JEE Physics System Of Particles and Rotational Motion MCQs Set 54

Answer: (b) \( \frac{5R}{4} \)

Explanation:
1. At the top the horizontal velocity is \( u \) and the point is at horizontal distance \( \frac R2 \).
2. Momentum conservation: \( 3mu=2mv \), so \( v=\frac{3u}{2} \) for the larger piece.
3. The fall time from the top is the same as before, so it travels a further \( \frac32\cdot\frac R2=\frac{3R}{4} \).
4. \( x=\frac R2+\frac{3R}{4}=\frac{5R}{4} \).

Teacher's Note:
1. After an explosion at the top of the path, the fall time is unchanged and only the horizontal speed changes.

Answer: (d) the kinetic energy but not the linear momentum of the system

Explanation:
1. Internal forces occur in equal and opposite pairs, so they cancel in the total momentum.
2. They can do net work (for example, in an explosion), so the kinetic energy can change.

Teacher's Note:
1. Momentum changes only through external forces; kinetic energy can change through internal work.

Answer: (a) II and III only

Explanation:
1. Speed is constant, so kinetic energy is constant.
2. Height changes, so potential energy changes.
3. The direction of velocity changes continuously, so the momentum vector changes.

Teacher's Note:
1. Momentum is a vector: a change in direction is a change in momentum even at constant speed.

Answer: (a) \( 10 \) m/s

Explanation:
1. \( v_{cm}=\frac{-5+10}{2}=2.5 \) m/s to the right. In the centre-of-mass frame the blocks have speeds \( 7.5 \) m/s.
2. Block velocities oscillate between \( 2.5+7.5=10 \) m/s and \( 2.5-7.5=-5 \) m/s, so the maximum speed is \( 10 \) m/s.

Teacher's Note:
1. The speeds are exchanged periodically when the spring returns to natural length.

Answer: (b) \( \sqrt{\frac{2gR}{1+\frac mM}} \)

Explanation:
1. Momentum: \( mv=MV \), so \( V=\frac{mv}{M} \).
2. Energy: \( mgR=\frac12mv^2+\frac12M\frac{m^2v^2}{M^2}=\frac12mv^2\left(1+\frac mM\right) \).
3. \( v=\sqrt{\frac{2gR}{1+m/M}} \).

Teacher's Note:
1. Combine horizontal momentum conservation with energy conservation.

Answer: (c) \( \left(\frac17,\frac{23}{14}\right) \)

Explanation:
1. Total mass \( =14 \) kg.
2. \( x_{cm}=\frac{5(1)+3(-1)+2(2)+4(-1)}{14}=\frac{2}{14}=\frac17 \).
3. \( y_{cm}=\frac{5(6)+3(5)+2(-3)+4(-4)}{14}=\frac{23}{14} \).

Teacher's Note:
1. Use \( x_{cm}=\frac{\sum m_ix_i}{\sum m_i} \) separately for each coordinate.

Answer: (b) \( 3\lambda gx \)

Explanation:
1. Weight of the portion on the ground is \( \lambda xg \).
2. The stopping force on the part landing is \( \lambda v^2 \), with \( v^2=2gx \), giving \( 2\lambda gx \).
3. \( N=\lambda gx+2\lambda gx=3\lambda gx \).

Teacher's Note:
1. The normal force equals the weight of the heap plus the rate of momentum destruction.

Answer: (d) two balls from the extreme right end move on with speed \( V \) each, the remaining balls stay at rest

Explanation:
1. Momentum \( 2mV \) and kinetic energy \( mV^2 \) must both be conserved.
2. Two balls leaving at \( V \) satisfy both, while one ball at \( 2V \) would give kinetic energy \( 2mV^2 \), which is too large.

Teacher's Note:
1. Elastic collisions between identical balls pass the momentum along the row, like a Newton's cradle.

Answer: (c) \( 2 \)

Explanation:
1. Equal masses exchange velocities in an elastic collision, so treat the particles as passing through each other.
2. Both 'ghost' particles are back at A together after the time \( \frac{2\pi r}{v} \) (the slow one has done one lap and the fast one two).
3. The relative speed is \( 3v \), so in this time the number of meetings is \( \frac{3v\times(2\pi r/v)}{2\pi r}=3 \). The last one is at A itself, so \( 3-1=2 \) collisions occur elsewhere.

