JEE Physics Centre of Mass MCQs Set 07

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

Explanation:
1. Each hit on a given wall gives an impulse \( 2mv_0 \).
2. Time between successive hits on the same wall: \( 2d/v_0 \).
3. \( \bar F=\frac{2mv_0}{2d/v_0}=\frac{mv_0^2}{d} \).

Teacher's Note:
1. Average force = impulse per collision times collisions per second.
2. The ball takes a round trip of 2d to return to the same wall.

Answer: (d) 0.5

Explanation:
1. Let the distance to the wall be d. Going: speed V, time \( d/V \); returning: speed eV, time \( d/(eV) \).
2. \( \text{average speed}=\frac{2d}{d/V+d/(eV)}=\frac{2e}{1+e}V=\frac23V \).
3. \( 3e=1+e\Rightarrow e=0.5 \).

Teacher's Note:
1. Average speed is total distance divided by total time, not the mean of the speeds.
2. Harmonic-mean-type expression: 2e/(1+e).

Answer: (a) \( \frac hd=\frac{1}{1-e^2} \)

Explanation:
1. From a peak h above a step the ball falls h and strikes with speed \( \sqrt{2gh} \).
2. It rebounds with \( e\sqrt{2gh} \) and rises \( e^2h \) above the same step.
3. That peak is h above the next lower step, i.e. \( h-d \) above this one: \( e^2h=h-d\Rightarrow\frac hd=\frac1{1-e^2} \).

Teacher's Note:
1. Measure all heights from the step on which the ball bounces.
2. The next step is lower by d, which reduces the required rise.

Answer: (c) \( 2m\cos\theta\sqrt{2gh} \)

Explanation:
1. Speed before impact \( v=\sqrt{2gh} \) (vertical). The component normal to the incline is \( v\cos\theta \).
2. Elastic and smooth: only the normal component reverses, so \( J=2mv\cos\theta=2m\cos\theta\sqrt{2gh} \).

Teacher's Note:
1. The impulse is perpendicular to the plane (smooth surface).
2. The angle between the vertical and the normal to the incline equals theta.

Answer: (d) \( \frac23 \)

Explanation:
1. For \( m_1 \) to move perpendicular to the line of centres, its component along that line after collision must be zero: \( v_1'=\frac{m_1-e\,m_2}{m_1+m_2}u_n=0 \).
2. \( e=\frac{m_1}{m_2}=\frac23 \).

Teacher's Note:
1. Only the component along the line of centres changes in an oblique collision between smooth spheres.
2. The tangential components remain unchanged.

Answer: (a) \( \frac23 \)

Explanation:
1. The line of centres from the striker to each ball makes 30 degrees with the line of motion.
2. Momentum along the line of motion: \( mv=2mV\cos30^\circ \), so \( V=\frac{v}{\sqrt3} \).
3. Each ball leaves along its line of centres with speed V while the striker stops; relative speed of separation \( =V \), of approach \( =v\cos30^\circ \): \( e=\frac{v/\sqrt3}{v\sqrt3/2}=\frac23 \).

Teacher's Note:
1. The striker stops completely, so its velocity along every direction is zero after impact.
2. Use the components along the line joining the centres for e.

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

Explanation:
1. At contact the line of centres makes angle \( \varphi \) with the velocity where \( \sin\varphi=\frac{R/2}{R}\Rightarrow\varphi=30^\circ \).
2. For equal masses in an elastic collision, the target receives the velocity component along the line of centres: \( v\cos30^\circ=\frac{\sqrt3}{2}v \).

Teacher's Note:
1. The impulse acts along the line of centres because the surfaces are smooth.
2. The striker keeps the component perpendicular to that line.

Answer: (c) \( F=(m_0-\lambda t)\frac{dv}{dt} \)

Explanation:
1. Sand leaks out with the same velocity as the wagon at that instant, so the relative velocity of the lost mass is zero.
2. Hence there is no thrust term: \( F=m\frac{dv}{dt} \) with \( m=m_0-\lambda t \).

Teacher's Note:
1. Variable-mass equations carry an extra term only when the ejected mass has a relative velocity.
2. Sand dropping vertically through the floor carries the horizontal velocity of the wagon.

