JEE Physics System Of Particles and Rotational Motion MCQs Set 15

Answer: (c) 0.9

Explanation:
1. The speed on reaching the plate is \( v = \sqrt{2gh} \) and the rebound speed is \( v' = \sqrt{2gh'} \).
2. \( e = \dfrac{v'}{v} = \sqrt{\dfrac{h'}{h}} = \sqrt{\dfrac{0.81}{1}} = 0.9 \).

Teacher's Note:
1. For a ball bouncing from a fixed surface, e is the square root of the height ratio.
2. Do not use the height ratio itself.

Answer: (a) \( \dfrac{4R}{3\pi} \)

Explanation:
1. Divide the disc into thin semicircular strips of radius r and thickness dr. Each has mass proportional to \( \pi r\,dr \) and its centre of mass is \( 2r/\pi \) from the centre.
2. \( y_{cm} = \dfrac{\int_0^R \frac{2r}{\pi}(\pi r\,dr)}{\int_0^R \pi r\,dr} = \dfrac{2R^3/3}{\pi R^2/2} \).
3. This simplifies to \( \dfrac{4R}{3\pi} \).

Teacher's Note:
1. Memorise: semicircular disc \( 4R/3\pi \), semicircular ring \( 2R/\pi \), hemisphere (solid) \( 3R/8 \).
2. The disc value is smaller than the ring value because the mass is nearer the centre.

Answer: (b) g/3

Explanation:
1. Before the collision the spring force balances the weight of A: \( F_{spring} = 2mg \).
2. The collision is instantaneous, so the spring does not change its length and the spring force is still 2mg upward.
3. Net force on the combined mass 3m is \( 3mg - 2mg = mg \) downward, so \( a = \dfrac{mg}{3m} = \dfrac{g}{3} \).

Teacher's Note:
1. During a very short collision, spring and gravity forces are negligible and the spring length does not change.
2. After the collision the new weight is no longer balanced.

Answer: (c) 3l/4

Explanation:
1. \( x_{cm} = \dfrac{\int_0^l x\,\lambda\,dx}{\int_0^l\lambda\,dx} = \dfrac{\int_0^l kx^3dx}{\int_0^l kx^2dx} \).
2. \( = \dfrac{l^4/4}{l^3/3} = \dfrac{3l}{4} \).

Teacher's Note:
1. The lighter end is at x = 0 because the density λ = kx² is smallest there.
2. The centre of mass lies closer to the heavier end.

Answer: (a) \( v_1 = \dfrac{M+m}{m}\sqrt{2gy} \)

Explanation:
1. Just after impact the combined mass moves with speed V and rises to height y, so \( \tfrac12(m+M)V^2 = (m+M)gy \), giving \( V = \sqrt{2gy} \).
2. Momentum is conserved in the collision: \( mv_1 = (m+M)V \).
3. So \( v_1 = \dfrac{M+m}{m}\sqrt{2gy} \).

Teacher's Note:
1. This is the ballistic pendulum: use momentum conservation for the collision and energy conservation for the swing.
2. Never use energy conservation across a perfectly inelastic collision.

Answer: (c) g/4

Explanation:
1. The acceleration of each block is \( a = \dfrac{(3m-m)g}{4m} = \dfrac{g}{2} \), 3m going down and m going up.
2. \( a_{cm} = \dfrac{3m\left(\frac g2\right) - m\left(\frac g2\right)}{4m} = \dfrac{g}{4} \) (downward).

Teacher's Note:
1. The centre of mass accelerates because external forces (gravity and the pulley support) do not balance.
2. Take downward as positive: the heavier block contributes +g/2 and the lighter block -g/2, each weighted by its mass.

Answer: (c) 55 m

Explanation:
1. The horizontal throw does not change the vertical motion, so both stones fall freely in the vertical direction.
2. Each falls \( \tfrac12gt^2 = \tfrac12\times10\times9 = 45\ \text{m} \) in 3 s.
3. The centre of mass is at height \( 100-45 = 55\ \text{m} \).

Teacher's Note:
1. Horizontal and vertical motions are independent.
2. Both stones have the same vertical acceleration, so the centre of mass falls the same distance.

Answer: (c) 0.5

Explanation:
1. Initial momentum \( = 1\times0.4 = 0.4 \) kg m/s along x.
2. After the collision the first ball has 0.3 kg m/s along y.
3. The momentum along x of the second ball is 0.4 and along y is -0.3, so its magnitude is \( \sqrt{0.4^2+0.3^2} = 0.5\ \text{kg m/s} \).

Teacher's Note:
1. Momentum is a vector: conserve each component separately.
2. The result forms a 3-4-5 right triangle.

