JEE Physics System Of Particles and Rotational Motion MCQs Set 52

Answer: (c) \( \frac12\vec I\cdot(\vec v_1+\vec v_2) \)

Explanation:
1. \( \vec I=m(\vec v_2-\vec v_1) \).
2. \( \Delta K=\frac12m(v_2^2-v_1^2)=\frac12m(\vec v_2-\vec v_1)\cdot(\vec v_2+\vec v_1)=\frac12\vec I\cdot(\vec v_1+\vec v_2) \).

Teacher's Note:
1. Factor the difference of squares using dot products.

Answer: (a) \( x_0\sqrt{\frac{k}{2m}} \)

Explanation:
1. Energy released: \( \frac12kx_0^2-\frac12k\frac{x_0^2}{4}=\frac38kx_0^2 \). Momentum: \( v_B=\frac{v_A}{2} \).
2. \( \frac12mv_A^2+\frac12(2m)\frac{v_A^2}{4}=\frac34mv_A^2=\frac38kx_0^2 \), so \( v_A=x_0\sqrt{\frac{k}{2m}} \).

Teacher's Note:
1. Use momentum conservation and the energy lost by the spring.

Answer: (d) is equal to \( g \)

Explanation:
1. Only gravity acts on each ball, so each has acceleration \( g \) downward.
2. \( a_{cm}=\frac{m_1g+m_2g}{m_1+m_2}=g \).

Teacher's Note:
1. The only external force is gravity, so the centre of mass accelerates at \( g \).

Answer: (b) \( \frac{2r}{3(4-\pi)} \)

Explanation:
1. Measure distances from O into the plate. Rectangle: area \( 2r^2 \), centre at \( \frac r2 \). Removed semicircle: area \( \frac{\pi r^2}{2} \), centre at \( \frac{4r}{3\pi} \).
2. \( x=\frac{2r^2\cdot\frac r2-\frac{\pi r^2}{2}\cdot\frac{4r}{3\pi}}{2r^2-\frac{\pi r^2}{2}}=\frac{r^3/3}{r^2(4-\pi)/2}=\frac{2r}{3(4-\pi)} \).

Teacher's Note:
1. Treat the cut-out as a body of negative mass.

Answer: (c) \( 2(m_1+m_2)gt_0 \)

Explanation:
1. The collision is internal, so the only external force is gravity \( (m_1+m_2)g \).
2. Change in total momentum over \( 2t_0 \) is \( (m_1+m_2)g\cdot2t_0 \).

Teacher's Note:
1. Change in momentum equals the external impulse, here due to gravity alone.

Answer: (a) \( \frac{mR}{M+m} \)

Explanation:
1. The horizontal centre of mass of the system does not move. The point mass moves a horizontal distance \( R-x \) relative to the ground while the block moves \( x \) the other way.
2. \( m(R-x)=Mx \) gives \( x=\frac{mR}{M+m} \).

Teacher's Note:
1. Relative horizontal displacement is \( R \).

Answer: (b) \( mg(5-4\cos\theta) \)

Explanation:
1. Horizontal momentum is zero, so the ball and block have equal and opposite speeds \( v \). Energy: \( mgL(1-\cos\theta)=2\cdot\frac12mv^2 \), so \( v^2=gL(1-\cos\theta) \).
2. The block has no acceleration at this instant (string vertical), so the block frame is inertial. Relative speed of the ball is \( 2v \).
3. \( T-mg=\frac{m(2v)^2}{L}=4mg(1-\cos\theta) \), so \( T=mg(5-4\cos\theta) \).

Teacher's Note:
1. At the vertical position the block does not accelerate, so use the circular-motion equation in its frame.

Answer: (a) \( \left(\frac{n-1}{n+1}\right)^2g \)

Explanation:
1. Each mass has acceleration \( a=\frac{n-1}{n+1}g \) in opposite directions.
2. \( a_{cm}=\frac{na-1\cdot a}{n+1}=\frac{(n-1)a}{n+1}=\left(\frac{n-1}{n+1}\right)^2g \).

Teacher's Note:
1. The centre-of-mass acceleration is the mass-weighted average of the two (opposite) accelerations.

