Laws of Motion Previous Year Questions — JEE Main Physics
34 past-year Laws of Motion questions from JEE Main (Physics) — page 3.
Q1368 (2024)
Consider a block and trolley system as shown in figure. If the coefficient of kinetic friction between the trolley and the surface is 0.04, the acceleration of the system in $\mathrm{ms}^{-2}$ is (Consider that the string is massless and unstretchable and the pulley is also massless and frictionless):
{{IMG1}}
- 3
- 4
- 2
- 1.2
### Related Formula
Kinetic friction force:
$$f_k = \mu_k N = \mu_k m_1 g$$
System acceleration:
$$a = \frac{F_{\text{pull}} - f_k}{m_{\text{total}}}$$
### Core Logic
Given values:
Trolley mass $m_1 = 20\mathrm{~kg}$, total system mass component in frame $m_{\text{total}} = 26\mathrm{~kg}$ (from solution fraction $\frac{60-8}{26}$).
Applied pulling force $F = 60\mathrm{~N}$.
$\mu_k = 0.04$.
Calculate the kinetic friction resisting the trolley:
$$f_k = 0.04 \times 20 \times 10 = 8\mathrm{~N}$$
### Step 1: Calculate Acceleration
Using the dynamic equation for connected translation systems:
$$a = \frac{60 - 8}{26} = \frac{52}{26} = 2\mathrm{~ms}^{-2}$$
### Pattern Recognition
Treat connected inline systems as a single collective mass block, balancing external driving forces against collective internal friction resistance.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Q37 (2024)
A ball of mass $0.5\mathrm{~kg}$ is attached to a string of length $50\mathrm{~cm}$. The ball is rotated on a horizontal circular path about its vertical axis. The maximum tension that the string can bear is $400\mathrm{~N}$. The maximum possible value of angular velocity of the ball in $\mathrm{rad/s}$ is:
- 1600
- 40
- 1000
- 20
### Related Formula
Centripetal force configuration for simplified horizontal rotation layout:
$T = m\omega^2 l$
### Core Logic
Given values:
$m = 0.5\mathrm{~kg}$, $l = 50\mathrm{~cm} = 0.5\mathrm{~m}$, $T_{\text{max}} = 400\mathrm{~N}$.
Equating max tension to centripetal requirement:
$$400 = 0.5 \times \omega^2 \times 0.5$$
### Step 1: Compute Angular Velocity
$$400 = 0.25 \omega^2$$
$$\omega^2 = \frac{400}{0.25} = 1600$$
$$\omega = \sqrt{1600} = 40\mathrm{~rad/s}$$
### Pattern Recognition
Ensure units are metric ($50\mathrm{~cm} \to 0.5\mathrm{~m}$). Direct mapping to horizontal string projection metrics.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Q36 (2024)
A stone of mass $900\text{ g}$ is tied to a string and moved in a vertical circle of radius $1\text{ m}$ making $10\text{ rpm}$. The tension in the string, when the stone is at the lowest point is (if $\pi^2 = 9.8$ and $g = 9.8\text{ m/s}^2$):
- $97\text{ N}$
- $9.8\text{ N}$
- $8.82\text{ N}$
- $17.8\text{ N}$
### Related Formula
At the lowest point of a vertical circle, the equation of motion for a mass $m$ is:
$$T - mg = m r \omega^2$$
Rearranging to solve for tension $T$:
$$T = mg + m r \omega^2$$
### Core Logic
Given data:
* Mass, $m = 900\text{ g} = 0.9\text{ kg}$
* Radius, $r = 1\text{ m}$
* Frequency, $N = 10\text{ rpm} = \frac{10}{60}\text{ rps} = \frac{1}{6}\text{ rps}$
* Angular velocity, $\omega = 2\pi N = 2\pi \left(\frac{1}{6}\right) = \frac{\pi}{3}\text{ rad/s}$
### Step 1: Calculate Force Values
Now we substitute our parameters into the tension equation:
$$T = (0.9)(9.8) + (0.9)(1)\left(\frac{\pi}{3}\right)^2$$
$$T = 8.82 + 0.9 \times \frac{\pi^2}{9}$$
Since $\pi^2 = 9.8$:
$$T = 8.82 + 0.1 \times 9.8$$
$$T = 8.82 + 0.98 = 9.80\text{ N}$$
{{SOL_IMG1}}
### Pattern Recognition
Always convert Mass to $\text{kg}$ and rotational speed to $\text{rad/s}$ first. Using the prompt constraint $\pi^2 = 9.8$ yields a perfect decimal addition match.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Q59 (2024)
A particle is moving in a circle of radius $50\text{ cm}$ in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at $t = 0$ is $4\text{ m/s}$, the time taken to complete the first revolution will be $\frac{1}{\alpha}\left[ 1 - e^{-2\pi} \right]\text{ s}$, where $\alpha =$ ________.