Answer: (a) 3/35 cm

Explanation:
1. Treat the hole as a negative mass proportional to its area: \( m_{hole} \propto \pi(1)^2 \) and \( m_{whole} \propto \pi(6)^2 \).
2. \( x_{cm} = \dfrac{-\pi(1)^2\times3}{\pi(6)^2-\pi(1)^2} = -\dfrac{3}{35}\ \text{cm} \).
3. The centre of mass is 3/35 cm from the disc centre, on the side opposite to the hole.

Teacher's Note:
1. Removed mass is treated as a negative mass at the centre of the removed part.
2. Mass of a uniform disc is proportional to the area, so the factor π cancels.

Answer: (a) (-3, -3, -3)

Explanation:
1. The first three masses (6 kg in total) act as if placed at (2, 2, 2).
2. For the new centre of mass at the origin: \( 6(2,2,2) + 4\vec r = 0 \).
3. So \( \vec r = -\dfrac{12}{4}(1,1,1) = (-3,-3,-3) \).

Teacher's Note:
1. A group of particles can be replaced by its total mass at its own centre of mass.
2. The new mass must lie on the opposite side of the origin.

Answer: (b) \( \sqrt{\dfrac{4gR}{3}} \)

Explanation:
1. There is no external horizontal force, so with speed v for the ball and V for the block: \( mv = 2mV \), so \( V = v/2 \).
2. Energy conservation: \( mgR = \tfrac12mv^2 + \tfrac12(2m)V^2 = \tfrac12mv^2 + \tfrac14mv^2 = \tfrac34mv^2 \).
3. So \( v = \sqrt{\dfrac{4gR}{3}} \).

Teacher's Note:
1. The block moves as the ball slides, so some energy goes into the block.
2. Combine horizontal momentum conservation with energy conservation.

Answer: (c) 0.6457 Å

Explanation:
1. Take the carbon atom (mass 12 u) at the origin and the oxygen atom (mass 16 u) at 1.130 Å.
2. \( x_{cm} = \dfrac{12\times0+16\times1.130}{12+16} = \dfrac{18.08}{28} \).
3. \( x_{cm} = 0.6457\ \text{Å} \) from the carbon atom.

Teacher's Note:
1. The centre of mass is nearer to the heavier atom.
2. Only the ratio of the masses matters, so the mass numbers can be used.

Answer: (a) M' = M

Explanation:
1. In a head-on elastic collision the fraction of kinetic energy transferred to the target is \( \dfrac{4MM'}{(M+M')^2} \).
2. This equals 1 when \( M' = M \), which is the maximum possible value.

Teacher's Note:
1. Maximum transfer takes place when the masses are equal.
2. Light targets and heavy targets both receive only a small fraction.

Answer: (b) 45°

Explanation:
1. The ball hits the incline with speed \( v = \sqrt{2gh} \) vertically downward. The normal to the incline makes an angle θ with the vertical.
2. For an elastic collision with a smooth surface the velocity is reflected about the normal, so the rebound direction makes an angle 2θ with the vertical. Its x-component is \( v\sin2\theta \).
3. Setting \( v\sin2\theta = v \) gives \( \sin2\theta = 1 \), so \( \theta = 45^\circ \).

Teacher's Note:
1. The component of velocity along the surface is unchanged and the normal component reverses.
2. At 45° a vertical ball is turned into a horizontal one.

Answer: (a) 1/3

Explanation:
1. For an elastic head-on collision with B at rest: \( v_A = \dfrac{m_A-m_B}{m_A+m_B}u \) and \( v_B = \dfrac{2m_A}{m_A+m_B}u \).
2. They move in opposite directions with equal speeds, so \( v_B = -v_A \): \( 2m_A = m_B - m_A \).
3. Hence \( m_B = 3m_A \) and \( \dfrac{m_A}{m_B} = \dfrac13 \).

Teacher's Note:
1. A negative \( v_A \) means the lighter particle bounces back.
2. Equal speeds of opposite sign give a quick equation for the mass ratio.

Answer: (b) less than or equal to R

Explanation:
1. The centre of mass is a weighted average of the position vectors, so \( |\vec r_{cm}| \le \dfrac{\sum m_i|\vec r_i|}{\sum m_i} = R \).
2. It equals R only if all the particles are at the same point and is smaller when they are spread around the origin.

Teacher's Note:
1. The centre of mass of a set of points always lies inside their convex hull.
2. A circle of particles has its centre of mass at the centre.

Answer: (d) \( \dfrac{8mv^2}{3L} \)

Explanation:
1. The centre of mass moves with velocity \( \dfrac{m(+v)+2m(-v)}{3m} = -\dfrac v3 \), so in the centre-of-mass frame the particle m moves at \( \dfrac{4v}{3} \) and 2m at \( \dfrac{2v}{3} \) in a circle about the centre of mass.
2. The particle m is at a distance \( \dfrac{2L}{3} \) from the centre of mass.
3. The tension provides its centripetal force: \( T = \dfrac{m(4v/3)^2}{2L/3} = \dfrac{8mv^2}{3L} \).