Answer: (b) 4 kg m/s

Explanation:
1. The train covers a semicircle: take it symmetric about the y-axis, so each element at angle \( \phi \) from the axis has velocity component along the axis \( v\cos\phi \).
2. \( p=\frac{M}{\pi}\int_{-\pi/2}^{\pi/2}v\cos\phi\,d\phi=\frac{2Mv}{\pi} \).
3. \( p=\frac{2\cdot\pi\cdot2}{\pi}=4 \) kg m/s.

Teacher's Note:
1. Velocity vectors point in different directions around the arc, so the momentum is less than Mv = 2 pi.
2. Integrate the component along the symmetry direction; the perpendicular components cancel.

Answer: (c) 60 m

Explanation:
1. The remaining fall of 60 m takes \( t=\sqrt{\frac{2(60)}{10}}=\sqrt{12} \) s for both man and bag (the throw is horizontal).
2. Man's horizontal speed: \( 2/\sqrt{12} \) m/s. Momentum conservation (no horizontal velocity initially): \( 60\cdot\frac{2}{\sqrt{12}}=2\,v_b \), so \( v_b=\frac{60}{\sqrt{12}} \) m/s.
3. Distance \( =v_bt=60 \) m.

Teacher's Note:
1. Both bodies share the same remaining fall time because the throw gives no vertical velocity.
2. The ratio of horizontal distances is the inverse of the mass ratio: 60 : 2 = 30 : 1.

Answer: (d) 4 m/s

Explanation:
1. First shot: \( 49V_1+1(V_1-100)=0\Rightarrow V_1=2 \) m/s.
2. Second shot: the remaining system is 49 kg: \( 49(2)=48V_2+1(V_2-100)\Rightarrow V_2=\frac{198}{49}\approx4.04 \) m/s.
3. To the nearest integer, 4 m/s.

Teacher's Note:
1. The shell's velocity in the ground frame is the car's velocity minus 100.
2. The second shot starts from the car's already non-zero speed.

Answer: (a) 6 m

Explanation:
1. Speed at the bottom: \( v=\sqrt{2gh}=\sqrt{120} \) m/s.
2. Common velocity: \( 3v=5V\Rightarrow V^2=\frac{9\cdot120}{25}=43.2 \).
3. Loss of KE \( =\tfrac12(3)(120)-\tfrac12(5)(43.2)=180-108=72 \) J \( =\mu m g\,x=12x \), so \( x=6 \) m.

Teacher's Note:
1. Friction work on the relative sliding distance equals the loss of kinetic energy.
2. Momentum is conserved horizontally since the cart is on a smooth surface.

Answer: (b) 8

Explanation:
1. Momentum: \( 0.02(400)=0.02(200)+5V\Rightarrow V=0.8 \) m/s.
2. \( V^2=2\mu gd\Rightarrow\mu=\frac{0.64}{2(10)(0.2)}=0.16 \).
3. \( \lambda=50\mu=8 \).

Teacher's Note:
1. The bullet's loss of momentum equals the block's gain, since friction during the short passage is negligible.
2. Then the block decelerates uniformly under friction.

Answer: (d) 70 N

Explanation:
1. Horizontal momentum zero: \( 0.5\,v_A+1\,v_B=0\Rightarrow v_A=-2v_B \).
2. Energy (B falls by the string length \( \ell \)): \( 1\,g\ell=\tfrac12(0.5)(4v_B^2)+\tfrac12v_B^2=1.5\,v_B^2 \).
3. The ring has no horizontal force at that instant, so it is an inertial frame: relative speed \( v_{rel}=3v_B \), \( v_{rel}^2=6g\ell \). Then \( T-mg=\frac{m\,v_{rel}^2}{\ell}\Rightarrow T=10+6(10)=70 \) N.

Teacher's Note:
1. When the string is vertical, the horizontal force on the ring is zero, so the ring does not accelerate.
2. Use the relative speed of B with respect to the ring for the circular-motion equation.

Answer: (a) 10 m/s

Explanation:
1. While \( m_2 \) slides down to the bottom, the wall holds \( m_1 \) in place: \( v_2=\sqrt{2gr}=10 \) m/s at the bottom.
2. Afterwards \( m_1 \) is pushed away from the wall; momentum and energy are both conserved, and for equal masses the speeds are exchanged when \( m_2 \) returns to the bottom.
3. So \( m_1 \) reaches a maximum speed of 10 m/s.