Teacher's Note:
1. Elastic collisions between equal masses are equivalent to the particles passing through each other.

Answer: (b) at \( \frac R5 \) from the centre of the bigger disc, towards the centre of the smaller disc

Explanation:
1. The disc masses are in the ratio of areas, \( 1:4 \). The distance between centres is \( 2R-R=R \).
2. Distance from the big disc's centre \( =\frac{1\times R}{1+4}=\frac R5 \) towards the small disc.

Teacher's Note:
1. The centre of mass divides the line of centres inversely to the masses.

Answer: (a) \( 2mv \)

Explanation:
1. Take towards the wall as positive: initial momentum \( +mv \), final \( -mv \).
2. Change \( =-mv-mv=-2mv \), magnitude \( 2mv \).

Teacher's Note:
1. Reversal of velocity doubles the momentum change.

Answer: (b) \( \sqrt{\frac{x_1^2+x_2^2}{2}} \)

Explanation:
1. \( kx_1=m(g+a) \) and \( kx_2=m(g-a) \).
2. Horizontal motion: the spring force has components \( mg \) (vertical) and \( ma \) (horizontal), so \( kx=m\sqrt{g^2+a^2} \).
3. \( x_1^2+x_2^2=\left(\frac mk\right)^2(2g^2+2a^2) \), so \( x=\sqrt{\frac{x_1^2+x_2^2}{2}} \).

Teacher's Note:
1. Resolve the spring force into the components that balance gravity and provide the horizontal acceleration.

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

Explanation:
1. Horizontal speed is unchanged and the vertical speed is multiplied by \( e \): \( \tan\theta'=\frac{\tan45^\circ}{e}=\frac1e\ge1 \).
2. So \( \theta'\ge45^\circ \), and \( 30^\circ \) is impossible.

Teacher's Note:
1. Since \( e\le1 \), the rebound angle with the vertical can never be smaller than the incident angle.

Answer: (d) \( aM_0+2a\mu t \)

Explanation:
1. Velocity \( v=at \) and mass \( m=M_0+\mu t \).
2. \( F=\frac{d(mv)}{dt}=ma+v\frac{dm}{dt}=(M_0+\mu t)a+at\mu=aM_0+2a\mu t \).

Teacher's Note:
1. Variable mass: \( F=m\frac{dv}{dt}+v\frac{dm}{dt} \) when the added mass is initially at rest.

Answer: (a) \( 4\sqrt5 \) N s

Explanation:
1. Speed before impact: \( v=\sqrt{2\times10\times4}=4\sqrt5 \) m/s.
2. The ball rebounds with the same speed: \( J=2mv=2\times0.5\times4\sqrt5=4\sqrt5 \) N s.

Teacher's Note:
1. Elastic rebound reverses the velocity, doubling the momentum change.

Answer: (c) \( 10 \) m/s\( ^2 \) upwards

Explanation:
1. Equilibrium shift \( =\frac{mg}{k_1+k_2}=\frac{20}{400}=0.05 \) m below the starting point, so the amplitude is \( 0.05 \) m.
2. At the lowest point the displacement from equilibrium is \( 0.05 \) m: \( a=\frac{(k_1+k_2)}{m}A=\frac{400}{2}\times0.05=10 \) m/s\( ^2 \) upward.

Teacher's Note:
1. The block executes SHM about the equilibrium point; maximum acceleration is at the extremes.

Answer: (b) \( \sqrt{\frac{2gR}{1+\frac mM}} \)

Explanation:
1. Momentum: \( mv=MV \).
2. Energy: \( mgR=\frac12mv^2+\frac12MV^2 \), so \( v=\sqrt{\frac{2gR}{1+m/M}} \).

Teacher's Note:
1. Both horizontal momentum and mechanical energy are conserved.

