JEE Physics System Of Particles and Rotational Motion MCQs Set 16

Answer: (b) \(\sqrt{\frac{2H}{g}}\left(\frac{1+e}{1-e}\right)\)

Explanation:
1. Time to fall the first time is \(t_0 = \sqrt{\frac{2H}{g}}\).
2. After each bounce the speed is multiplied by e, so each flight (up and down) lasts \(2e^n t_0\).
3. Total time \(= t_0\left(1 + 2e + 2e^2 + \dots\right) = t_0\left(1 + \frac{2e}{1-e}\right)\).
4. \(= t_0\,\frac{1+e}{1-e}\).

Teacher's Note:
1. Flight times form a geometric series with ratio e.
2. Add the first fall separately because it is not doubled.

Answer: (c) \(120^\circ\)

Explanation:
1. Let each have speed v and let the angle between the velocities be \(\theta\). Momentum conservation: \(m\vec v_1 + m\vec v_2 = 2m\vec V\).
2. \(|\vec V| = \frac{|\vec v_1 + \vec v_2|}{2} = \frac{v\sqrt{2 + 2\cos\theta}}{2} = v\cos\frac\theta2\).
3. Setting \(V = \frac v2\) gives \(\cos\frac\theta2 = \frac12\), so \(\frac\theta2 = 60^\circ\) and \(\theta = 120^\circ\).

Teacher's Note:
1. Add the two momentum vectors using the parallelogram law.
2. The final velocity is the vector sum divided by the total mass.

Answer: (c) 120 m/s

Explanation:
1. The block rises 0.10 m, so its speed just after the collision is \(V = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.1} = 1.4\) m/s.
2. Momentum conservation: \(0.01 \times 400 = 2 \times 1.4 + 0.01\,v\).
3. \(4 = 2.8 + 0.01v\), so \(v = 120\) m/s.

Teacher's Note:
1. The string length is not needed.
2. Use the rise in height to get the block's speed, then conserve momentum for the collision.

Answer: (b) \(\frac12\)

Explanation:
1. \(e = \frac{\text{speed of separation}}{\text{speed of approach}}\).
2. Speed of approach \(= 8 - 2 = 6\) m/s and speed of separation \(= 5 - 2 = 3\) m/s.
3. \(e = \frac36 = \frac12\).

Teacher's Note:
1. Use the relative speeds along the line of impact.
2. Momentum is conserved here because the masses are equal and the totals match.

Answer: (c) 10 m

Explanation:
1. Momentum conservation: \(m V = 2m \times 5\), so the skater recoils at \(V = 10\) m/s.
2. Retardation due to friction \(= \mu g = 5\) m/s².
3. Distance \(= \frac{V^2}{2\mu g} = \frac{100}{10} = 10\) m.

Teacher's Note:
1. The skater and stone start at rest, so their momenta are equal and opposite after the throw.
2. Then use the kinematics of uniform retardation.

Answer: (c) 0.51

Explanation:
1. Momentum conservation: \(0.02 \times 250 = 0.25 V\), so \(V = 20\) m/s.
2. Friction brings the combined body to rest in 40 m: \(V^2 = 2\mu g s\).
3. \(\mu = \frac{400}{2 \times 9.8 \times 40} = 0.51\).

Teacher's Note:
1. First find the common speed after the perfectly inelastic impact.
2. Then use work-energy or kinematics for the sliding.

Answer: (a) \(R/4\)

Explanation:
1. Mass is proportional to area. The full disc has area \(9\pi R^2\) and the removed hole has area \(\pi R^2\).
2. Treat the hole as a negative mass at distance 2R from O: \(x_{cm} = \frac{9\pi R^2(0) - \pi R^2(2R)}{9\pi R^2 - \pi R^2} = -\frac{2R}{8}\).
3. The centre of mass is \(\frac R4\) from O, on the side opposite to the hole.

Teacher's Note:
1. A hole is treated as a negative mass in centre-of-mass problems.
2. The result is always on the side away from the hole.

Answer: (b) \(\frac{5v_0}{2}\)

Explanation:
1. At the top the velocity is horizontal, \(v_0\cos60^\circ = \frac{v_0}{2}\). Let each fragment have mass m.
2. Momentum before: \(3m\cdot\frac{v_0}{2}\hat i\). First fragment (retraces path): \(-m\frac{v_0}{2}\hat i\). Second: \(m\frac{3v_0}{2}\hat j\).
3. Third fragment: \(m\vec v_3 = \frac{3mv_0}{2}\hat i + \frac{mv_0}{2}\hat i - \frac{3mv_0}{2}\hat j = 2mv_0\hat i - \frac{3mv_0}{2}\hat j\).
4. \(v_3 = \sqrt{4 + \frac94}\,v_0 = \frac{5v_0}{2}\).

Teacher's Note:
1. Momentum is conserved in the explosion, component by component.
2. Retracing the path means velocity reversed along the horizontal.

