oscillates with angular frequency. The spring is then stretched by block A. 

Therefore,

∞ = √ k / m + m = √ k / 2m

Therefore, maximum acceleration in SHM = ∞2 A

Am = Ka/ 2m ….. (i)

Now consider the lower block.

Let the maximum force of friction = fm

Therefore,

Fm = mam or fm = m X Ka / 2m

Q. 8 Two particles of mass m each are tied at the., ends of a light string of length 2a. The whole system is kept on a frictionless ‘ a ‘ a horizontal surface with the string held tight so that each mass is at a distance a from the center P (as shown in the figure). Now, the mid-point of the string is pulled vertically upwards with a small but constant force F. As a result, the particles move towards each other on the surface. The magnitude of acceleration, when the separation between them

Solution.

Q. 9 Two blocks A and B of masses 2m and m, respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in the figure. The magnitudes of acceleration of A and B, immediately after the string is cut, are respectively

1. g, g / 2

2. g / 2, g

3. g, g

4. g / 2, g / 2

Solution.

Q. 10 If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that ? 

1. linear momentum of the system does not change in time

2. kinetic energy of the system does not change in time

3. angular momentum of the system does not change in time

4. potential energy of the system does not change in time

Solution: (a) Linear momentum remains constant if net external force on the system of parlicle is zero.

Q. 11 A piece of wire is bent in the shape of a parabola y = kx2 ( y - axis vertical ) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x - axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y - axis is

Solution.

Q. 12 A reference frame attached to the earth

1. is an inertial frame by definition.

2. cannot be an inertial frame because the earth is revolving round the sun.

3. is an inertial frame because Newton’s laws are applicable in this frame.

4. cannot be an inertial frame because the earth is rotating about its own axis.

Solution:(b), (d) (a) It is incorrect.

 Earth is an accelerated frame, obviously it cannot be an inertial frame.
b) It is correct
c) It is incorrect
Newton’s laws are applicable to it only as a limiting case.
d) It is correct
In fact, a rotating or a revolving frame is accelerating and hence non-inertial. Options (b) and (d) are correct.

Q. 13 Suppose a bike with a rider on it having a total mass of 63 kg brakes and reduces its velocity from 8.5 m/s to 0 m/s in 3.0 second. What is the magnitude of the braking force?

Solution:

The combined mass of the rider and the bike = 63 kg
Initial Velocity = 8.5 m/s
Final Velocity = 0 m/s
The time in which the bike stops = 3 s

The net force acting on the body is equal to the rate of change of momentum of the object.

F=Δp /Δt

The momentum of a body with mass m and velocity v is given by p = mv

Hence, the change in momentum of the car is given by

Δp = mv – mu = m (v−u)

Hence, the net force acting on the car is given by 

F = m (v−u) Δt

Substituting the value, we get

F = 63kg × (0−8.5m/s) 3.0s = 1.8 × 102N

F = 1.8 × 102N

Q. 14 Calculate the net force required to give an automobile of mass 1600 kg an acceleration of 4.5 m/s2

Solution:

We calculate the force using the following formula

F = ma

Substituting the values in the equation, we get

F = 100kg × 4.5m / s2 = 7200N

F = 7200N

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JEE Main Laws of Motion: Quick Formulas

Below mentioned are some of the important formulas based on the topic of Laws of Motion which are necessary for the robust and effective preparation of the topic in JEE Main examination.

• First Law of Motion - The first law of motion implies that things cannot start, stop, or change direction all by themselves. It requires some force from the outside to cause such a change. This property of massive bodies to resist changes in their state of motion is called inertia. Newton’s first law is also known as the law of inertia.

• Second Law of Motion - the second law of motion describes what happens to the massive body when acted upon by an external force. The 2nd law of motion states that the force acting on the body is equal to the product of its mass and acceleration. Mathematically, we express the second law of motion as follows:

F ∝ dP / dt

⇒f ∝ mv−mu / t

⇒f ∝ m(v−u)t

⇒f ∝ ma

⇒f = k ma

• Third Law of Motion - The third law of motion describes what happens to the body when it exerts a force on another body. When two bodies interact with each other, they tend to apply the force on each other and that are also equal in magnitude and are opposite in direction. 

To understand Newton's third law of motion with the help of an example, let us consider a pen resting on a table. The book will now apply a downward force that is equal to its weight on the table. According to the third law of motion, the table applies an equal and opposite force on the book. This force occurs because the book slightly deforms the table. So, as a result, the table pushes back on the book like a coiled spring. Newton’s third law of motion implies the conservation of momentum (i.e., the momentum of a system is constant if no external forces are acting on the system).

Applications of Newton’s laws of motion 

• First Law of Motion - The motion of a ball falling through the atmosphere, or a model rocket being launched up into the atmosphere.

• Second Law of Motion - Riding a bicycle is a good example to this. In this example, the bicycle is the mass. The leg muscles pushing on the pedals of the bicycle is the force.

• Third Law of Motion - When you hit a wall with a certain amount of force and the wall returns that same amount of force is an example of Newton’s 3rd law of motion.

Check JEE Main Study Notes of Laws of Motion