### Related Formula
For a particle in circular motion:
* Normal (centripetal) acceleration: $a_c = \frac{v^2}{r}$
* Tangential acceleration: $a_t = \frac{dv}{dt}$
### Core Logic
Given $a_c = a_t$:
$$\frac{v^2}{r} = \frac{dv}{dt}$$
$$\int_{v_0}^{v} \frac{dv}{v^2} = \int_{0}^{t} \frac{dt}{r}$$
$$\left[ -\frac{1}{v} \right]_{v_0}^{v} = \frac{t}{r}$$
$$-\frac{1}{v} + \frac{1}{v_0} = \frac{t}{r} \implies \frac{1}{v} = \frac{1}{v_0} - \frac{t}{r}$$
$$v = \frac{v_0}{1 - \frac{v_0 t}{r}}$$
### Step 1: Relate Velocity to Position and Integrate
Substitute the parameters $v_0 = 4\text{ m/s}$ and $r = 50\text{ cm} = 0.5\text{ m}$:
$$v = \frac{4}{1 - 8t} = \frac{ds}{dt}$$
Integrating this to find the position $s(t)$:
$$\int_{0}^{s} ds = \int_{0}^{t} \frac{4}{1 - 8t} dt$$
$$s = 4 \left[ \frac{\ln(1 - 8t)}{-8} \right]_0^t = -\frac{1}{2} \ln(1 - 8t)$$
### Step 2: Solve for Time of First Revolution
To complete the first revolution, the distance covered is:
$$s = 2\pi r = 2\pi (0.5) = \pi\text{ m}$$
Equating the distance:
$$\pi = -\frac{1}{2} \ln(1 - 8t)$$
$$-2\pi = \ln(1 - 8t)$$
$$1 - 8t = e^{-2\pi}$$
$$8t = 1 - e^{-2\pi} \implies t = \frac{1}{8} \left[ 1 - e^{-2\pi} \right]\text{ s}$$
Comparing this to $\frac{1}{\alpha}\left[ 1 - e^{-2\pi} \right]\text{ s}$, we get:
$\alpha = 8$
### Pattern Recognition
The condition $a_t = a_c \implies v \frac{dv}{ds} = \frac{v^2}{r} \implies \frac{dv}{v} = \frac{ds}{r}$. Integrating directly gives $v = v_0 e^{s/r}$. Substituting this back into $v = ds/dt$ makes the final time integral much more intuitive.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Q36 (2024)
A train is moving with a speed of $12\text{ m/s}$ on rails which are $1.5\text{ m}$ apart. To negotiate a curve of radius $400\text{ m}$, the height by which the outer rail should be raised with respect to the inner rail is (Given, $g = 10\text{ m/s}^{2}$):
- $6.0\text{ cm}$
- $5.4\text{ cm}$
- $4.8\text{ cm}$
- $4.2\text{ cm}$
### Related Formula
$$\tan\theta = \frac{v^2}{Rg}$$
For small angles, $\tan\theta \approx \sin\theta = \frac{h}{d}$, where $h$ is the raised height and $d$ is the separation between the tracks.
### Core Logic
Equating the two relationships:
$$\frac{h}{d} = \frac{v^2}{Rg}$$
$$\frac{h}{1.5} = \frac{12 \times 12}{400 \times 10}$$
### Step 1: Compute height value
$$h = 1.5 \times \frac{144}{4000} = 1.5 \times 0.036 = 0.054\text{ m}$$
Converting to centimeters:
$$h = 0.054 \times 100 = 5.4\text{ cm}$$
### Pattern Recognition
Whenever $\theta$ is small, geometry permits approximating $\tan\theta$ with $\frac{h}{\text{width}}$, vastly reducing computational transcendental overhead.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
