Laws of Motion Previous Year Questions — JEE Main Physics
34 past-year Laws of Motion questions from JEE Main (Physics) — page 6.
Q31 (2026)
A 4 kg mass moves under the influence of a force $\vec{\mathrm{F}} = (4\mathrm{t}^3\hat{\mathrm{i}} - 3\mathrm{t}\hat{\mathrm{j}})\text{ N}$ where t is the time in second. If mass starts from origin at $t = 0$, the velocity and position after $t = 2\text{s}$ will be:
- $\vec{\mathrm{v}} = 3\hat{\mathrm{i}} +\frac{3}{2}\hat{\mathrm{j}}, \quad \vec{\mathrm{r}} = \frac{6}{5}\hat{\mathrm{i}} +\hat{\mathrm{j}}$
- $\vec{\mathrm{v}} = 4\hat{\mathrm{i}} -\frac{3}{2}\hat{\mathrm{j}}, \quad \vec{\mathrm{r}} = \frac{8}{5}\hat{\mathrm{i}} -\hat{\mathrm{j}}$
- $\vec{\mathrm{v}} = 4\hat{\mathrm{i}} +\frac{5}{2}\hat{\mathrm{j}}, \quad \vec{\mathrm{r}} = \frac{8}{5}\hat{\mathrm{i}} +2\hat{\mathrm{j}}$
- $\vec{\mathrm{v}} = 4\hat{\mathrm{i}} -\frac{3}{2}\hat{\mathrm{j}}, \quad \vec{\mathrm{r}} = \frac{6}{5}\hat{\mathrm{i}} -\hat{\mathrm{j}}$
### Related Formula
$$\vec{a} = \frac{\vec{F}}{m}$$
$$\vec{v} = \int \vec{a} \, dt$$
$$\vec{r} = \int \vec{v} \, dt$$
### Core Logic
Given $\vec{F} = 4t^3\hat{i} - 3t\hat{j}$ and mass $m = 4\text{ kg}$.
Acceleration:
$$\vec{a} = \frac{\vec{F}}{m} = \frac{4t^3\hat{i} - 3t\hat{j}}{4} = t^3\hat{i} - \frac{3}{4}t\hat{j}$$
### Step 1: Calculating Velocity
Velocity is the integral of acceleration with $v(0) = 0$:
$$\vec{v} = \int \left( t^3\hat{i} - \frac{3}{4}t\hat{j} \right) dt = \frac{t^4}{4}\hat{i} - \frac{3}{8}t^2\hat{j}$$
At $t = 2\text{ s}$:
$$\vec{v}(2) = \frac{(2)^4}{4}\hat{i} - \frac{3}{8}(2)^2\hat{j} = 4\hat{i} - \frac{3}{2}\hat{j}$$
### Step 2: Calculating Position
Position is the integral of velocity with $\vec{r}(0) = \vec{0}$:
$$\vec{r} = \int \vec{v} \, dt = \int \left( \frac{t^4}{4}\hat{i} - \frac{3}{8}t^2\hat{j} \right) dt$$
$$\vec{r} = \frac{t^5}{20}\hat{i} - \frac{3}{24}t^3\hat{j} = \frac{t^5}{20}\hat{i} - \frac{t^3}{8}\hat{j}$$
At $t = 2\text{ s}$:
$$\vec{r}(2) = \frac{(2)^5}{20}\hat{i} - \frac{(2)^3}{8}\hat{j} = \frac{32}{20}\hat{i} - \frac{8}{8}\hat{j} = \frac{8}{5}\hat{i} - \hat{j}$$
### Pattern Recognition
Recognize $F \to a \to v \to r$ means consecutive integrations. Because the initial state is from rest at origin, we don't have to worry about constants of integration.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Class 11 Physics: Kinematics
Q43 (2026)
A large drum having radius R is spinning around its axis with angular velocity $\omega$, as shown in figure. The minimum value of $\omega$ so that a body of mass M remains stuck to the inner wall of the drum, taking the coefficient of friction between the drum surface and mass M is $\mu$, is :
{{IMG1}}
- $\sqrt{\frac{\mu g}{R}}$
- $\sqrt{\frac{2g}{\mu R}}$
- $\sqrt{\frac{g}{2\mu R}}$
- $\sqrt{\frac{g}{\mu R}}$
### Related Formula
$N = M\omega^2 R$
$$f_{\text{max}} = \mu N$$
$f = Mg$
### Core Logic
{{SOL_IMG1}}
For the mass $M$ to remain stuck to the inner wall and not slide down, the upward frictional force must balance the downward gravitational force.
