Laws of Motion JEE Main Questions 2027: Practice Questions with Solutions

Weightage of Laws of Motion in JEE Main 2026

Based on the 2026 exam analysis, Laws of Motion contribute 3.79% weightage in the JEE Main Physics paper. In practical terms, 1 question appears in most of the shifts of the exam and around 18 overall in both sessions, worth roughly 4 to 8 marks out of the 100 marks allotted to Physics.

Topic-Wise Weightage and Number of Questions

Not every sub-topic within laws of motion carries equal weightage. On the basis of analysis 2026 JEE Main papers, the topic-wise breakdown of questions is given below:

Also Read: JEE Mains 2027: Physics Set of 5 Sample Paper with Solution

Last 10 Years' Weightage of Laws of Motion in JEE Main

The Laws of Motion chapter question is appearing constantly over the past 10 years has remained the important topic of JEE Main, with questions frequently appearing from friction, circular motion, pulley systems, Newton's laws of motion, and equilibrium concepts.

Last 10 Years' Previous Year Questions with Solutions

Laws of Motion JEE Main Previous Year Questions

Question 1: A car of mass $m$ drives on a banked road having radius ' r ' and banking angle $\theta$. To avoid slipping from the banked road, the maximum permissible speed of the car is $v_0$. The coefficient of friction $\mu$ between the wheels of the car and the banked road is:-

1) $\mu=\frac{\mathrm{v}_0^2+\mathrm{rg} \tan \theta}{\mathrm{rg}-\mathrm{v}_0^2 \tan \theta}$

2) $\mu=\frac{v_0^2+r g \tan \theta}{r g+v_0^2 \tan \theta}$

3) $\mu=\frac{v_0^2-r g \tan \theta}{r g+v_0^2 \tan \theta}$

4) $\mu=\frac{v_0^2-r g \tan \theta}{r g-v_0^2 \tan \theta}$

Correct answer: (3)

Solution:

$\begin{aligned} & N \sin \theta+f \cos \theta=\frac{m v^2}{R} \\ & N \cos \theta-f \sin \theta=m g \\ & \frac{\sin \theta+\mu \cos \theta}{\cos \theta-\mu \sin \theta}=\frac{v^2}{R g} \\ & R g \tan \theta+\mu R g=v^2-v^2 \mu \tan \theta\\ & \mu=\frac{v^2-R g \tan \theta}{R g+v^2 \tan \theta}\end{aligned}$

Question 2: A cubic block of mass $m$ is sliding down on an inclined plane at $60^{\circ}$ with an acceleration of $\frac{g}{2}$, the value of coefficient of kinetic friction is

1) (correct) $\sqrt{3}-1$

2) $\frac{\sqrt{3}}{2}$

3) $\frac{\sqrt{2}}{3}$

4)$1-\frac{\sqrt{3}}{2}$

Correct Answer: (1)

Solution:

$\begin{aligned} & \mathrm{mg} \sin 60^{\circ}-\mu \mathrm{mg} \cos 60^{\circ}=\mathrm{ma} \\ & g \sin 60-\mu g \cos 60=\frac{g}{2} \\ & \frac{\sqrt{3}}{2}-\frac{\mu}{2}=\frac{1}{2} \\ & \mu=\sqrt{3}-1\end{aligned}$

Question 3: A block of mass 2 kg is attached to one end of a massless spring whose other end is fixed at a wall. The spring-mass system moves on a frictionless horizontal table. The spring's natural length is 2 m and spring constant is $200 \mathrm{~N} / \mathrm{m}$. The block is pushed such that the length of the spring becomes 1 m and then released. At distance $\mathrm{x} \mathrm{m}(\mathrm{x}<2)$ from the wall. the speed of the block will be :

1) $10[1-(2-x)]^{3 / 2} \mathrm{~m} / \mathrm{s}$

2) $10\left[1-(2-x)^2\right]^{1 / 2} \mathrm{~m} / \mathrm{s}$

3) $10\left[1-(2-x)^2\right] \mathrm{m} / \mathrm{s}$

4) $10\left[1-(2-x)^2\right]^2 \mathrm{~m} / \mathrm{s}$

Correct Answer: (2)

Solution:

Given,

Natural length of spring $=2 \mathrm{~m}$
Initial compression in spring $\left(\mathrm{x}_{\mathrm{i}}\right)=1 \mathrm{~m}$
Final compression in spring $\left(x_f\right)=(2-x) m$
Using energy conservation

