JEE Physics System Of Particles and Rotational Motion MCQs Set 51

Answer: (c) \( 0.8 \) N s

Explanation:
1. The speed is \( \frac22=1 \) m/s, and the velocity reverses from \( +1 \) m/s to \( -1 \) m/s at each turning point.
2. Impulse \( =m\Delta v=0.4\times2=0.8 \) N s.

Teacher's Note:
1. The impulse equals the change of momentum at each turning point.

Answer: (d) \( \frac{v_0}{2\sqrt5} \)

Explanation:
1. Vertical momentum of the small ball: \( M\left(v_0-\frac{v_0}{2}\right)=2J\cos\phi \), so \( J=\frac{Mv_0}{4\cos\phi} \).
2. Horizontal speed of a bigger ball: \( v=\frac{J\sin\phi}{M}=\frac{v_0}{4}\tan\phi=\frac{v_0}{2\sqrt5} \) (using \( \tan\phi=\frac{2}{\sqrt5} \)).

Teacher's Note:
1. Same geometry as before with half the impulse.

Answer: (d) the trolley moves towards the right, with velocity \( \frac{m_1u_{rel}}{m_1+m_2+M} \), and the centre of mass of the system stays at rest

Explanation:
1. No external horizontal force: \( m_1(v-u_{rel})+(m_2+M)v=0 \).
2. \( v=\frac{m_1u_{rel}}{m_1+m_2+M} \), positive (towards the right), and the centre of mass stays at rest.

Teacher's Note:
1. The rest of the system recoils to the right with momentum equal to that of the jumper.

Answer: (b) elastic

Explanation:
1. Momentum: \( 2m v=2m\frac v3+mv_2 \), so \( v_2=\frac{4v}{3} \).
2. \( e=\frac{v_2-v_1}{v}=\frac{\frac{4v}{3}-\frac v3}{v}=1 \), so the collision is elastic.

Teacher's Note:
1. A coefficient of restitution of \( 1 \) means an elastic collision.

Answer: (d) (i), (ii) and (iii)

Explanation:
1. Equal impulse on equal masses gives the same centre-of-mass velocity, so (i) holds.
2. B also rotates, so its total kinetic energy is larger and depends on the distance of the line of action from its centre. Thus (ii) and (iii) hold; (iv) is false.

Teacher's Note:
1. Linear speed depends on the impulse alone; rotational kinetic energy depends on the point of application.

Answer: (a) \( v\sqrt{\frac mk} \)

Explanation:
1. In the frame of end A, the block starts with speed \( v \) and the spring unstretched.
2. \( \frac12mv^2=\frac12kx^2 \), so \( x=v\sqrt{\frac mk} \).

Teacher's Note:
1. Energy conservation in the frame of the moving end.

Answer: (c) \( \frac{m_1v}{m_1-m_2} \)

Explanation:
1. Momentum conservation: \( m_1v=m_2(0)+(m_1-m_2)v' \).
2. \( v'=\frac{m_1v}{m_1-m_2} \).

Teacher's Note:
1. The other fragment has mass \( m_1-m_2 \); apply momentum conservation.

Answer: (b) \( \frac{5L}{9} \)

Explanation:
1. \( \lambda(x)=\lambda_0\left(1+\frac xL\right) \) with \( x \) measured from A.
2. \( x_{cm}=\frac{\int_0^Lx\lambda\,dx}{\int_0^L\lambda\,dx}=\frac{\frac{L^2}{2}+\frac{L^2}{3}}{L+\frac L2}=\frac{5L}{9} \).

Teacher's Note:
1. The centre of mass shifts towards the denser end.

Answer: (b) \( \frac{kx_0^2}{4} \)

Explanation:
1. From the previous result, \( v_A^2=\frac{kx_0^2}{2m} \).
2. Work done on A \( =\frac12mv_A^2=\frac{kx_0^2}{4} \).

Teacher's Note:
1. By the work-energy theorem, work equals the kinetic energy gained from rest.

Answer: (a) \( \frac{mMu^2}{2d(M+m)} \)

Explanation:
1. Common velocity: \( V=\frac{mu}{M+m} \).
2. Loss of kinetic energy \( =\frac12mu^2-\frac12(M+m)V^2=\frac{mMu^2}{2(M+m)} \).
3. This equals \( F\cdot d \), so \( F=\frac{mMu^2}{2d(M+m)} \).

Teacher's Note:
1. Work done by the resistive force over the relative distance equals the loss of kinetic energy.

Answer: (b) twice the original impulse

Explanation:
1. Impulse \( =\int F\,dt=F\,\Delta t \) for the same force.
2. Doubling the contact time doubles the impulse.

Teacher's Note:
1. Impulse is force multiplied by contact time.

Answer: (d) (i), (ii) and (iii)

Explanation:
1. Momentum gives \( v_B=6 \) m/s, reversed relative to its original direction, so (i) is true.
2. \( e=\frac{6-1}{25}=0.2 \) so (ii) is true.
3. \( K_i=\frac12(441)+\frac12(2)(16)=236.5 \) J and \( K_f=\frac12+\frac12(2)(36)=36.5 \) J, so the loss is \( 200 \) J: (iii) true.
4. Impulse on B \( =2(6+4)=20 \) N s, not \( 40 \), so (iv) is false.

Teacher's Note:
1. Always check each statement numerically; the impulse here is \( 20 \) N s.

Answer: (a) \( \frac{mv^2}{4} \)

Explanation:
1. Elastic collision between identical blocks: C stops and A moves with speed \( v \).
2. At maximum compression A and B move together with \( \frac v2 \).
3. \( K=\frac12(2m)\left(\frac v2\right)^2=\frac{mv^2}{4} \).

