Laws of Motion Previous Year Questions — JEE Main Physics
34 past-year Laws of Motion questions from JEE Main (Physics).
Q16 (2025)
A body of mass $1\mathrm{kg}$ is suspended with the help of two strings making angles as shown in figure. Magnitude of tensions $\mathbf{T}_1$ and $\mathbf{T}_2$ , respectively, are (in N):
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- $5, 5\sqrt{3}$
- $5\sqrt{3}, 5$
- $5\sqrt{3}, 5\sqrt{3}$
- $5, 5$
### Related Formula
For a system in static equilibrium:
$$\sum F_x = 0 \quad \text{and} \quad \sum F_y = 0$$
### Core Logic
Let's resolve the tension forces $\vec{T}_1$ and $\vec{T}_2$ into horizontal and vertical components:
- $T_1$ makes $60^\circ$ with the horizontal.
- $T_2$ makes $30^\circ$ with the horizontal.
- Downward gravitational force: $W = m g = 1 \times 10 = 10 \ \mathrm{N}$.
1. **Horizontal Equilibrium ($\sum F_x = 0$):**
$$T_1 \cos 60^\circ = T_2 \cos 30^\circ$$
$$T_1 \cdot \frac{1}{2} = T_2 \cdot \frac{\sqrt{3}}{2} \implies T_1 = T_2 \sqrt{3}$$
2. **Vertical Equilibrium ($\sum F_y = 0$):**
$$T_1 \sin 60^\circ + T_2 \sin 30^\circ = m g = 10$$
$$T_1 \cdot \frac{\sqrt{3}}{2} + T_2 \cdot \frac{1}{2} = 10$$
### Step 1: Solve for Tensions
Substitute $T_1 = T_2 \sqrt{3}$ into the vertical equilibrium equation:
$$(T_2 \sqrt{3}) \frac{\sqrt{3}}{2} + \frac{T_2}{2} = 10$$
$$\frac{3 T_2}{2} + \frac{T_2}{2} = 10 \implies 2 T_2 = 10 \implies T_2 = 5 \ \mathrm{N}$$
Substitute $T_2$ back to obtain $T_1$:
$$T_1 = 5 \sqrt{3} \ \mathrm{N}$$
Thus, the tension magnitudes are $T_1 = 5\sqrt{3} \ \mathrm{N}$ and $T_2 = 5 \ \mathrm{N}$.
### Pattern Recognition
Sees: Suspending particle static equilibrium with asymmetric strings.
Trap: Associating components with incorrect trigonometry axes or swapping $T_1$ and $T_2$ in options.
Shortcut: Since the incline of $T_1$ ($60^\circ$) is steeper than that of $T_2$ ($30^\circ$), $T_1$ must carry a larger portion of the load, meaning $T_1 > T_2$. From the choices, only (2) satisfies this hierarchy.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Q14 (2025)
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
- $\sqrt{3} - 1$
- $\frac{\sqrt{3}}{2}$
- $\frac{\sqrt{2}}{3}$
- $1 - \frac{\sqrt{3}}{2}$
### Related Formula
For a block sliding down an inclined plane of inclination $\theta$:
$$mg \sin\theta - f_k = ma$$
Where normal reaction is $N = mg \cos\theta$ and kinetic friction is:
$$f_k = \mu_k N = \mu_k mg \cos\theta$$
### Core Logic
Substitute $f_k$ into the equation of motion:
$$mg \sin\theta - \mu_k mg \cos\theta = ma$$
Divide by $m$:
$$g \sin\theta - \mu_k g \cos\theta = a$$
### Step 1: Substitute Values
Given $a = \frac{g}{2}$ and $\theta = 60^\circ$:
$$g \sin 60^\circ - \mu_k g \cos 60^\circ = \frac{g}{2}$$
$$\frac{\sqrt{3}}{2} - \frac{\mu_k}{2} = \frac{1}{2}$$
$$\sqrt{3} - \mu_k = 1 \implies \mu_k = \sqrt{3} - 1$$
### Pattern Recognition
Sees: Block sliding down with acceleration on an incline.