$
\begin{aligned}
& \mathrm{K}_{\mathrm{i}}+\mathrm{U}_{\mathrm{i}}=\mathrm{K}_{\mathrm{f}}+\mathrm{U}_{\mathrm{f}} \\
& 0+\frac{1}{2} \mathrm{Kx}_{\mathrm{i}}^2=\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{Kx}_{\mathrm{f}}^2 \\
& \frac{1}{2} \mathrm{mv}^2=\frac{1}{2} \mathrm{~K}\left(\mathrm{x}_{\mathrm{i}}^2-\mathrm{x}_{\mathrm{f}}^2\right) \\
& \frac{1}{2} \times 2 \times \mathrm{v}^2=\frac{1}{2} \times 200 \times\left(1^2-(2-\mathrm{x})^2\right) \\
& \mathrm{v}^2=100\left[1-(2-\mathrm{x})^2\right] \\
& \mathrm{v}=10\left[1-(2-\mathrm{x})^2\right]^{1 / 2}
\end{aligned}
$

Question 4:

A string of length $L$ is fixed at one end and carries a mass of M at the other end. The mass makes $\left(\frac{3}{\pi}\right)$ rotations per second about the vertical axis passing through the end of the string as shown. The tension in the string is ___ ML.

Correct Answer: 36

Solution:

$\mathrm{T} \cos \theta=\mathrm{mg} \quad \ldots(1)$

$\mathrm{T} \sin \theta=\mathrm{M} \omega^2 \mathrm{R} \quad \ldots(2)$

Using equation (2)

$
\begin{aligned}
& T \sin \theta=M \omega^2(L \sin \theta) \\
& T=M \omega^2 L=M\left(\frac{3}{\pi} \times 2 \pi\right)^2 \mathrm{L}
\end{aligned}
$
$
\mathrm{T}=36 \mathrm{ML}
$
Hence, the answer is 36.

Question 5: An object with mass 500 g moves along x -axis with speed $v=4 \sqrt{x} \mathrm{~m} / \mathrm{s}$. The force acting on the object is:

1) 8N

2) 5N

3) 6N

4) 4N

Correct answer: (4)

Solution:

$\begin{aligned} v & =4 \sqrt{x} \\ a & =\frac{v d v}{d x}=4 \sqrt{x}\left(-4 \cdot \frac{1}{2 \sqrt{x}}\right) \\ & =8 \\ F & =\frac{1}{2} \times 8=4 N\end{aligned}$

Question 6: A body of mass 2 kg moving with velocity of $\overrightarrow{\mathrm{v}}_{\mathrm{in}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}} \mathrm{ms}^{-1}$ 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 the velocity of the body when it emerges from the force field is

1) $4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$

2) $3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$

3) $3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}$

4) $3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\sqrt{5} \hat{\mathrm{k}}$

Correct Answer: (2)

Solution:

$\begin{aligned} & \mathrm{m}=2 \mathrm{~kg} \\ & V_{\text {in }}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}} \mathrm{m} / \mathrm{sec} . \\ & \mathrm{F}=6 \hat{\mathrm{k}} \mathrm{N} \\ & \mathrm{t}=\frac{5}{3} \mathrm{sec}\end{aligned}$

Velocity after t time

$
\begin{aligned}
& \stackrel{\rightharpoonup}{v}=\mathrm{u}+\overrightarrow{\mathrm{a}} \mathrm{t} \\
& V=(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}})+\frac{\mathrm{F}}{\mathrm{~m}} \mathrm{t} \quad\{\mathrm{~F}=\mathrm{m} \hat{\mathrm{a}}\} \\
& V=(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}})+\left(\frac{6}{2}\right) \hat{\mathrm{k}} \times \frac{5}{3} \\
& V=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}
\end{aligned}
$

Question 7: A balloon and its contents having mass M are moving up with an acceleration ' $a$ '. The mass that must be released from the content so that the balloon starts moving up with an acceleration '3a' will be : (Take ' g ' as acceleration due to gravity)

1) $\frac{3 \mathrm{Ma}}{2 \mathrm{a}-\mathrm{g}}$

2) $\frac{3 \mathrm{Ma}}{2 \mathrm{a}+\mathrm{g}}$

3) $\frac{2 \mathrm{Ma}}{3 \mathrm{a}+\mathrm{g}}$

4) $\frac{2 \mathrm{Ma}}{3 \mathrm{a}-\mathrm{g}}$

Correct Answer: (3)

