JEE Physics System Of Particles and Rotational Motion MCQs Set 21

Answer: (b) 0.4 N

Explanation:
1. The sand gains horizontal momentum from zero to the belt's speed.
2. Force \( F=v\frac{dm}{dt}=0.2\times2=0.4\ \text{N} \).

Teacher's Note:
1. Force equals the rate of change of momentum.
2. The sand has no horizontal momentum before it lands.

Answer: (a) \( \frac{Adv^3}{2} \)

Explanation:
1. Mass flowing per second \( \frac{dm}{dt}=Adv \).
2. Kinetic energy given per second \( =\tfrac12\frac{dm}{dt}v^2=\frac{Adv^3}{2} \).

Teacher's Note:
1. Each kilogram leaves with kinetic energy ½v².
2. Multiply by the mass flow rate.

Answer: (d) \( \frac{m_1}{m_1+m_2} \)

Explanation:
1. No horizontal external force acts, so the centre of mass stays fixed. Let the boat move back by x.
2. The boy moves forward by \( 1-x \) relative to the ground: \( m_1(1-x)=m_2x \).
3. \( x=\frac{m_1}{m_1+m_2} \) m.

Teacher's Note:
1. The centre of mass of the system does not move.
2. Displacement of the boy relative to the ground is less than 1 m.

Answer: (c) \( \frac{2M}{m}\sqrt{5g\ell} \)

Explanation:
1. Momentum conservation: \( mu=m\frac u2+MV\Rightarrow V=\frac{mu}{2M} \).
2. To complete a vertical circle on a string the bob needs at least \( V=\sqrt{5g\ell} \) at the bottom.
3. \( u=\frac{2M}{m}\sqrt{5g\ell} \).

Teacher's Note:
1. The minimum speed at the bottom for a complete circle on a string is √(5gℓ).
2. Use momentum conservation for the very short collision.

Answer: (b) 0.2 m/s

Explanation:
1. Rain has no horizontal momentum, so horizontal momentum is conserved: \( 5000\times1.2=6000v \).
2. \( v=1.0\ \text{m/s} \).
3. Change in velocity \( =1.2-1.0=0.2\ \text{m/s} \) (a decrease).

Teacher's Note:
1. Only horizontal momentum is conserved because the rain adds no horizontal momentum.
2. The wagon slows down as its mass increases.

Answer: (a) h/4

Explanation:
1. Speed just before collision \( v=\sqrt{2gh} \).
2. Perfectly inelastic collision of equal masses: \( mv=2mv'\Rightarrow v'=\frac v2 \).
3. Height risen \( h'=\frac{v'^2}{2g}=\frac{v^2}{8g}=\frac h4 \).

Teacher's Note:
1. The velocity halves, so the kinetic energy and the height fall to one quarter.
2. Use energy conservation before and after the collision.

Answer: (d) \( \sqrt3 \) m/s at 30° to the original direction

Explanation:
1. Momentum along the original line: \( 2=1\cos60^\circ+v\cos\phi\Rightarrow v\cos\phi=1.5 \).
2. Momentum perpendicular: \( 0=1\sin60^\circ-v\sin\phi\Rightarrow v\sin\phi=0.866 \).
3. \( v=\sqrt{1.5^2+0.866^2}=\sqrt3\ \text{m/s} \), \( \tan\phi=\frac{0.866}{1.5}=0.577\Rightarrow\phi=30^\circ \) on the other side.

Teacher's Note:
1. Resolve momentum along and perpendicular to the original direction.
2. The other ball moves on the opposite side of the original line.

Answer: (c) moves with a speed v along BG

Explanation:
1. The velocities along the medians are at 120° to each other, so the total momentum is zero before the collision.
2. After the collision the total momentum is still zero. A has none, and B moves with speed v from G towards B (momentum mv along GB).
3. So C must have momentum mv in the opposite direction, along BG: it moves with speed v along BG.

Teacher's Note:
1. Three equal vectors at 120° sum to zero.
2. Momentum conservation fixes the last particle's velocity.

Answer: (b) 5.7 m/s

Explanation:
1. Along the original line: \( 6.6=3.3\cos60^\circ+v\cos\phi\Rightarrow v\cos\phi=4.95 \).
2. Perpendicular: \( v\sin\phi=3.3\sin60^\circ=2.86 \).
3. \( v=\sqrt{4.95^2+2.86^2}=\sqrt{32.7}\approx5.7\ \text{m/s} \).

Teacher's Note:
1. The momentum components along and across the original line are separately conserved.
2. Combine the components with Pythagoras' theorem.

Answer: (a) 250 m

Explanation:
1. Let the horizontal speed at the top be u. Momentum: \( 3mu=m(-u)+0+mv_3\Rightarrow v_3=4u \).
2. The top is at 50 m horizontal distance and the fall time equals the time to reach the top, so the other piece covers \( 4\times50=200\ \text{m} \) after the explosion.
3. Distance from the point of projection \( =50+200=250\ \text{m} \).

Teacher's Note:
1. The explosion happens at the top, where the velocity is horizontal.
2. Time of fall equals half the total time of flight.

