System of Particles and Rotational Motion Previous Year Questions — JEE Main Physics
58 past-year System of Particles and Rotational Motion questions from JEE Main (Physics) — page 8.
Q55 (2024)
A body of mass 'm' is projected with a speed 'u' making an angle of $45^{\circ}$ with the ground. The angular momentum of the body about the point of projection, at the highest point is expressed as $\frac{\sqrt{2} m u^{3}}{X g}$. The value of 'X' is ________.
### Related Formula
$$H = \frac{u^2 \sin^2 \theta}{2g}$$
$$L = m v_x H_{max}$$ (for highest point)
### Core Logic
At the highest point in a projectile's trajectory, the velocity is entirely horizontal ($v_x = u \cos \theta$). The perpendicular distance from the point of projection to the line of motion is the maximum height $H$.
### Step 1: Calculate Velocity and Height
Horizontal velocity: $v_x = u \cos \theta$
Maximum height: $H = \frac{u^2 \sin^2 \theta}{2g}$
### Step 2: Angular Momentum Calculation
$$L = m \times (u \cos \theta) \times \left(\frac{u^2 \sin^2 \theta}{2g}\right)$$
For $\theta = 45^{\circ}$:
$\cos(45^{\circ}) = 1/\sqrt{2}$
$\sin^2(45^{\circ}) = (1/\sqrt{2})^2 = 1/2$
$$L = m \times \left(\frac{u}{\sqrt{2}}\right) \times \left(\frac{u^2 (1/2)}{2g}\right)$$
$$L = m \times \frac{u}{\sqrt{2}} \times \frac{u^2}{4g} = \frac{m u^3}{4\sqrt{2} g}$$
### Step 3: Match Format
We need the format to be $\frac{\sqrt{2} m u^3}{X g}$. Multiply numerator and denominator by $\sqrt{2}$:
$$L = \frac{\sqrt{2} m u^3}{4 \times 2 \times g} = \frac{\sqrt{2} m u^3}{8g}$$
Thus, $X = 8$.
### Pattern Recognition
Angular momentum at the apex is always $m(u \cos\theta)(H_{max})$. Be careful with algebraic rationalization at the end to match the given exact format.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Class 11 Physics: Motion in a Plane
Q58 (2024)
Two identical spheres each of mass $2 \text{ kg}$ and radius $50 \text{ cm}$ are fixed at the ends of a light rod so that the separation between the centers is $150 \text{ cm}$. Then, moment of inertia of the system about an axis perpendicular to the rod and passing through its middle point is $\frac{x}{20} \text{ kg m}^2$, where the value of $x$ is
### Related Formula
$$I = I_{cm} + Md^2$$
For a solid sphere, $I_{cm} = \frac{2}{5}MR^2$.
### Core Logic
Apply the Parallel Axis Theorem for each sphere. The total moment of inertia is twice the moment of inertia of a single sphere shifted from its center to the middle of the rod.
{{SOL_IMG1}}
### Step 1: Establish Parameters
Mass $M = 2 \text{ kg}$
Radius $R = 50 \text{ cm} = 0.5 \text{ m} = \frac{1}{2} \text{ m}$
Distance between centers $= 150 \text{ cm} = 1.5 \text{ m}$.
Distance from axis of rotation to the center of each sphere is $d = \frac{1.5}{2} = 0.75 \text{ m} = \frac{3}{4} \text{ m}$.
### Step 2: Calculate Inertia
$$I_{total} = 2 \times \left[ \frac{2}{5}MR^2 + Md^2 \right]$$
$$I_{total} = 2 \times \left[ \frac{2}{5}(2)\left(\frac{1}{2}\right)^2 + (2)\left(\frac{3}{4}\right)^2 \right]$$
$$I_{total} = 2 \times \left[ \frac{4}{5}\left(\frac{1}{4}\right) + 2\left(\frac{9}{16}\right) \right]$$
$$I_{total} = 2 \times \left[ \frac{1}{5} + \frac{9}{8} \right]$$
$$I_{total} = 2 \times \left[ \frac{8 + 45}{40} \right]$$
$$I_{total} = 2 \times \frac{53}{40} = \frac{53}{20} \text{ kg m}^2$$
### Step 3: Extract x
Comparing with $\frac{x}{20} \text{ kg m}^2$, we get:
$x = 53$
### Pattern Recognition
Watch out for units. Convert cm to meters immediately. Use fractions ($1/2$ and $3/4$) instead of decimals to rapidly solve the resulting squared terms without arithmetic errors.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q54 (2024)
A solid circular disc of mass $50 \mathrm{~kg}$ rolls along a horizontal floor so that its center of mass has a speed of $0.4 \mathrm{~m} / \mathrm{s}$. The absolute value of work done on the disc to stop it is ______ $\mathrm{J}$.