Answer: (a) \(y = -5\) cm

Explanation:
1. The explosion involves only internal forces, so the centre of mass keeps moving along the x-axis and its y-coordinate stays 0.
2. \(\frac m4(15) + \frac{3m}{4}\,y = 0\).
3. \(y = -5\) cm.

Teacher's Note:
1. Internal forces cannot move the centre of mass.
2. The fragments lie on opposite sides of the centre of mass line.

Answer: (d) \(\frac{(n-k)}{n}g\)

Explanation:
1. Let each fragment have mass m, so the total mass is nm.
2. After the k fragments have landed (and are held by the ground), the only unbalanced external force on the system is gravity on the remaining \(n - k\) fragments: \((n - k)mg\).
3. \(a_{cm} = \frac{(n-k)mg}{nm} = \frac{(n-k)}{n}g\).

Teacher's Note:
1. The acceleration of the centre of mass depends only on the net external force.
2. The ground force on the landed fragments cancels their weight.

Answer: (c) a circle of radius \(l/2\) with centre at the origin

Explanation:
1. Let the rod make angle \(\theta\) with the floor. A is at \((0, l\sin\theta)\) and B at \((l\cos\theta, 0)\).
2. The midpoint is \(\left(\frac l2\cos\theta, \frac l2\sin\theta\right)\).
3. \(x^2 + y^2 = \frac{l^2}{4}\), a circle of radius \(\frac l2\) centred at the origin.

Teacher's Note:
1. The midpoint of the hypotenuse of a right triangle is at equal distance from all three vertices.
2. That distance is half the rod length, so it traces a circle about the corner.

Answer: (c) 55 m

Explanation:
1. Both stones have the same vertical motion: zero initial vertical velocity and acceleration g.
2. Each falls \(\frac12 g t^2 = \frac12 \times 10 \times 9 = 45\) m in 3 s.
3. So both are 55 m above the ground and so is their centre of mass.

Teacher's Note:
1. The horizontal motion does not affect the vertical position.
2. Equal vertical positions mean the centre of mass height equals that common height.

Answer: (b) 2.22 m

Explanation:
1. Acceleration of each block \(a = \frac{2 - 1}{2 + 1}g = \frac g3\).
2. \(a_{cm} = \frac{2a - 1a}{3} = \frac a3 = \frac g9 = \frac{10}{9}\) m/s² (downwards).
3. Distance \(= \frac12 a_{cm}t^2 = \frac12 \times \frac{10}{9} \times 4 = 2.22\) m.

Teacher's Note:
1. One block goes down and the other goes up, so their accelerations enter with opposite signs.
2. The centre of mass starts at rest.

Answer: (b) 3

Explanation:
1. The centre of mass of all fragments follows the original projectile path, so it lands at the range \(R = \frac{u^2\sin60^\circ}{g} = \frac{100 \times \frac{\sqrt3}{2}}{10} = 5\sqrt3\) m.
2. Taking moments of the landing positions: \(n \cdot 5\sqrt3 = \sqrt3 + (n - 1)\cdot 7\sqrt3\).
3. \(5n = 1 + 7n - 7\), so \(2n = 6\) and \(n = 3\).

Teacher's Note:
1. Explosion forces are internal, so the centre of mass keeps its original trajectory.
2. All fragments land together, so the centre of mass formula applies to landing positions.

Answer: (d) The centre of mass of the system is moving with a velocity of 1 m/s

Explanation:
1. The spring force is internal, so the velocity of the centre of mass stays constant.
2. Taking right as positive, \(v_{cm} = \frac{1(-3) + 1(5)}{2} = 1\) m/s towards the right.
3. Unequal speeds are possible if the system had some initial momentum, so only the 1 m/s statement is correct.

Teacher's Note:
1. The centre of mass velocity depends on the total momentum, not on the spring.
2. The statement that it stays stationary holds only if the total momentum is zero.

Answer: (d) \(\frac{4n}{(1+n)^2}\)

Explanation:
1. For a head-on elastic collision with the second ball at rest, \(v_2 = \frac{2m}{m + nm}u = \frac{2u}{1+n}\).
2. Energy gained by the heavier ball \(= \frac12 (nm)\frac{4u^2}{(1+n)^2} = \frac{2nmu^2}{(1+n)^2}\).
3. Fraction \(= \frac{2nmu^2/(1+n)^2}{\frac12 mu^2} = \frac{4n}{(1+n)^2}\).

Teacher's Note:
1. The transfer is largest when the masses are equal, giving a fraction of 1.
2. Check by putting \(n = 1\): \(\frac{4}{4} = 1\).

Answer: (b) \(v\sqrt5\)

Explanation:
1. Along the initial direction: \(m(2v) = m v_{2x}\), so \(v_{2x} = 2v\).
2. Perpendicular direction: \(0 = m(v) + m v_{2y}\), so \(v_{2y} = -v\).
3. Speed \(= \sqrt{(2v)^2 + v^2} = v\sqrt5\).

