Laws of Motion JEE Main PYQs (2020–2026)

Solved Laws of Motion PYQs

3 recent questions with answers. The full set of 113 is free to practise.

1. JEE Main 2025MediumNewton's Laws of Motion, Momentum, and Impulse
   Q1

   A body of mass 2 kg moving with velocity of ${\overrightarrow{v}}_{in}=3\hat{i}+4\hat{j}{\text{ ms}}^{-1}$ enters into a constant force field of 6N directed along positive z-axis. If the body remains in the field for a period of $\frac{5}{3}$ seconds, then velocity of the body when it emerges from force field is.

   • A$3\hat{i}+4\hat{j}+\sqrt{5}\hat{k}$
   • B$4\hat{i}+3\hat{j}+5\hat{k}$
   • C$3\hat{i}+4\hat{j}-5\hat{k}$
   • D$3\hat{i}+4\hat{j}+5\hat{k}$

   Show answer & solution

   Correct answer: (D)

   To determine the velocity of the body as it emerges from the force field, we can break this process down as follows:
   
   Force and Acceleration Calculation:
   
   The body is subjected to a constant force of 6 N, which acts along the positive z-axis. The mass of the body is 2 kg.
   
   Using Newton’s second law, $\overrightarrow{F}=m\overrightarrow{a}$ , the acceleration $\overrightarrow{a}$ can be calculated as:
   
   $$\overrightarrow{a}=\frac{\overrightarrow{F}}{m}=\frac{6\hat{k}}{2}=3\hat{k}{\text{ ms}}^{-2}$$
   
   Initial Velocity:
   
   The initial velocity of the body is given by:
   
   $$\overrightarrow{u}=3\hat{i}+4\hat{j}{\text{ ms}}^{-1}$$
   
   Time of Motion:
   
   The time the body spends in the force field is:
   
   $$t=\frac{5}{3}\text{ seconds}$$
   
   Final Velocity Calculation:
   
   The final velocity $\overrightarrow{v}$ is given by the equation of motion:
   
   $$\overrightarrow{v}=\overrightarrow{u}+\overrightarrow{a}t$$
   
   Substituting the known values:
   
   $$\overrightarrow{v}=(3\hat{i}+4\hat{j})+3\hat{k}\left(\frac{5}{3}\right)$$
   
   Simplifying the expression gives:
   
   $$\overrightarrow{v}=3\hat{i}+4\hat{j}+5\hat{k}$$
   
   Therefore, the velocity of the body as it exits the force field is $3\hat{i}+4\hat{j}+5\hat{k}{\text{ ms}}^{-1}$ .
2. JEE Main 2024EasyEquilibrium of Concurrent Forces and String-Pulley Systems
   Q2

   A body of weight $200N$ is suspended from a tree branch through a chain of mass $10kg$ . The branch pulls the chain by a force equal to (if $g=10m/s^2$ ) :

   • A300 N
   • B100 N
   • C150 N
   • D200 N

   Show answer & solution

   Correct answer: (A)

   To determine the force that the branch pulls on the chain, we need to consider the combined weight of the body and the chain, as this is the total force that the branch must support due to gravity.
   
   The weight of the body is given as $200N$ .
   
   To find the weight of the chain, we use the formula for weight, which is the mass of an object multiplied by the acceleration due to gravity ( $g$ ). The mass of the chain is given as $10kg$ and $g=10{m/s}^2$ , so:
   
   $$\text{Weight of the chain}=\text{Mass of the chain}\times g$$
   
   $$=10kg\times 10{m/s}^2$$
   
   $$=100N$$
   
   The total force that the branch must support is the sum of the weight of the body and the weight of the chain:
   
   $$\text{Total force}=\text{Weight of the body}+\text{Weight of the chain}$$
   
   $$=200N+100N$$
   
   $$=300N$$
   
   Therefore, the branch pulls the chain by a force of $300N$ , which corresponds to Option A.
3. JEE Main 2024EasyFriction on Horizontal Surfaces and Multi-Block Systems
   Q3

   A heavy box of mass $50kg$ is moving on a horizontal surface. If co-efficient of kinetic friction between the box and horizontal surface is 0.3 then force of kinetic friction is :

   • A1470 N
   • B147 N
   • C1.47 N
   • D14.7 N

   Show answer & solution

   Correct answer: (B)

   To find the force of kinetic friction acting on the box, we can use the formula for kinetic friction, which is given by:
   
   $$F_{\text{friction}}={\mu }_k\cdot N$$
   
   where:
   
   $F_{\text{friction}}$ is the force of friction,
   
   ${\mu }_k$ is the coefficient of kinetic friction,
   
   $N$ is the normal force exerted by the surface onto the box.
   
   Since the box is moving on a horizontal surface, the normal force $N$ would be equal to the weight of the box, which is calculated by $g$ , where $m$ is the mass of the box and $g$ is the acceleration due to gravity. For most calculations, $g$ is approximated as $9.8{\text{m/s}}^2$ .
   
   Substituting the given values:
   
   $m=50\text{kg}$ ,
   
   ${\mu }_k=0.3$ ,
   
   $g=9.8{\text{m/s}}^2$ .
   
   First, calculate the weight of the box, which is the normal force:
   
   $$N=mg=50\text{kg}\times 9.8{\text{m/s}}^2=490\text{N}$$
   
   Then, use this value to find the force of kinetic friction:
   
   $$F_{\text{friction}}={\mu }_k\cdot N=0.3\times 490\text{N}=147\text{N}$$
   
   Thus, the force of kinetic friction acting on the box is 147 N. The correct answer is Option B.

Topic-wise question count

Where JEE Main has asked from Laws of Motion, most frequent first.

• Newton's Laws of Motion, Momentum, and Impulse31
• Equilibrium of Concurrent Forces and String-Pulley Systems26
• Dynamics on Inclined Planes and Curved Surfaces24
• Friction on Horizontal Surfaces and Multi-Block Systems21
• Non-Inertial Frames, Pseudo Forces, and Elevators10
• Equilibrium of Concurrent Forces and String-Pulley System1

Year-wise Laws of Motion questions in JEE Main

Number of questions from this chapter in each year's paper(s).