Teacher's Note:
1. At maximum compression the two blocks share the centre-of-mass velocity.

Answer: (c) \( \frac L2(1-\cos\theta) \) vertically downward

Explanation:
1. The horizontal position of the centre of mass is unchanged.
2. The ball falls \( L(1-\cos\theta) \) and the block stays at the same level, so the centre of mass falls \( \frac L2(1-\cos\theta) \).

Teacher's Note:
1. Only the ball changes height; the centre of mass drops by half of that.

Answer: (d) the rubber ball suffers a greater change in momentum

Explanation:
1. Metal ball: \( \Delta p=mv \). Rubber ball: \( \Delta p=mv-(-mv') \), which is larger.
2. Hence the rubber ball changes momentum more.

Teacher's Note:
1. A rebound reverses the momentum, giving a bigger change.

Answer: (c) \( 0.60\,v \) to the left

Explanation:
1. A and B have equal masses, so A stops and B moves with \( v \).
2. B hits C: \( v_B=\frac{m-4m}{5m}v=-0.6v \) and \( v_C=0.4v \).
3. B (moving left at \( 0.6v \)) hits the stationary A: velocities exchange, so A moves left at \( 0.6v \) and B stops. No more collisions.

Teacher's Note:
1. Equal masses exchange velocities; use \( \frac{m_1-m_2}{m_1+m_2} \) for the unequal pair.

Answer: (c) It could be non-zero, but it must be constant

Explanation:
1. With no external force, \( \frac{d\vec P}{dt}=0 \), so \( \vec P \) is constant.
2. The constant need not be zero.

Teacher's Note:
1. Total momentum is conserved, but its constant value can be anything.

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

Explanation:
1. \( x_{cm}=\frac{\int_0^Lx\cdot x^n\,dx}{\int_0^Lx^n\,dx}=\frac{n+1}{n+2}L \).
2. For \( n=2 \): \( \frac34L \).

Teacher's Note:
1. The general result is \( x_{cm}=\frac{n+1}{n+2}L \).

Answer: (d) \( \frac\pi4 \)

Explanation:
1. At the top the first particle has velocity \( u_0\cos\alpha \) horizontally.
2. The second particle has risen to the same height \( \frac{u_0^2\sin^2\alpha}{2g} \), so its vertical speed there is \( \sqrt{u_0^2-u_0^2\sin^2\alpha}=u_0\cos\alpha \).
3. Equal horizontal and vertical momentum components: the composite moves at \( 45^\circ=\frac\pi4 \).

Teacher's Note:
1. The two velocity components turn out to be equal.

Answer: (b) \( \frac{\pi}{56} \) s

Explanation:
1. \( \omega=\sqrt{\frac k\mu}=\sqrt{\frac{1120\times7}{10}}=28 \) rad/s, so \( T=\frac{2\pi}{28} \) s.
2. The spring starts at natural length and stretches first: maximum extension after \( \frac T4=\frac\pi{56} \) s.

Teacher's Note:
1. The extension varies as \( \sin\omega t \) starting from zero.

Answer: (a) \( \frac{2}{5\sqrt3} \) m

Explanation:
1. \( v_{rel}=4-2=2 \) m/s, \( \mu=\frac{2\times4}{6}=\frac43 \) kg.
2. \( x=v_{rel}\sqrt{\frac{\mu}{k}}=2\sqrt{\frac{4}{300}}=\frac{2}{5\sqrt3}\text{ m} \)

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

Answer: (a) \( \sqrt{gL(1-\cos\theta)} \)

Explanation:
1. Block and ball have equal and opposite speeds when the string is vertical.
2. \( mgL(1-\cos\theta)=2\cdot\frac12mv^2 \) gives \( v=\sqrt{gL(1-\cos\theta)} \).

Teacher's Note:
1. The speed of the block is greatest when the ball passes the lowest point.

Answer: (b) \( \sqrt{105} \) m/s

Explanation:
1. Speed after B is \( \sqrt{45} \) m/s (frictionless).
2. \( v_C^2=45+2\times10\times3=105 \), so \( v_C=\sqrt{105} \) m/s.

Teacher's Note:
1. On a smooth incline the speed changes only through the height fallen.

Answer: (a) \( 0.2 \)

Explanation:
1. Momentum: \( 1(21)+2(-4)=1(1)+2v_B \), so \( v_B=6 \) m/s.
2. \( e=\frac{v_B-v_A}{u_A-u_B}=\frac{6-1}{21-(-4)}=\frac{5}{25}=0.2 \).

Teacher's Note:
1. \( e \) is the ratio of the speed of separation to the speed of approach.

Answer: (a) the total momentum of the ball and the earth is conserved

Explanation:
1. The ball's momentum reverses and shrinks, so its own momentum is not conserved.
2. The ball and the earth together form a system with no external force during the collision, so their total momentum is conserved.

Teacher's Note:
1. Momentum is conserved for the combined system of ball and earth.

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

Explanation:
1. Total momentum is zero: \( \vec v_1+\vec v_2+\vec v_3=0 \) for equal masses.
2. \( \vec v_3=-[(3-1)\hat i+(2-4)\hat j]=-2\hat i+2\hat j \).

Teacher's Note:
1. Zero initial momentum means the vector sum of the fragment momenta is zero.

Answer: (c) \( 65 \) m/s towards east

Explanation:
1. The truck is so heavy that its velocity is unchanged. In the truck's frame the ball approaches at \( 20+25=45 \) m/s and rebounds at \( 45 \) m/s.
2. Ground velocity of the ball \( =20+45=65 \) m/s towards east.

Teacher's Note:
1. Elastic collision with a very heavy body: relative velocity reverses.