Teacher's Note:
1. Always shift to the centre-of-mass frame, which is inertial here, before applying circular motion.
2. Check with the other particle: \( \dfrac{2m(2v/3)^2}{L/3} = \dfrac{8mv^2}{3L} \).

Answer: (b) If the velocity of the centre of mass remains zero, then the net external force acting on the system must be zero

Explanation:
1. \( \vec F_{ext} = M\vec a_{cm} \). If the velocity of the centre of mass stays zero, then \( \vec a_{cm} = 0 \), so the net external force is zero.
2. (a) is false because internal forces can do work even when the centre of mass is at rest. (c) is false because internal forces cannot change the speed of the centre of mass.
3. (d) is false: the third particle's momentum must cancel the sum of the other two, which is non-zero for non-collinear motion, so it must move.

Teacher's Note:
1. Only external forces change the motion of the centre of mass; internal forces may change the kinetic energy.
2. Zero centre-of-mass velocity means zero total momentum.

Answer: (a) \( \dfrac{2}{\sqrt3}v \)

Explanation:
1. Conserve momentum along the original direction (x): \( mv = mv_{2x} \), so \( v_{2x} = v \).
2. Along the perpendicular direction (y): \( 0 = m\dfrac{v}{\sqrt3} + mv_{2y} \), so \( v_{2y} = -\dfrac{v}{\sqrt3} \).
3. Speed \( v_2 = \sqrt{v^2+\dfrac{v^2}{3}} = \dfrac{2v}{\sqrt3} \).

Teacher's Note:
1. In a two-dimensional collision, conserve momentum along both axes.
2. The second mass goes off at 30° below the original direction.

Answer: (b) 30°

Explanation:
1. The ground is smooth, so the horizontal component is unchanged and the vertical component is multiplied by e (0 ≤ e ≤ 1).
2. \( \tan\theta' = \dfrac{v_x}{e\,v_y} = \dfrac{\tan45^\circ}{e} \ge 1 \), so \( \theta' \ge 45^\circ \).
3. An angle of 30° would need \( e > 1 \), which is impossible.

Teacher's Note:
1. Inelasticity makes the ball leave at a flatter angle, never a steeper one.
2. e = 1 gives the same angle as incidence.

Answer: (b) 12 m/s upwards

Explanation:
1. The speed of the ball on reaching the lift is \( \sqrt{2\times10\times5} = 10\ \text{m/s} \) downward (taking g = 10 m/s²).
2. In the frame of the lift, the ball approaches at \( 10+1 = 11\ \text{m/s} \) and, since the collision is elastic, leaves at 11 m/s upward.
3. Back in the ground frame, the velocity is \( 11+1 = 12\ \text{m/s} \) upward.

Teacher's Note:
1. For a very heavy moving surface use its frame: the ball simply reverses its relative velocity.
2. Add the surface velocity back to return to the ground frame.

Answer: (c) m/3 kg

Explanation:
1. For the elastic collision with the target at rest, \( v_1 = \dfrac{m-M}{m+M}u \).
2. Setting \( v_1 = \dfrac u2 \): \( 2(m-M) = m+M \), so \( m = 3M \).
3. The mass of the target is \( M = \dfrac m3 \).

Teacher's Note:
1. A positive final velocity means the striking body is heavier than the target.
2. The target ends up moving at \( \dfrac{2m}{m+M}u = \dfrac{3u}{2} \).

Answer: (a) \( \dfrac{3a}{2} \)

Explanation:
1. Volume is conserved, so each small cube has edge \( a/2 \).
2. Original centre of mass height \( = \dfrac a2 \). The stack of 8 small cubes has height \( 8\times\dfrac a2 = 4a \), so its centre of mass is at \( 2a \).
3. Vertical shift \( = 2a - \dfrac a2 = \dfrac{3a}{2} \).

Teacher's Note:
1. Equal cubes stacked in a column have their centre of mass at the middle of the column.
2. Edge scales as the cube root of volume.

Answer: (a) \( he^{2n} \)

Explanation:
1. After each impact the speed is multiplied by e, so the height (proportional to the speed squared) is multiplied by \( e^2 \).
2. After n rebounds, \( h_n = he^{2n} \).

Teacher's Note:
1. Heights form a geometric progression with ratio \( e^2 \).
2. Speeds form one with ratio e.

Answer: (b) \( e = \dfrac mM \le 1 \)

Explanation:
1. Momentum conservation with the first particle at rest afterwards: \( mu = Mv_2 \), so \( v_2 = \dfrac{m}{M}u \).
2. The coefficient of restitution is \( e = \dfrac{v_2 - 0}{u - 0} = \dfrac{m}{M} \).
3. Since \( 0 \le e \le 1 \), this needs \( m \le M \).

Teacher's Note:
1. Combine momentum conservation with the definition of e.
2. The incoming particle can stop only if it is not heavier than the target.