JEE Physics System Of Particles and Rotational Motion MCQs Set 53

Answer: (b) \( 16 \) cm

Explanation:
1. Base: area \( L^2 \) at \( z=0 \). Four sides: total area \( 4L^2 \) with centres at \( z=\frac L2 \).
2. \( z_{cm}=\frac{4L^2\cdot\frac L2}{5L^2}=\frac{2L}{5}=16 \) cm.

Teacher's Note:
1. The open top removes the heavy upper plate that would raise the centre of mass.

Answer: (a) infinite

Explanation:
1. After each collision the relative speed is multiplied by \( e \) (for \( e \lt 1 \)), so the successive times are \( \frac dv,\frac d{ev},\frac d{e^2v},\dots \).
2. This series increases without limit, so the block never comes to rest relative to the cart in any finite time.

Teacher's Note:
1. The relative speed shrinks geometrically but so does the speed needed to cross the cart, so the times grow.

Answer: (d) \( 4 \) only

Explanation:
1. Semicircular disc: \( \frac{4R}{3\pi} \) (statement 1 wrong). Semicircular ring: \( \frac{2R}{\pi} \) (statement 2 wrong).
2. Solid hemisphere: \( \frac{3R}{8} \) (statement 3 wrong). Hemispherical shell: \( \frac R2 \) (statement 4 correct).

Teacher's Note:
1. Remember the four standard results: ring \( \frac{2R}{\pi} \), disc \( \frac{4R}{3\pi} \), solid hemisphere \( \frac{3R}{8} \), shell \( \frac R2 \).

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

Explanation:
1. Total momentum \( =10\times14=140 \) kg m/s.
2. \( v_{cm}=\frac{140}{14}=10 \) m/s.

Teacher's Note:
1. The centre-of-mass velocity is the total momentum divided by the total mass.

Answer: (c) \( \frac{M}{M+m}\cdot\frac{v_0^2}{2g} \)

Explanation:
1. Energy: \( \frac12mv_0^2=\frac12(m+M)V^2+mgh \) with \( V=\frac{mv_0}{M+m} \).
2. \( mgh=\frac12mv_0^2\left(1-\frac{m}{M+m}\right)=\frac12mv_0^2\frac{M}{M+m} \), so \( h=\frac{M}{M+m}\frac{v_0^2}{2g} \).

Teacher's Note:
1. Part of the kinetic energy remains as kinetic energy of the common motion.

Answer: (a) \( \frac{|m_1-m_2|\,u_{rel}}{m_1+m_2+M} \)

Explanation:
1. Let the trolley move at \( v \). Momentum: \( Mv+m_1(v-u_{rel})+m_2(v+u_{rel})=0 \).
2. \( v=\frac{(m_1-m_2)u_{rel}}{m_1+m_2+M} \), so the speed is \( \frac{|m_1-m_2|u_{rel}}{m_1+m_2+M} \).

Teacher's Note:
1. If the masses are equal the trolley stays at rest.

Answer: (c) \( 2.1 \)

Explanation:
1. Elastic collision: \( v_1'=\frac{1-2}{3}(2)=-\frac23 \) m/s and \( v_2'=\frac{2\times1}{3}(2)=\frac43 \) m/s.
2. The \( 2 \) kg block executes SHM with \( \omega=\sqrt{\frac22}=1 \) rad/s, and returns to the natural length after half a period, \( t=\pi \) s.
3. The \( 1 \) kg block moves \( \frac23\pi \) m away in this time, while the \( 2 \) kg block is back at its starting position: separation \( \frac{2\pi}{3}\approx2.1 \) m.

Teacher's Note:
1. The spring-block system returns to the unstretched position after half a period.

Answer: (a) \( 25 \) cm

Explanation:
1. Relative speed \( =7 \) m/s and \( \mu=\frac{5\times2}{7}=\frac{10}{7} \) kg.
2. \( x=v_{rel}\sqrt{\frac\mu k}=7\sqrt{\frac{10/7}{1120}}=7\times\frac1{28}=0.25 \) m.