Answer: (c) \( 25 \) cm

Explanation:
1. \( v_{rel}=1.8+2.2=4 \) m/s, \( \mu=\frac{3\times6}{9}=2 \) kg.
2. \( x=v_{rel}\sqrt{\frac\mu k}=4\sqrt{\frac{2}{512}}=0.25 \) m.

Teacher's Note:
1. Use reduced mass and relative speed.

Answer: (d) \( 6 \) cm

Explanation:
1. \( k=10^4 \) N/m. In the frame of the spring's free end (inertial), the block's initial speed is \( 2+1=3 \) m/s.
2. \( \frac12kx^2=\frac12mv^2 \) gives \( x=3\sqrt{\frac{4}{10^4}}=0.06 \) m \( =6 \) cm.

Teacher's Note:
1. Work in the frame moving with the free end, where it is at rest.

Answer: (b) \( \left(\frac{5a}{6},\frac{5a}{6}\right) \)

Explanation:
1. Full square: mass \( 4 \) units at \( (a,a) \). Removed square: mass \( 1 \) unit at \( \left(\frac{3a}{2},\frac{3a}{2}\right) \).
2. \( x_{cm}=\frac{4a-\frac{3a}{2}}{3}=\frac{5a}{6} \), and \( y_{cm}=\frac{5a}{6} \) by symmetry.

Teacher's Note:
1. Treat the cut-out as negative mass.

Answer: (b) \( \frac L4 \)

Explanation:
1. No horizontal external force, so the centre of mass stays fixed. Let the man move \( x \) and the plank move \( L-x \) the other way.
2. \( Mx=\frac M3(L-x) \) gives \( 3x=L-x \), so \( x=\frac L4 \).

Teacher's Note:
1. With no external horizontal force, the displacements of the parts are inversely proportional to their masses.

Answer: (d) \( v_0\sqrt{\frac{m}{2k}} \)

Explanation:
1. Block 1 stopping means block 2 acquires \( v_0 \) (coefficient of restitution \( 1 \)).
2. Blocks 2 and 3 with reduced mass \( \frac m2 \): \( \frac12kx^2=\frac12\cdot\frac m2v_0^2 \).
3. \( x=v_0\sqrt{\frac m{2k}} \).

Teacher's Note:
1. Use the reduced mass of the two blocks joined by the spring.

Answer: (a) \( \frac{7}{16}M\sqrt{g\ell} \)

Explanation:
1. The chain falls freely, so its speed is \( v=\sqrt{2g\cdot\frac\ell8}=\frac{\sqrt{g\ell}}{2} \).
2. The portion above the table has mass \( \frac78M \), so \( p=\frac78M\cdot\frac{\sqrt{g\ell}}{2}=\frac7{16}M\sqrt{g\ell} \).

Teacher's Note:
1. Only the part still in the air carries momentum.

Answer: (a) \( 100 \) m

Explanation:
1. Initial height of the centre of mass \( =\frac{0+40}{2}=20 \) m, initial velocity \( \frac{50+30}{2}=40 \) m/s, acceleration \( -g \).
2. \( H=20+\frac{40^2}{2\times10}=20+80=100 \) m.

Teacher's Note:
1. Treat the centre of mass as a single particle under gravity.

Answer: (b) \( 56\% \)

Explanation:
1. Momentum after: \( (2mv,\ 2mv) \) with magnitude \( 2\sqrt2\,mv \), so \( V=\frac{2\sqrt2}{3}v \).
2. \( K_i=\frac12m(4v^2)+\frac12(2m)v^2=3mv^2 \) and \( K_f=\frac12(3m)\frac{8v^2}{9}=\frac{4mv^2}{3} \).
3. Loss \( =1-\frac49=\frac59\approx56\% \).

Teacher's Note:
1. Work in components for perpendicular momenta.

Answer: (c) \( \frac{Ft}{M_0+\mu t} \)

Explanation:
1. The sand falls with zero horizontal momentum, so \( F=\frac{d(mv)}{dt} \) and \( mv=Ft \).
2. \( (M_0+\mu t)v=Ft \), so \( v=\frac{Ft}{M_0+\mu t} \).

Teacher's Note:
1. Apply \( F=\frac{dp}{dt} \) with the mass changing.