Answer: (c) \( \frac{2}{\sqrt3}v \)

Explanation:
1. Along the initial direction: \( mv=mv_{2x} \), so \( v_{2x}=v \). Perpendicular: \( 0=m\frac{v}{\sqrt3}+mv_{2y} \), so \( |v_{2y}|=\frac v{\sqrt3} \).
2. \( v_2=\sqrt{v^2+\frac{v^2}{3}}=\frac{2v}{\sqrt3} \).

Teacher's Note:
1. Conserve momentum separately along two perpendicular directions.

Answer: (a) \( 90^\circ \)

Explanation:
1. \( \vec F=\frac{d\vec p}{dt}=-Ak\left[\hat i\sin kt+\hat j\cos kt\right] \).
2. \( \vec F\cdot\vec p=-A^2k(\sin kt\cos kt-\cos kt\sin kt)=0 \), so the angle is \( 90^\circ \).

Teacher's Note:
1. A force perpendicular to the momentum changes only the direction, not the magnitude.

Answer: (d) \( 0.64h \)

Explanation:
1. Rebound speed \( =0.8v \), with \( v^2=2gh \).
2. \( h'=\frac{(0.8v)^2}{2g}=0.64h \).

Teacher's Note:
1. Height varies as the square of the speed.

Answer: (c) \( \frac1{\sqrt2} \)

Explanation:
1. Speed at the ground: \( v^2=2gh+2gh=4gh \), so \( v=2\sqrt{gh} \).
2. To rise to \( h \) the rebound speed must be \( \sqrt{2gh} \).
3. \( e=\frac{\sqrt{2gh}}{2\sqrt{gh}}=\frac1{\sqrt2} \).

Teacher's Note:
1. Compare rebound speed with impact speed.

Answer: (b) \( \sqrt2\,v_0 \)

Explanation:
1. \( v_1+v_2=v_0 \) and \( v_1^2+v_2^2=1.5v_0^2 \), so \( 2v_1v_2=-0.5v_0^2 \).
2. \( (v_2-v_1)^2=1.5v_0^2+0.5v_0^2=2v_0^2 \), so \( |v_{rel}|=\sqrt2v_0 \).

Teacher's Note:
1. Kinetic energy has increased, so the collision is explosive; use the two equations to find \( v_1-v_2 \).

Answer: (c) \( 17.6 \) kg m/s

Explanation:
1. \( p=3.513\times5.00=17.565 \) kg m/s.
2. The speed has three significant figures, so the product is quoted to three: \( 17.6 \) kg m/s.

Teacher's Note:
1. A product cannot have more significant figures than the least precise factor.

Answer: (a) \( 30 \) m

Explanation:
1. \( v_x=v_y=15\sqrt2 \) m/s; time to wall \( =\frac{50}{15\sqrt2} \) s; total flight time \( =\frac{2\times15\sqrt2}{10}=3\sqrt2 \) s.
2. Remaining time after the wall: \( 3\sqrt2-\frac{10}{3\sqrt2}=\frac{4\sqrt2}{3}\text{ s} \)
3. Horizontal speed after the wall is \( \frac12\times15\sqrt2 \), so it returns \( \frac{15\sqrt2}{2}\times\frac{4\sqrt2}{3}=20 \) m. Landing point \( =50-20=30 \) m.

Teacher's Note:
1. After hitting the wall the horizontal velocity is multiplied by \( -e \).

Answer: (d) \( 4 \) m/s

Explanation:
1. Elastic collision of A with B: \( v_B=\frac{2m}{m+2m}\times9=6 \) m/s.
2. B and C stick: \( 2m\times6=3mV \), so \( V=4 \) m/s.

Teacher's Note:
1. Apply the elastic formula first, then conservation of momentum for the sticking collision.

Answer: (d) \( 360 \) m

Explanation:
1. Velocity at the top is \( 50\cos53^\circ=30 \) m/s horizontal. The top is at \( x=120 \) m and height \( 80 \) m (fall time \( 4 \) s).
2. Momentum: \( 24\times30=12v \), so the second fragment moves at \( 60 \) m/s.
3. It lands \( 60\times4=240 \) m beyond the top: \( 120+240=360 \) m.

Teacher's Note:
1. After the explosion the second fragment's horizontal speed doubles.