Laws of Motion Previous Year Questions — JEE Main Physics
34 past-year Laws of Motion questions from JEE Main (Physics) — page 5.
Q32 (2024)
All surfaces shown in figure are assumed to be frictionless and the pulleys and the string are light. The acceleration of the block of mass $2 \mathrm{~kg}$ is:
{{IMG1}}
- $g$
- $\frac{g}{3}$
- $\frac{g}{2}$
- $\frac{g}{4}$
### Related Formula
$$\sum F = ma$$
$$a_1 = \text{Constraint relation}\times a_2$$
### Core Logic
{{SOL_IMG1}}
Let the tension in the string attached to the $2 \mathrm{~kg}$ block be $T$. By tracing the string around the movable pulley, the tension supporting the $4 \mathrm{~kg}$ mass becomes $2T$.
From the principle of virtual work (or string constraints), if the $4 \mathrm{~kg}$ block moves down with an acceleration $a$, the string shortens by $2x$ on the incline side, meaning the $2 \mathrm{~kg}$ block moves up the incline with an acceleration of $2a$.
### Step 1: Write Equations of Motion
For the $4 \mathrm{~kg}$ block (moving downwards):
$4g - 2T = 4a$
$$40 - 2T = 4a \quad \dots (i)$$
For the $2 \mathrm{~kg}$ block (moving up the incline):
$$T - 2g \sin(30^\circ) = 2(2a)$$
$$T - 2(10)\left(\frac{1}{2}\right) = 4a$$
$$T - 10 = 4a \quad \dots (ii)$$
### Step 2: Solve for Acceleration
From equation (ii), we have $4a = T - 10$. Substitute this directly into equation (i) or add the appropriately multiplied equations:
$$40 - 2T = T - 10$$
$$3T = 50 \Rightarrow T = \frac{50}{3} \mathrm{~N}$$
Substitute $T$ back into (ii):
$$\frac{50}{3} - 10 = 4a$$
$$4a = \frac{20}{3} \Rightarrow a = \frac{5}{3} \mathrm{~m/s^2}$$
We need the acceleration of the $2 \mathrm{~kg}$ block, which is $2a$:
$$2a = 2 \times \left(\frac{5}{3}\right) = \frac{10}{3} \mathrm{~m/s^2}$$
Since $g = 10 \mathrm{~m/s^2}$, this acceleration is exactly $\frac{g}{3}$.
### Pattern Recognition
Movable pulleys double the force but halve the displacement/acceleration on the supported side. Tension on the movable pulley side is twice the tension on the single string side. Remember to explicitly solve for the requested block's specific acceleration, not just 'a'.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Q31 (2024)
A light string passing over a smooth light fixed pulley connects two blocks of masses $m_1$ and $m_2$. If the acceleration of the system is $g/8$, then the ratio of masses is
{{IMG1}}
- $\frac{9}{7}$
- $\frac{8}{1}$
- $\frac{4}{3}$
- $\frac{5}{3}$
### Related Formula
$$a = \frac{(m_1 - m_2)g}{(m_1 + m_2)}$$
### Core Logic
Assuming $m_1 > m_2$, the net pulling force is $(m_1 - m_2)g$ and the total mass to be accelerated is $(m_1 + m_2)$.
Given that the acceleration of the system is $a = \frac{g}{8}$.
### Step 1: Algebraic Manipulation
$$\frac{g}{8} = \frac{(m_1 - m_2)g}{(m_1 + m_2)}$$
$$m_1 + m_2 = 8m_1 - 8m_2$$
$$8m_2 + m_2 = 8m_1 - m_1$$
$9m_2 = 7m_1$
$$\frac{m_1}{m_2} = \frac{9}{7}$$
### Pattern Recognition
For standard Atwood machines, $a = g \times \frac{\text{Difference in mass}}{\text{Sum of mass}}$. If $a/g = 1/8$, then $(m_1-m_2)/(m_1+m_2) = 1/8$, which can be solved using componendo and dividendo: $m_1/m_2 = (8+1)/(8-1) = 9/7$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Q42 (2024)
A block of mass $5 \text{ kg}$ is placed on a rough inclined surface as shown in the figure.
{{IMG1}}
If $\vec{F}_1$ is the force required to just move the block up the inclined plane and $\vec{F}_2$ is the force required to just prevent the block from sliding down, then the value of $|\vec{F}_1| - |\vec{F}_2|$ is: [Use $g = 10 \text{ m/s}^2$]
- $25\sqrt{3}\text{ N}$
- $50\sqrt{3}\text{ N}$
- $\frac{5 \sqrt{3}}{2} \text{ N}$
- $10 \text{ N}$
### Related Formula
$$f_k = \mu mg \cos \theta$$
$$F_1 = mg \sin \theta + f_k$$
$$F_2 = mg \sin \theta - f_k$$
### Core Logic
To move the block up, the applied force $F_1$ must overcome both the downward gravitational component and the downward frictional force.