Teacher's Note:
1. Conserve momentum separately along and perpendicular to the original line of motion.
2. The perpendicular components must cancel.

Answer: (d) \(30^\circ\)

Explanation:
1. Let the angles with the normal be \(\theta\) before and \(\theta'\) after. The tangential component is unchanged: \(u\sin\theta = v\sin\theta'\).
2. The normal component is reduced: \(v\cos\theta' = eu\cos\theta\). So \(\tan\theta' = \frac{\tan\theta}{e}\).
3. The directions before and after are at right angles when \(\theta + \theta' = 90^\circ\), i.e. \(\tan\theta' = \cot\theta\).
4. \(\cot\theta = \frac{\tan\theta}{e}\), so \(\tan^2\theta = e = \frac13\) and \(\theta = 30^\circ\).

Teacher's Note:
1. Split the velocity into components along and perpendicular to the surface.
2. Only the normal component is affected by e.

Answer: (d) zero

Explanation:
1. Take the cart's direction as positive. The man's velocity relative to the ground is \(2 - 3 = -1\) m/s.
2. Total momentum \(= 80(-1) + 40(2) = 0\).
3. So the speed of the centre of mass is zero.

Teacher's Note:
1. Convert the man's velocity to the ground frame before computing momentum.
2. Zero total momentum means a stationary centre of mass.

Answer: (a) \(-\frac{\hat i + \hat j + \hat k}{2}\)

Explanation:
1. \(\vec r_{cm} = \frac{10(\hat i + \hat j + \hat k) + 30(-\hat i - \hat j - \hat k)}{40}\).
2. \(= \frac{-20(\hat i + \hat j + \hat k)}{40} = -\frac{\hat i + \hat j + \hat k}{2}\).

Teacher's Note:
1. The centre of mass lies nearer the heavier particle.
2. Weight each position vector by its mass.

Answer: (a) 2 kg

Explanation:
1. Total momentum after the blast is zero. First piece: \(2\hat i\). Second piece: \(2 \times 1 = 2\) along y.
2. The third piece must carry \(-(2\hat i + 2\hat j)\), of magnitude \(2\sqrt2\).
3. \(m\sqrt2 = 2\sqrt2\), so \(m = 2\) kg.

Teacher's Note:
1. The momenta of all fragments add to zero if the object was at rest.
2. The third momentum is the negative of the vector sum of the other two.

Answer: (c) \(\frac{4a}{4+\pi}\) right to the centre of the disc

Explanation:
1. Mass is proportional to area: plate \(a^2\) and disc \(\frac{\pi a^2}{4}\).
2. The centre of the disc is at distance \(\frac a2 + \frac a2 = a\) from the centre of the plate.
3. Measured from the disc's centre, \(x_{cm} = \frac{a^2 \cdot a}{a^2 + \frac{\pi a^2}{4}} = \frac{4a}{4 + \pi}\), towards the plate (right).

Teacher's Note:
1. Use the disc centre as the origin.
2. The centre of mass lies on the line joining the two centres, nearer the heavier plate.

Answer: (b) \(x = 7\) m

Explanation:
1. Initial centre of mass: \(x_{cm} = \frac{1(0) + 1(10)}{2} = 5\) m.
2. \(v_{cm} = \frac{1(5) + 1(-3)}{2} = 1\) m/s.
3. Mutual forces are internal, so the centre of mass moves uniformly: \(x = 5 + 1 \times 2 = 7\) m.

Teacher's Note:
1. The mutual gravitational attraction is an internal force pair.
2. The centre of mass keeps its constant velocity.

Answer: (d) of magnitude \(\sqrt5\,p\) inclined at \(\pi + \tan^{-1}\frac12\) with the x-axis

Explanation:
1. Total momentum is zero, so \(\vec p_3 = -(2p\hat i + p\hat j)\).
2. Magnitude \(= \sqrt{4p^2 + p^2} = \sqrt5\,p\).
3. It points into the third quadrant, with \(\tan\phi = \frac{p}{2p} = \frac12\), so its angle with the +x axis is \(\pi + \tan^{-1}\frac12\).

Teacher's Note:
1. Fragments of an object at rest have momenta that sum to zero.
2. Measure the angle anticlockwise from the +x axis.

Answer: (b) \(0.25\,mv_0^2\)

Explanation:
1. Components of the initial velocity: \(v_0\cos45^\circ\) perpendicular to the walls and \(v_0\sin45^\circ\) along them.
2. Each wall collision multiplies the component perpendicular to the walls by \(e = 0.5\) and leaves the component along the walls unchanged.
3. After many collisions the perpendicular component tends to zero, so only \(v_0\sin45^\circ = \frac{v_0}{\sqrt2}\) is left.
4. Final KE \(= \frac12 m\frac{v_0^2}{2} = 0.25\,mv_0^2\).

Teacher's Note:
1. With no gravity, the component along the walls never changes.
2. The collisions only remove the component perpendicular to the walls.