Answer: (b) \( \frac Lt\left(\frac m{M+m}\right) \)

Explanation:
1. Speed of man relative to the plank \( =\frac Lt \).
2. Let the plank move at V (opposite direction). The man's speed relative to the ground is \( \frac Lt-V \).
3. Momentum conservation: \( m\left(\frac Lt-V\right)=MV\Rightarrow V=\frac Lt\cdot\frac{m}{M+m} \).

Teacher's Note:
1. The relative speed is L/t, but the ground speeds are smaller.
2. Initial momentum of the system is zero.

Answer: (d) \( \sqrt{2gh\left(1+\frac1n\right)} \)

Explanation:
1. At the top (minimum case) block and wedge share the same horizontal velocity v: \( mu=(m+nm)v\Rightarrow v=\frac{u}{1+n} \).
2. Energy: \( \tfrac12mu^2=\tfrac12(m+nm)v^2+mgh \).
3. \( \tfrac12u^2\left(1-\frac1{1+n}\right)=gh\Rightarrow u^2=2gh\,\frac{1+n}{n} \), so \( u=\sqrt{2gh\left(1+\frac1n\right)} \).

Teacher's Note:
1. The wedge moves as the block climbs, so some kinetic energy remains at the top.
2. Use both momentum and energy conservation.

Answer: (a) B will reach the top of W again

Explanation:
1. Before the collision no external horizontal force acts, so the centre of mass has no horizontal motion (b is correct) and \( mv=MV \) for the wedge.
2. After the elastic collision B moves away from the wall at speed v and W also moves away at V: \( v_{cm}=\frac{mv+MV}{m+M}=\frac{2mv}{m+M} \) (c is correct). At the highest point both move together with this velocity (d is correct).
3. Energy check: the kinetic energy of the centre of mass motion cannot be converted back into potential energy, so B rises to \( h'=h\left(\frac{M-m}{M+m}\right)^2 \), which is less than h unless M = m. So (a) is incorrect.

Teacher's Note:
1. Part of the kinetic energy always remains in the motion of the centre of mass.
2. Compare the heights using energy conservation.

Answer: (c) \( -\hat i+2\hat j \)

Explanation:
1. The wall is parallel to j, so the normal is along i. The j component is unchanged: 2.
2. The normal component reverses and is multiplied by e: \( -e\times2=-1 \).
3. New velocity \( =-\hat i+2\hat j \).

Teacher's Note:
1. Only the component along the normal changes in a smooth collision.
2. Normal velocity of separation = e × normal velocity of approach.

Answer: (b) 8

Explanation:
1. For a collision the positions must coincide at t = 2 s: \( \vec r_1+\vec v_1t=\vec r_2+\vec v_2t \).
2. x: \( 3+8=-5+2a\Rightarrow a=8 \). y: \( 5+6=-3+14 \) (consistent).

Teacher's Note:
1. The two particles meet when their position vectors are equal.
2. Check the y equation to confirm the data are consistent.

Answer: (a) \( \sqrt{\frac{kM}{m(M+m)}}\,x \)

Explanation:
1. Momentum: \( mv_1=Mv_2\Rightarrow v_2=\frac{m}{M}v_1 \).
2. Energy: \( \tfrac12kx^2=\tfrac12mv_1^2+\tfrac12Mv_2^2=\tfrac12mv_1^2\left(1+\frac mM\right) \).
3. \( v_1=x\sqrt{\frac{kM}{m(M+m)}} \).

Teacher's Note:
1. The initial momentum is zero, so the momenta are equal and opposite.
2. The spring's energy is shared between the two blocks.

Answer: (d) centre of the circle

Explanation:
1. Angular momentum is conserved only when the torque about the point is zero.
2. Only about the centre is the force (centripetal) directed through the point, giving zero torque.
3. About any other point the torque is non-zero, so the angular momentum changes.

Teacher's Note:
1. Torque = r × F, which is zero only if F passes through the reference point.
2. The centripetal force always passes through the centre.

Answer: (c) K/2

Explanation:
1. No external torque, so angular momentum is conserved: \( I\omega_0=2I\omega\Rightarrow\omega=\frac{\omega_0}{2} \).
2. Final kinetic energy \( =\tfrac12(2I)\left(\frac{\omega_0}{2}\right)^2=\frac14I\omega_0^2=\frac K2 \).

Teacher's Note:
1. Angular momentum is conserved but kinetic energy is not.
2. KE = L²/(2I), so doubling I halves the kinetic energy.

Answer: (b) straight line

Explanation:
1. The surface is smooth, so the only external forces (gravity and normal reaction) are vertical.
2. There is no horizontal force, so the horizontal position of the centre of mass does not change: it moves only vertically.
3. Hence the path of the centre of mass is a straight vertical line.

Teacher's Note:
1. No horizontal external force means no horizontal acceleration of the centre of mass.
2. Other points of the rod (such as the ends) trace ellipses.

Answer: (d) \( \sqrt{\frac{8g}{3R}} \)

Explanation:
1. Moment of inertia of the system: \( I=\tfrac12MR^2+MR^2=\tfrac32MR^2 \).
2. The object falls through \( 2R \): \( Mg(2R)=\tfrac12I\omega^2 \).
3. \( \omega^2=\frac{4MgR}{\frac32MR^2}=\frac{8g}{3R} \).

Teacher's Note:
1. The attached object is treated as a point mass at distance R from the pivot.
2. Loss of potential energy becomes rotational kinetic energy.