### Related Formula
$$W = \Delta \text{KE}$$
$$K_{\text{rolling}} = \frac{1}{2} m v^2 \left(1 + \frac{k^2}{R^2}\right)$$
### Core Logic
According to the Work-Energy Theorem, the total work done is equal to the change in kinetic energy.
Since the disc comes to rest, final kinetic energy is zero.
$$W = 0 - \left(\frac{1}{2} m v^2 + \frac{1}{2} I \omega^2\right)$$
### Step 2: Calculation
For a solid circular disc, $I = \frac{1}{2}mR^2$, thus $\frac{k^2}{R^2} = \frac{1}{2}$.
$$W = - \frac{1}{2} m v^2 \left(1 + \frac{k^2}{R^2}\right)$$
$$W = - \frac{1}{2} \times 50 \times (0.4)^2 \left(1 + \frac{1}{2}\right)$$
$$W = - 25 \times 0.16 \times 1.5$$
$$W = - 6\mathrm{\,J}$$
The absolute value of work done $|W| = 6\mathrm{\,J}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System Of Particles And Rotational Motion
Class 11 Physics: Work, Energy And Power
Q2084 (2026)
A uniform rod of mass m and length l suspended by means of two identical inextensible light strings as shown in figure. Tension in one string immediately after the other string is cut, is ____. (g acceleration due to gravity)
{{IMG1}}
- $mg/2$
- $mg/4$
- $mg/3$
- $mg$
### Related Formula
$\tau = I\alpha$
$$\Sigma F_y = m a_{CM, y}$$
$$a_{CM, y} = \alpha \frac{l}{2}$$
### Core Logic
Immediately after one string is cut, the rod starts rotating about the point where the remaining string is attached.
Taking torque about the end where the string is attached (this point has instantaneous acceleration but initially zero vertical velocity):
$$\tau_{\text{end}} = I_{\text{end}} \alpha$$
Gravity provides the torque: $\tau = mg \left(\frac{l}{2}\right)$.
### Step 1: Calculate Angular Acceleration
Moment of inertia about the end is $I = \frac{ml^2}{3}$.
$$mg \frac{l}{2} = \frac{ml^2}{3} \alpha$$
$$\alpha = \frac{3g}{2l}$$
{{SOL_IMG1}}
### Step 2: Calculate Force and Tension
The acceleration of the center of mass (CM) is downwards:
$$a_{c} = \alpha \frac{l}{2} = \left(\frac{3g}{2l}\right) \left(\frac{l}{2}\right) = \frac{3g}{4}$$
Applying Newton's second law for translational motion of the CM in vertical direction:
$$mg - T = m a_{c}$$
$$T = mg - m a_{c} = mg - m \left(\frac{3g}{4}\right) = \frac{mg}{4}$$
### Pattern Recognition
Classic 'cut string' rigid body problem. Always take torque about the pivot/hinge point to find $\alpha$, then relate the center of mass linear acceleration $a = r_{cm} \alpha$ to find the unknown tension using $F_{\text{net}} = ma$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q2093 (2026)
Two identical thin rods of mass M kg and length L m are connected as shown in figure. Moment of inertia of the combined rod system about an axis passing through point P and perpendicular to the plane of the rods is $\frac{x}{2} ML^{2}\text{ kg m}^{2}$. The value of x is
{{IMG1}}
### Related Formula
$$I_{\text{end}} = \frac{ML^2}{3}$$
$$I_{\text{parallel axis}} = I_{\text{cm}} + Md^2 = \frac{ML^2}{12} + Md^2$$
### Core Logic
Let the rods be Rod 1 (vertical, passing through P at its end) and Rod 2 (horizontal, attached at the other end of Rod 1).
{{SOL_IMG1}}
For Rod 1 (length $L$, mass $M$):
The axis passes through its end perpendicular to its length.
$I_1 = \frac{ML^2}{3}$
For Rod 2 (length $L$, mass $M$):
The axis passes parallel to Rod 2's center of mass axis, at a distance $L$ from it (since it's attached to the bottom end of Rod 1).
$I_2 = I_{\text{cm}} + M d^2 = \frac{ML^2}{12} + M(L)^2$
### Step 1: Total Moment of Inertia
$$I = I_1 + I_2 = \frac{ML^2}{3} + \left(\frac{ML^2}{12} + ML^2\right)$$
$$I = \frac{4ML^2 + ML^2 + 12ML^2}{12}$$
$$I = \frac{17}{12} ML^2$$
We are given that $I = \frac{x}{12} ML^2$ (Correction from source PDF text: the source question text says $\frac{x}{2} ML^2$, but the solution uses $\frac{x}{12} ML^2$. Following the solution steps: $x=17$ is consistent if the denominator is 12. Let's assume the question asked for $\frac{x}{12}$ or $x=17/6$, but the official answer gives 17. Our output will state 17).
### Pattern Recognition
For composite shapes, calculate $I$ for each simple shape separately about the desired axis using Parallel Axis Theorem, then sum them up.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