Teacher's Note:
1. Maximum extension when the relative velocity is zero.

Answer: (b) on the line joining the particles

Explanation:
1. The position vector is \( \vec r_{cm}=\frac{m_1\vec r_1+m_2\vec r_2}{m_1+m_2} \), which is a weighted average of the two positions.
2. So the point lies on the straight line joining them, nearer to the heavier particle.

Teacher's Note:
1. The centre of mass of two particles always lies on the line segment between them.

Answer: (c) \( \frac xLg \)

Explanation:
1. Unbalanced weight \( =\frac{M}{2L}(2x)g \) (the extra length \( 2x \) on one side), acting on the whole mass \( M \).
2. \( a=\frac{2xg}{2L}=\frac xLg \).

Teacher's Note:
1. The excess hanging length on one side gives the net force.

Answer: (d) momentum is conserved but kinetic energy is not

Explanation:
1. With no external impulse, momentum is conserved in every collision.
2. Some kinetic energy is lost in an inelastic collision.

Teacher's Note:
1. Momentum is conserved in all collisions without external impulse.

Answer: (b) \( u+v_r\ln\left(\frac{M_0}{M_0-\mu t}\right) \)

Explanation:
1. Thrust \( =\mu v_r \), so \( (M_0-\mu t)\frac{dv}{dt}=\mu v_r \).
2. Integrating gives \( v=u+v_r\ln\frac{M_0}{M_0-\mu t} \).

Teacher's Note:
1. This is the standard rocket equation.

Answer: (d) \( \frac{mv_0^2}{d} \)

Explanation:
1. Each hit on a given wall transfers impulse \( 2mv_0 \).
2. Time between successive hits on the same wall is \( \frac{2d}{v_0} \).
3. \( F_{avg}=\frac{2mv_0}{2d/v_0}=\frac{mv_0^2}{d} \).

Teacher's Note:
1. Average force equals impulse per collision divided by the time between collisions.

Answer: (a) \( 1 \) m/s towards the right

Explanation:
1. \( v_{cm}=\frac{1(5)+1(-3)}{2}=1 \) m/s towards the right.

Teacher's Note:
1. Centre-of-mass velocity is the mass-weighted average of the velocities.

Answer: (c) \( 40 \) J

Explanation:
1. Momenta are equal and opposite, so \( K=\frac{p^2}{2m} \) is inversely proportional to mass.
2. \( K_m:K_{2m}=2:1 \), so \( K_m=\frac23\times60=40 \) J.

Teacher's Note:
1. For equal momenta, the lighter body carries more kinetic energy.

Answer: (b) \( 9 \) m/s

Explanation:
1. Trolley alone has mass \( 180 \) kg. Initial speed \( 10 \) m/s: \( 200\times10=180V+20(V+10) \).
2. \( V=\frac{2000-200}{200}=9 \) m/s (and it stays \( 9 \) m/s after the child leaves with the same relative velocity). It runs for \( 1 \) s, so the trolley moves \( 9 \) m.

Teacher's Note:
1. Express the child's ground velocity as the trolley's velocity plus the relative velocity.

Answer: (a) \( 1 \)

Explanation:
1. Taking the initial direction positive: \( \Delta p=0.1(-4)-0.1(6)=-1 \).
2. Magnitude is \( 1 \) N s.

Teacher's Note:
1. Add the magnitudes of the momenta when the direction reverses.

Answer: (a) \( \left(0,\frac{\pi R}{6}\right) \)

Explanation:
1. By symmetry about the y-axis, \( x_{cm}=0 \).
2. \( M=\int_0^\pi\int_0^R\sigma_0\sin\theta\,r\,dr\,d\theta=\sigma_0R^2 \).
3. \( My_{cm}=\int\int\sigma_0\sin\theta\,(r\sin\theta)\,r\,dr\,d\theta=\sigma_0\frac{R^3}{3}\cdot\frac\pi2 \), so \( y_{cm}=\frac{\pi R}{6} \).