$f = Mg$
The normal force $N$ providing the friction is generated by the centrifugal effect (or provides the centripetal acceleration):
$N = M\omega^2 R$
The condition for no sliding is that required friction cannot exceed the maximum static friction:
$Mg \le \mu N$
### Step 1: Final Conclusion
Substitute $N$ into the inequality:
$$Mg \le \mu (M\omega^2 R)$$
$$g \le \mu \omega^2 R$$
$$\omega^2 \ge \frac{g}{\mu R}$$
For minimum angular velocity:
$$\omega_{\text{min}} = \sqrt{\frac{g}{\mu R}}$$
### Pattern Recognition
In a "rotor" ride setup, gravity is opposed by friction, and the normal force is solely centripetal. Equating $\mu F_c$ to $mg$ always yields the minimum spin limit.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Q28 (2026)
A block is sliding down on an inclined plane of slope $\theta$ and at an instant $t = 0$ this block is given an upward momentum so that it starts moving up on the inclined surface with velocity $u$ . The distance (S) travelled by the block before its velocity become zero, is ____. (g = gravitational acceleration)
- $\frac{u^2}{4g\sin\theta}$
- $\frac{2u^2}{g\cos\theta}$
- $\frac{u^2}{\sqrt{2}g\cos\theta}$
- $\frac{u^2}{2g\sin\theta}$
### Related Formula
$v^2 = u^2 + 2as$
### Core Logic
{{SOL_IMG1}}
Since the surface is frictionless (implied as no friction coefficient is given), the only force along the incline opposing the upward motion is the component of gravity, $mg \sin \theta$.
Acceleration $a = -g \sin \theta$.
### Step 1: Apply Kinematic Equation
Using the third equation of motion:
$V^2 = U^2 + 2as$
Set final velocity $V = 0$:
$$0 = u^2 - 2(g \sin \theta) S$$
$$S = \frac{u^2}{2g \sin \theta}$$
### Pattern Recognition
For upward motion on a smooth incline, deceleration is purely $g \sin \theta$. The stopping distance is always $u^2 / 2a$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Class 11 Physics: Kinematics
Q36 (2026)
A spring of force constant 15 N/m is cut into two pieces. If the ratio of their length is 1:3, then the force constant of smaller piece is ____ N/m
- $15$
- $20$
- $60$
- $45$
### Related Formula
$$K \ell = \text{constant}$$
$$K \propto \frac{1}{\ell}$$
### Core Logic
{{SOL_IMG1}}
The spring of total length $\ell$ is cut into two parts with a length ratio of 1:3.
The lengths of the two pieces are:
$\ell_1 = \frac{1}{4} \ell$ (smaller piece)
$\ell_2 = \frac{3}{4} \ell$ (larger piece)
### Step 1: Calculate Spring Constant of Smaller Piece
Since $K \ell = K' \ell'$, for the smaller piece:
$$K \ell = K' \left(\frac{\ell}{4}\right)$$
$K' = 4K$
Given $K = 15 \text{ N/m}$:
$$K' = 4 \times 15 = 60 \text{ N/m}$$
### Pattern Recognition
Spring constant is inversely proportional to spring length. Cutting a spring to $1/n$ of its length increases its stiffness by a factor of $n$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Q49 (2026)
In the given figure the blocks A, B and C weigh 4 kg, 6 kg and 8 kg respectively. The co-efficient of sliding friction between any two surfaces is 0.5. The force $\vec{\mathrm{F}}$ required to slide the block C with constant speed is ____ N. (Used $g = 10 \text{ m/s}^{2}$)
{{IMG1}}
### Related Formula
$f_k = \mu_k N$
$$F_{\text{net}} = 0 \quad (\text{for constant speed})$$
### Core Logic
{{SOL_IMG1}}
Since C moves with constant speed, $a=0$, meaning net force is zero.
For block A ($4\text{ kg}$) over B, sliding friction $f_A = \mu (m_A)g = 0.5 \times 4 \times 10 = 20\text{ N}$.
For block B ($6\text{ kg}$) over C, the normal force includes block A. $N_B = (m_A + m_B)g = 100\text{ N}$.
Friction on B from C is $f_B = 0.5 \times 100 = 50\text{ N}$.
For block C ($8\text{ kg}$) over ground, normal force is total weight $N_C = (m_A + m_B + m_C)g = 180\text{ N}$.
Friction on C from ground is $f_C = 0.5 \times 180 = 90\text{ N}$.
### Step 1: Force Balance Equations
For block B to move relative to A and C (assuming standard string configuration where pulling C left causes B to be dragged right relative to C via pulley):
The tension $T$ on B balances friction from A and C:
$$T = f_A + f_B = 20 + 50 = 70\text{ N}$$
For block C moving left, the force $F$ must overcome friction from ground, friction from B, and the tension $T$ attached to it (depending on pulley setup. If pulley is on wall, string pulls C right):
Assuming standard setup where string connects B and C around wall pulley:
$$F = f_C + f_B + T$$
$$F = 90 + 50 + 70 = 210\text{ N}$$
### Pattern Recognition
In multi-block pulley systems, pulling the bottom block accumulates all nested friction forces. The bottom block fights the ground, the block above it, AND the tension of the string anchoring the block above it.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