To prevent it from sliding down, the applied force $F_2$ acts upwards and is aided by friction which acts upwards to oppose impending downward slip.
{{SOL_IMG1}}
{{SOL_IMG2}}
### Step 1: Calculate Friction
$$f_k = \mu mg \cos \theta$$
$$f_k = 0.1 \times 5 \times 10 \times \cos(30^{\circ})$$
$$f_k = 5 \times \frac{\sqrt{3}}{2} = 2.5\sqrt{3} \text{ N}$$
### Step 2: Force Equations
Moving up:
$$F_1 = mg \sin \theta + f_k = 50 \sin(30^{\circ}) + 2.5\sqrt{3} = 25 + 2.5\sqrt{3}$$
Preventing slip down:
$$F_2 = mg \sin \theta - f_k = 50 \sin(30^{\circ}) - 2.5\sqrt{3} = 25 - 2.5\sqrt{3}$$
### Step 3: Difference calculation
$$|F_1| - |F_2| = (25 + 2.5\sqrt{3}) - (25 - 2.5\sqrt{3})$$
$$|F_1| - |F_2| = 2 \times 2.5\sqrt{3} = 5\sqrt{3} \text{ N}$$
*Note: The official options had an anomaly where $5\sqrt{3} \text{ N}$ was missing or evaluated as a bonus. Option 2 was listed as $50\sqrt{3}$ in the primary text. We track the closest logic path indicating Bonus.*
### Pattern Recognition
The difference between 'push up' and 'hold from sliding' forces on an incline is always precisely $2 f_k$ ($2 \mu mg \cos \theta$). Bypass calculating the $mg \sin \theta$ terms entirely.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Q1993 (2024)
In the given arrangement of a doubly inclined plane two blocks of masses $M$ and $m$ are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is $0.25$. The value of $m$, for which $M = 10\mathrm{\ kg}$ will move down with an acceleration of $2\mathrm{\ m/s^2}$ is : (take $g = 10\mathrm{\ m/s^2}$ and $\tan 37^{\circ} = 3 / 4$)
{{IMG1}}
- $9\mathrm{\ kg}$
- $4.5\mathrm{\ kg}$
- $6.5\mathrm{\ kg}$
- $2.25\mathrm{\ kg}$
### Related Formula
$$\sum F = ma$$
$$f_k = \mu_k N = \mu_k mg \cos\theta$$
### Core Logic
{{SOL_IMG1}}
Since block $M$ moves down the incline, kinetic friction opposes its motion (acts upwards).
Block $m$ is pulled up the incline, so kinetic friction opposes its motion (acts downwards).
For $M$ block ($53^\circ$ slope):
$$Mg \sin 53^{\circ} - \mu Mg \cos 53^{\circ} - T = Ma$$
### Step 1: Tension Calculation
Given $M = 10\mathrm{\,kg}$, $a = 2\mathrm{\,m/s^2}$, $\mu = 0.25$, $g = 10\mathrm{\,m/s^2}$.
$\sin 53^\circ = 4/5 = 0.8$, $\cos 53^\circ = 3/5 = 0.6$.
$$10(10)(0.8) - 0.25(10)(10)(0.6) - T = 10(2)$$
$$80 - 15 - T = 20$$
$$65 - T = 20 \Rightarrow T = 45\mathrm{\,N}$$
### Step 2: Evaluate mass m
For $m$ block ($37^\circ$ slope) moving upward:
$$T - mg \sin 37^{\circ} - \mu mg \cos 37^{\circ} = ma$$
$\sin 37^\circ = 3/5 = 0.6$, $\cos 37^\circ = 4/5 = 0.8$.
$$45 - m(10)(0.6) - 0.25(m)(10)(0.8) = m(2)$$
$$45 - 6m - 2m = 2m$$
$45 - 8m = 2m$
$$10m = 45 \Rightarrow m = 4.5\mathrm{\,kg}$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws Of Motion
Q1997 (2024)
A coin is placed on a disc. The coefficient of friction between the coin and the disc is $\mu$. If the distance of the coin from the center of the disc is $r$, the maximum angular velocity which can be given to the disc, so that the coin does not slip away, is:
- $\frac{\mu g}{r}$
- $\sqrt{\frac{r}{\mu g}}$
- $\sqrt{\frac{\mu g}{r}}$
- $\frac{\mu}{\sqrt{rg}}$
### Related Formula
$$f_s \leq \mu_s N$$
$$F_c = mr\omega^2$$
### Core Logic
{{SOL_IMG1}}
{{SOL_IMG2}}
To prevent the coin from slipping, the static friction must provide the necessary centripetal force for circular motion.
$f = m\omega^2 r$
The normal force on the flat disc is $N = mg$.
The maximum static friction is $f_{\text{max}} = \mu N = \mu mg$.
For no slipping:
$$m r \omega^2 \leq \mu mg$$
$$\omega^2 \leq \frac{\mu g}{r}$$
$$\omega_{\text{max}} = \sqrt{\frac{\mu g}{r}}$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws Of Motion