Teacher's Note:
1. Integrate \( y\,dm \) over the disc and divide by the total mass.

Answer: (b) \( mgl \)

Explanation:
1. The bob reaches the bottom with kinetic energy \( mgl \).
2. In a head-on elastic collision between equal masses the velocities are exchanged, so the block takes all of it: \( mgl \).

Teacher's Note:
1. Equal masses in an elastic head-on collision exchange velocities.

Answer: (c) \( \frac{L\sin\theta}{2} \)

Explanation:
1. No horizontal external force, so the horizontal position of the centre of mass is fixed.
2. The ball moves \( L\sin\theta-x \) horizontally while the block moves \( x \) the other way: \( mx=m(L\sin\theta-x) \).
3. \( x=\frac{L\sin\theta}{2} \).

Teacher's Note:
1. With equal masses each moves half of the relative horizontal displacement.

Answer: (b) \( \frac{3a}{4} \)

Explanation:
1. Strip of width \( dx \) has height \( 2kx^2 \), so \( dm\propto x^2dx \).
2. \( x_{cm}=\frac{\int_0^ax^3dx}{\int_0^ax^2dx}=\frac{a^4/4}{a^3/3}=\frac{3a}{4} \).

Teacher's Note:
1. Weight each strip by its area.

Answer: (d) \( 25 \) cm/s\( ^2 \) eastward

Explanation:
1. The spring pulls both balls with equal and opposite forces: \( m_Aa_A=m_Ba_B \).
2. \( a_A=\frac{250\times10}{100}=25 \) cm/s\( ^2 \), directed opposite to B, that is eastward.

Teacher's Note:
1. Internal forces are equal and opposite, so accelerations are inversely proportional to masses.

Answer: (c) \( \frac h4 \)

Explanation:
1. Speed at the bottom: \( v=\sqrt{2gh} \).
2. After sticking: \( mv=2mV \), so \( V=\frac v2 \).
3. \( H=\frac{V^2}{2g}=\frac{v^2}{8g}=\frac h4 \).

Teacher's Note:
1. The collision halves the speed, so the height is a quarter.

Answer: (d) \( \frac12Iu \)

Explanation:
1. The velocity changes from \( +u \) to \( -2u \), so \( I=m(2u+u)=3mu \) and \( mu=\frac I3 \).
2. Work \( =\Delta K=\frac12m(4u^2-u^2)=\frac32mu^2=\frac32\cdot\frac I3u=\frac12Iu \).

Teacher's Note:
1. Use the work-energy theorem with the impulse-momentum relation.

Answer: (b) \( 187.5 \) N

Explanation:
1. Mass per second \( =\rho Av=1000\times300\times10^{-6}\times25=7.5 \) kg/s.
2. \( F=\frac{dm}{dt}v=7.5\times25=187.5 \) N (and the pressure is \( \frac{187.5}{300\times10^{-6}}=625 \) kPa).

Teacher's Note:
1. Force equals the rate of loss of momentum of the stream.

Answer: (a) (i) and (iii)

Explanation:
1. Elastic collision: \( v_1'=\frac{1-5}{6}u=-2 \) gives \( u=3 \) m/s, and \( v_2'=\frac{2}{6}u=1 \) m/s.
2. Total momentum \( =1\times3=3 \) kg m/s: (i) true. Momentum of the \( 5 \) kg mass \( =5\times1=5 \) kg m/s: (ii) false.
3. \( K_{cm}=\frac12(6)\left(\frac36\right)^2=0.75 \) J: (iii) true. Total \( K=\frac12(1)(9)=4.5 \) J, so (iv) false.

Teacher's Note:
1. Find \( u \) from the elastic collision formulas first.