Shortcut: The acceleration on an incline is $a = g(\sin\theta - \mu_k \cos\theta)$. For $\theta = 60^\circ$, this is $a = g\left(\frac{\sqrt{3}}{2} - \frac{\mu_k}{2}\right)$. Equating to $g/2$ gives $\mu_k = sqrt{3} - 1$ directly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Q15 (2025)
A body of mass $2\mathrm{~kg}$ moving with velocity of $\vec{v}_{\mathrm{in}} = 3\hat{i} +4\hat{j}\mathrm{~m/s}$ enters into a constant force field of $6\mathrm{~N}$ 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:
- $4\hat{i} +3\hat{j} +5\hat{k}$
- $3\hat{i} +4\hat{j} +5\hat{k}$
- $3\hat{i} + 4\hat{j} - 5\hat{k}$
- $3\hat{i} + 4\hat{j} + \sqrt{5}\hat{k}$
### Related Formula
$$\vec{F} = m \vec{a}$$
$$\vec{v} = \vec{u} + \vec{a}t$$
where,
$\vec{F}$ = constant force vector
$m$ = mass of body
$\vec{a}$ = acceleration vector
$\vec{u}$ = initial velocity vector
$\vec{v}$ = final velocity vector
### Core Logic
Given parameters:
- Mass, $m = 2\mathrm{~kg}$
- Initial velocity, $\vec{u} = 3\hat{i} + 4\hat{j}\mathrm{~m/s}$
- Force, $\vec{F} = 6\hat{k}\mathrm{~N}$ (directed along positive z-axis)
- Time interval, $t = \frac{5}{3}\mathrm{~s}$
Calculate the acceleration vector $\vec{a}$:
$$\vec{a} = \frac{\vec{F}}{m} = \frac{6\hat{k}}{2} = 3\hat{k}\mathrm{~m/s}^{2}$$
### Step 1: Compute Final Velocity
Using the kinematic equation of motion:
$$\vec{v} = \vec{u} + \vec{a}t$$
$$\vec{v} = (3\hat{i} + 4\hat{j}) + (3\hat{k}) \left(\frac{5}{3}\right)$$
$$\vec{v} = 3\hat{i} + 4\hat{j} + 5\hat{k}\mathrm{~m/s}$$
Thus, the emerging velocity of the body is $3\hat{i} + 4\hat{j} + 5\hat{k}\mathrm{~m/s}$.
### Pattern Recognition
Sees: Orthogonal initial velocity and force field direction.
Shortcut: Since the force acts entirely along the z-axis, the x and y components of the velocity remain unchanged ($3\hat{i} + 4\hat{j}$). Simply compute the z-component change: $v_z = a_z t = \left(\frac{6}{2}\right) \left(\frac{5}{3}\right) = 5$. Result: $3\hat{i} + 4\hat{j} + 5\hat{k}$. ✓
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Class 11 Physics: Kinematics
Q12 (2025)
A sand dropper drops sand of mass $m(t)$ on a conveyer belt at a rate proportional to the square root of speed ($v$) of the belt, i.e. $\frac{dm}{dt} \propto \sqrt{v}$ . If $P$ is the power delivered to run the belt at constant speed then which of the following relationship is true?