Teacher's Note:
1. The total energy after the groove leaves the wall remains 50 J of kinetic energy.
2. The speed of m1 is greatest when m2 is back at the bottom of the groove with zero speed.

Answer: (b) 30 cm

Explanation:
1. The centre of mass is at rest (momentum \( 3(1)-6(2) \) gives \( v_{cm}=-1 \) m/s) and the relative speed is 3 m/s.
2. Reduced mass \( \mu=\frac{3\cdot6}{9}=2 \) kg.
3. \( \tfrac12\mu v_{rel}^2=\tfrac12kx^2\Rightarrow x^2=\frac{2(9)}{200}=0.09 \), so \( x=0.3 \) m = 30 cm.

Teacher's Note:
1. The kinetic energy relative to the centre of mass is the only energy that can be stored in the spring.
2. Reduced mass times relative speed squared gives the stored energy quickly.

Answer: (c) 10 m

Explanation:
1. Relative to the centre of mass, the effective force on the separation coordinate is \( F_{eff}=\frac{m_2F_1+m_1F_2}{m_1+m_2}=\frac{2(4)+1(2)}{3}=\frac{10}{3} \) N.
2. Maximum elongation occurs when the relative velocity vanishes: \( F_{eff}x=\tfrac12kx^2\Rightarrow x=\frac{2F_{eff}}{k} \).
3. \( x=\frac{2(10/3)}{2/3}=10 \) m.

Teacher's Note:
1. The extension oscillates about the equilibrium extension F_eff/k and reaches twice that at its maximum.
2. Work done by the constant forces on the relative coordinate equals the stored spring energy.

Answer: (a) 1

Explanation:
1. Maximum loss (perfectly inelastic): \( \tfrac12\mu u^2=\tfrac12(1)(4)=2 \) J. For a general e: \( \text{loss}=2(1-e^2) \).
2. \( 2(1-e^2)=1\Rightarrow e^2=\tfrac12\Rightarrow e=\frac1{\sqrt2} \).
3. \( \lambda=1 \).

Teacher's Note:
1. Loss of KE = (1/2) mu u^2 (1 - e^2), where mu is the reduced mass.
2. Half the maximum loss means (1 - e squared) = 1/2.

Answer: (c) 35

Explanation:
1. Initial KE \( =\tfrac12(5)(4)+\tfrac12(10)(3)=25 \) J.
2. Momentum after: \( p^2=10^2+(10\sqrt3)^2=400 \), so KE \( =\frac{400}{2\times15}=\frac{40}{3} \) J.
3. Heat \( =25-\frac{40}{3}=\frac{35}{3} \) J, so \( \lambda=35 \).

Teacher's Note:
1. The momenta are perpendicular, so add their squares to find the final momentum.
2. Heat liberated = initial KE minus final KE.

Answer: (a) 15 N

Explanation:
1. Linear density \( \lambda=1 \) kg/m. Weight of the part already on the table: \( \lambda x\,g=5 \) N.
2. Rate of momentum destruction of the arriving links: \( \lambda v^2 \) with \( v^2=2gx=10 \), so \( 10 \) N.
3. Total force \( N=\lambda xg+\lambda v^2=3\lambda gx=15 \) N.

Teacher's Note:
1. The force has a static part (weight of the heap) and a dynamic part (stopping the falling links).
2. The dynamic part is twice the static part for a freely falling chain.

Answer: (a) (i) and (iv) only

Explanation:
1. Internal explosion forces conserve momentum: (i) true, (iii) false.
2. Chemical or stored energy is converted to kinetic energy, so the kinetic energy increases: (iv) true, (ii) false.

Teacher's Note:
1. An explosion adds kinetic energy but cannot alter the total momentum.
2. This is the reverse of a perfectly inelastic collision.

Answer: (d) (iii) and (iv) only

Explanation:
1. Horizontal momentum is conserved during the impact: the momentum is mv.
2. Common speed \( V=\frac{mv}{M+m} \), so \( KE=\tfrac12(M+m)V^2=\frac{m^2v^2}{2(M+m)} \).

Teacher's Note:
1. During the short impact the thread tension is vertical and cannot change the horizontal momentum.
2. KE after impact is smaller than the bullet's original KE.