Answer: (d) \( \frac{4v}{A-4} \)

Explanation:
1. The daughter has mass number \( A-4 \) and the alpha particle has mass number \( 4 \).
2. \( 4v=(A-4)V \), so \( V=\frac{4v}{A-4} \).

Teacher's Note:
1. Use momentum conservation with mass numbers as relative masses.

Answer: (c) \( \frac{v_0}{\sqrt5} \)

Explanation:
1. Centre distance from the small ball to each big ball is \( \frac{3R}{2} \) and the horizontal offset is \( R \), so the line of centres makes angle \( \phi \) with the vertical where \( \sin\phi=\frac23 \), \( \cos\phi=\frac{\sqrt5}{3} \).
2. Vertical momentum of the small ball: \( Mv_0=2J\cos\phi \), so \( J=\frac{Mv_0}{2\cos\phi} \).
3. The ground cancels vertical motion of the big balls, so \( Mv=J\sin\phi \) and \( v=\frac{v_0}{2}\tan\phi=\frac{v_0}{\sqrt5} \).

Teacher's Note:
1. The impulse acts along the line of centres; the frictionless ground takes up the vertical part.

Answer: (a) \( 30 \) N

Explanation:
1. \( F=\frac{\Delta p}{\Delta t}=\frac{0.15\times20}{0.1}=30 \) N.

Teacher's Note:
1. Force equals change of momentum per unit time.

Answer: (c) \( \frac{3h}{4} \)

Explanation:
1. Take discs of radius \( r=\frac{R}{h}z \) at distance \( z \) from the vertex: \( dm\propto z^2dz \).
2. \( z_0=\frac{\int_0^hz^3dz}{\int_0^hz^2dz}=\frac{h^4/4}{h^3/3}=\frac{3h}{4} \).

Teacher's Note:
1. The centre of mass of a cone is one quarter of the height from the base.

Answer: (a) \( \sqrt{45} \) m/s

Explanation:
1. Speed just before B: \( \sqrt{2\times10\times3}=\sqrt{60} \) m/s, directed \( 60^\circ \) below horizontal.
2. Totally inelastic: the component normal to the second incline is lost, leaving the component along it. The angle between directions is \( 30^\circ \).
3. \( v=\sqrt{60}\cos30^\circ=\sqrt{45} \) m/s.

Teacher's Note:
1. Keep only the velocity component along the new surface.

Answer: (b) \( \frac{mL}{M+m} \)

Explanation:
1. The centre of mass of insect plus straw does not move horizontally.
2. Insect moves \( L-x \) relative to the ground and the straw moves \( x \) in the opposite direction: \( m(L-x)=Mx \).
3. \( x=\frac{mL}{M+m} \).

Teacher's Note:
1. Relative displacement splits inversely with mass.

Answer: (d) \( 288 \) J

Explanation:
1. Momentum: \( 4v=12\times4 \), so \( v=12 \) m/s.
2. \( K=\frac12(4)(12)^2=288 \) J.

Teacher's Note:
1. Equal and opposite momenta; the lighter piece gets more energy.

Answer: (d) \( 0.67 \) J

Explanation:
1. \( V=\frac{0.5\times2}{1.5}=0.667 \) m/s.
2. \( K_i=\frac12(0.5)(4)=1 \) J, \( K_f=\frac12(1.5)(0.667)^2=0.333 \) J. Loss \( =0.67 \) J.

Teacher's Note:
1. Loss in a perfectly inelastic collision is \( \frac12\mu v_{rel}^2 \).

Answer: (b) \( -0.2 \) kg m/s

Explanation:
1. Velocity before \( t=2 \) s is \( \frac42=2 \) m/s; after it is \( 0 \).
2. Impulse \( =m(0-2)=0.1\times(-2)=-0.2 \) kg m/s.

Teacher's Note:
1. The slope of an \( x \)-\( t \) graph gives velocity; impulse is the change in momentum.

Answer: (c) \( \frac{md}{M+m} \)

Explanation:
1. Resistive force \( F=\frac{mMu^2}{2d(M+m)} \) acts on the block.
2. \( Fs=\frac12MV^2=\frac{Mm^2u^2}{2(M+m)^2} \), so \( s=\frac{md}{M+m} \).

Teacher's Note:
1. Apply the work-energy theorem to the block alone.