Solution:

l

$\begin{aligned} & \mathrm{F}-\mathrm{mg}=\mathrm{ma} \\ & \mathrm{F}=\mathrm{ma}+\mathrm{mg} \\ & F-(m-x) g=(m-x) 3 a \\ & \text { Put } \mathrm{F} \\ & \mathrm{ma}+\mathrm{mg}-\mathrm{mg}+\mathrm{xg}=3 \mathrm{ma}-3 \mathrm{xa} \\ & x=\frac{2 m a}{g+3 a}\end{aligned}$

Question 8:

A body of mass 1 kg is suspended with the help of two strings making angles as shown in figure. Magnitude of tensions $T_1$ and $T_2$, respectively, are (in N ):

1) $5,5 \sqrt{3}$

2) $5 \sqrt{3}, 5$

3) $5 \sqrt{3}, 5 \sqrt{3}$

4) 5,5

Correct Answer: (2)

Solution:

$
\begin{aligned}
& \overrightarrow{F_x}=0 \\
& T_1 \cos 60=T_2 \cos 30 \\
& \frac{1}{2} T_1=\frac{\sqrt{3}}{2} T_2 \\
& T_1=\sqrt{3} T_2 ---(1) \\
& T_1 \sin 60+T_2 \sin 30=10 \\
& \Rightarrow \sqrt{3} T_2 \times \frac{\sqrt{3}}{2}+T_2 \times \frac{1}{2}=10 \\
& \Rightarrow \frac{3 T_2}{2}+\frac{T_2}{2}=10 \\
& \Rightarrow \frac{4 T_2}{2}=10 \\
& \Rightarrow T_2=\frac{10}{2}N \\
& T_2=5 \mathrm{~N}
\end{aligned}
$

from equation (1)

$
\begin{aligned}
& T_1=\sqrt{3} \times 5 \\
& T_1=5 \sqrt{3}
\end{aligned}
$

Question 9: A body of mass $m$ is suspended by two strings making angles $\theta_1$ and $\theta_2$ with the horizontal ceiling with tensions $\mathrm{T}_1$ and $\mathrm{T}_2$ simultaneously. $\mathrm{T}_1$ and $\mathrm{T}_2$ are related by $T_1=\sqrt{3} T_2$. the angles $\theta_1$ and $\theta_2$ are

1) $\theta_1=30^{\circ}, \theta_2=60^{\circ}$ with $\mathrm{T}_2=\frac{3 \mathrm{mg}}{4}$

2) $\theta_1=30^{\circ}, \theta_2=60^{\circ}$ with $\mathrm{T}_2=\frac{3 \mathrm{mg}}{4}$

3) $\theta_1=45^{\circ}, \theta_2=45^{\circ}$ with $\mathrm{T}_2=\frac{3 \mathrm{mg}}{4}$

4) $\theta_1=30^{\circ}, \theta_2=60^{\circ}$ with $\mathrm{T}_2=\frac{4 \mathrm{mg}}{5}$

Correct Answer: (2)

Solution:

$\mathrm{T}_1 \sin \theta_1+\mathrm{T}_2 \sin \theta_2=\mathrm{mg}$

$\mathrm{~T}_1=\sqrt{3} \mathrm{~T}_2 $

$ \Rightarrow \mathrm{~T}_2\left[\sqrt{3} \sin \theta_1+\sin \theta_2\right]=\mathrm{mg} $

$ \text { for } \theta_1=60^{\circ} \& \theta_2=30^{\circ} $

$ \mathrm{T}_2=\frac{\mathrm{mg}}{2}$

Question 10: Three blocks A, B, C of masses $3 \mathrm{~kg}, 4 \mathrm{~kg}, 7 \mathrm{~kg}$ are stacked as shown, with the ground under $C$ frictionless. The coefficient of friction is $\mu_1=0.4$ and $\mu_2=0.2$. A horizontal force $F$ is applied to block A. Taking $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$, find the maximum F for which all three blocks move together (no slipping at either contact).