- $\mathrm{P}^2\propto \mathrm{v}^3$
- $\mathrm{P} \propto \sqrt{\mathrm{v}}$
- $\mathrm{P} \propto \mathrm{v}$
- $\mathbf{P}^2\propto \mathbf{v}^5$
### Related Formula
$$F_{\text{thrust}} = \left(\frac{dm}{dt}\right)v$$
$P = F \cdot v$
### Core Logic
Given the sand dropping rate rule:
$$\frac{dm}{dt} = C \sqrt{v} \quad (\text{where } C \text{ is a constant})$$
To maintain a constant velocity $v$, the continuous force applied by the conveyor system must balance the rate of gain of momentum of the dropped sand:
$$F = \left(\frac{dm}{dt}\right) v = (C \sqrt{v}) \cdot v = C v^{3/2}$$
Power delivered is the product of force and speed:
$$P = F \cdot v = (C v^{3/2}) \cdot v = C v^{5/2}$$
Squaring both sides of the expression:
$P^2 \propto v^5$
### Pattern Recognition
In variable mass problems involving dropping dust/sand at rest onto a moving frame, the thrust force always simplifies to $v \cdot \frac{dm}{dt}$, which makes power scale as $v^2 \cdot \frac{dm}{dt}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Q6 (2025)
Two blocks of masses m and M, $(\mathbf{M} > \mathbf{m})$ , are placed on a frictionless table as shown in figure. A massless spring with spring constant k is attached with the lower block. If the system is slightly displaced and released then
$(\mu =\text{coefficient of friction between the two blocks})$
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(A) The time period of small oscillation of the two blocks is $\mathrm{T} = 2\pi \sqrt{\frac{(\mathrm{m} + \mathrm{M})}{\mathrm{k}}}$
(B) The acceleration of the blocks is $a = \frac{kx}{M + m}$ (x = displacement of the blocks from the mean position)
(C) The magnitude of the frictional force on the upper block is $\frac{\mathrm{m}k|\mathrm{x}|}{\mathrm{M} + \mathrm{m}}$
(D) The maximum amplitude of the upper block, if it does not slip, is $\frac{\mu(\mathbf{M} + \mathbf{m})\mathbf{g}}{\mathbf{k}}$
(E) Maximum frictional force can be $\mu (\mathbf{M} + \mathbf{m})\mathbf{g}$.
Choose the correct answer from the options given below:
- A, B, D Only
- B, C, D Only
- C, D, E Only
- A, B, C Only
### Related Formula
For combined system performing simple harmonic motion without relative slipping:
$$T = 2\pi \sqrt{\frac{m_{\text{total}}}{k}}$$
$$a = -\omega^2 x = -\frac{k}{M+m} x$$
### Core Logic
Let's analyze each statement:
- **Statement (A)**: Since both blocks perform SHM together, the combined mass is $(M + m)$. The spring constant is $k$. Thus, the time period of small oscillation is:
$$T = 2\pi \sqrt{\frac{M+m}{k}}$$
This is correct. (A is True)
- **Statement (B)**: When the system is displaced by $x$, the restoring spring force on the combined system is $F = -kx$. The common acceleration of the combined mass is:
$$a = -\frac{kx}{M+m} \implies |a| = \frac{k|x|}{M+m}$$
This matches the expression (taking magnitude). (B is True)
- **Statement (C)**: The upper block of mass $m$ moves solely due to the static frictional force $f$ acting on it. Thus:
$$f = m a = m \left( \frac{kx}{M+m} \right) = \frac{mkx}{M+m}$$
Statement (C) claims the frictional force is $\frac{m\mu|x|}{M+m}$, which is incorrect because friction is determined by acceleration, not by coefficient of friction $\mu$ during static grip. (C is False)
- **Statement (D)**: For no slipping to occur, the maximum frictional force required at peak amplitude $A$ must be less than or equal to the limiting static friction $f_L = \mu mg$:
$$f_{\text{max}} = \frac{mkA}{M+m} \le \mu mg$$
$$\frac{kA}{M+m} \le \mu g \implies A \le \frac{\mu(M+m)g}{k}$$
Thus, the maximum amplitude is $\frac{\mu(M+m)g}{k}$. (D is True)
- **Statement (E)**: The maximum static frictional force between the blocks is $f_L = \mu mg$, not $\mu (M+m)g$. (E is False)
### Step 1: Conclusion
Only statements A, B, and D are correct. Hence, the correct option is (1).
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{{SOL_IMG2}}
### Pattern Recognition
In stacked blocks with springs, always identify the force driving the non-spring-loaded block. Here, mass $m$ is driven purely by friction, so $f = m \cdot a$. Slipping begins when this required force exceeds $f_{\text{limit}} = \mu m g$. This simple boundary matches the derivation of maximum amplitude perfectly!
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion: Friction
Class 11 Physics: Oscillations: Simple Harmonic Motion