1) 12N

2) 14N

3) 7N

4) 21N

Correct answer: (2)

Solution:

For all the blocks to be moving together

$a_{\text {common }}=\frac{\text { Net external force }}{\text { mass }}=\frac{F}{14}$

$\begin{gathered}
\left(f_1\right)_L=\mu_1 N_1=0.4 \times 30=12 \mathrm{~N} \\
\left(a_1\right)_{\max }=4 \mathrm{~m} / \mathrm{s}^2
\end{gathered}$

For block 2, $\left(a_2\right)_{\text {max }}=1.25 \mathrm{~m} / \mathrm{s}^2$
For block $3,\left(a_3\right)_{\text {max }}=1 \mathrm{~m} / \mathrm{s}^2$

$a_{{ common }}=\frac{F}{14} \leqslant 1 \mathrm{~m} / \mathrm{s}^2 \quad F=14 \mathrm{~N}$

Question 11: A massless spring gets elongated by an amount $\mathrm{x}_1$ under a tension of 5 N. Its elongation is $\mathrm{x}_2$ under the tension of 7 N. For the elongation of $\left(5 x_1-2 x_2\right)$, the tension in the spring will be,

1) 15N

2) 20N

3) 11N

4) 39N

Correct Answer: (3)

Solution:

$\begin{aligned} & \mathrm{kx}_1=5 \mathrm{~N} \\ & \mathrm{kx}_2=7 \mathrm{~N} \\ & \mathrm{k}\left(5 \mathrm{x}_1-2 \mathrm{x}_2\right)=5 \mathrm{kx}_1-2 \mathrm{kx}_2 \\ & =5 \times 5-2 \times 7=11 \mathrm{~N}\end{aligned}$

Question 12: A vehicle of mass 200 kg is moving along a levelled curved road of radius 70 m with an angular velocity of 0.2 rad/s. The centripetal force acting on the vehicle is:

1) 2800 N

2) (correct) 560 N

3) 2240 N

4) 14 N

Correct Answer: (2)

Solution:

$\begin{aligned} & F_c=m \omega^2 r=200 \times(0.2)^2 \times 70 \\ & =560 \mathrm{~N}\end{aligned}$

Question 13: An average force of 125 N is applied on a machine gun firing bullets each of mass 10 g at the speed of 250 m/s to keep it in position. The number of bullets fired per second by the machine gun is :

1) 25

2) 5

3)100

4) (correct) 50

Correct Answer: (4)

Solution:

$\begin{aligned} & \mathrm{F}=125 \mathrm{~N} \\ & \mathrm{~F}=\frac{\mathrm{dp}}{\mathrm{dt}} \quad \mathrm{n} \rightarrow \text { No. of bullets } \\ & \mathrm{F}=\frac{\mathrm{d}(\mathrm{nmv})}{\mathrm{dt}}=\mathrm{mv} \frac{\mathrm{dn}}{\mathrm{dt}} \\ & 125=\frac{10 \mathrm{n}}{1000} \times 250 \times \frac{\mathrm{dn}}{\mathrm{dt}} \\ & \frac{125 \times 1000}{2500}=\frac{\mathrm{dn}}{\mathrm{dt}} \\ & \frac{\mathrm{dn}}{\mathrm{dt}}=50\end{aligned}$

Question 14: The position vector of a particle related to time t is given by

$\overrightarrow{\mathrm{r}}=\left(10 \mathrm{t} \hat{\mathrm{i}}+15 \mathrm{t}^2 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}\right) \mathrm{m}$

The direction of the net force experienced by the particle is:

1) Positive z-axis

2) In x – y plane

3) (correct) Positive y–axis

4) Positive x–axis

Correct Answer: (3)

Solution:

Given, $\overrightarrow{\mathrm{r}}=10 \hat{t} \hat{i}+15 \mathrm{t}^2 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$

$\begin{aligned}
& \overrightarrow{\mathrm{v}}=\frac{\mathrm{dr}}{\mathrm{dt}}=10 \hat{\mathrm{j}}+30 \mathrm{t} \hat{\mathrm{j}} \\
& \overrightarrow{\mathrm{a}}=\frac{\mathrm{d} \overrightarrow{\mathrm{v}}}{\mathrm{dt}}=30 \hat{\mathrm{j}} \\
& \overrightarrow{\mathrm{~F}}=\mathrm{ma} \rightarrow \text { along }(+) y \text {-axis }
\end{aligned}$

Question 15: A machine gun of mass 10 kg fires 20 g bullets at the rate of 180 bullets per minute with a speed of $100 \mathrm{~m} \mathrm{~s}^{-1}$ each. The recoil velocity of the gun is

1) $1.5 \mathrm{~m} / \mathrm{s}$

2) $0.6 \mathrm{~m} / \mathrm{s}$

3) $2.5 \mathrm{~m} / \mathrm{s}$

4) $0.02 \mathrm{~m} / \mathrm{s}$

Correct Answer: (2)

Solution:

$\begin{aligned} & 20 \times 10^{-3} \times \frac{180}{60} \times 100=10 \mathrm{~V} \\ & \mathrm{~V}=0.6 \mathrm{~ms}^{-1}\end{aligned}$