JEE Challenger
More from System of Particles and Rotational Motion
PhyHardAdv 2017 • Paper 1

Motion of Block and Point Mass on Frictionless Surface

Consider a block of mass MM with a circular cut featuring a frictionless surface as shown. The block rests on a horizontal frictionless surface of a fixed table. Initially, the right edge of the block is at x=0x = 0, in a coordinate system fixed to the table. A point mass mm is released from rest at the topmost point of the path as shown and slides down. When the mass loses contact with the block, its position is xx and its velocity is vv. At that instant, which of the following options is/are correct?

Question Diagram 1

Options

A

The position of the point mass mm is given by x=−2mRM+mx = -\sqrt{2} \frac{mR}{M+m}

B

The speed of the point mass mm is given by v=2gR1+mMv = \sqrt{\frac{2gR}{1+\frac{m}{M}}}

Correct
C

The xx-component of displacement of the center of mass of the block MM is −mRM+m-\frac{mR}{M+m}

Correct
D

The velocity of the block MM is V=−mM2gRV = -\frac{m}{M}\sqrt{2gR}

Topics & Concepts

System of Particles and Rotational MotionCenter of MassRelative MotionKinematics

Step-by-Step Solution

To determine which options are correct, we analyze the motion of the system consisting of the block of mass MM and the point mass mm using the principles of conservation of momentum and conservation of mechanical energy.

1. Analysis of Horizontal Displacement

Since no external force acts on the system along the horizontal (xx) direction, the horizontal component of the position of the center of mass remains unchanged, and the total linear momentum along the xx-axis is conserved.

Let ΔxB\Delta x_B be the xx-component of displacement of the block MM, and Δxm\Delta x_m be the xx-component of displacement of the point mass mm. The condition for zero horizontal displacement of the center of mass gives: MΔxB+mΔxm=0M \Delta x_B + m \Delta x_m = 0

From the given figure:

  • Initially, the point mass mm is at x=−Rx = -R relative to the right edge of the block.
  • When mm slides down to the lowest point of the circular arc and loses contact, its horizontal position relative to the block has moved to the right by a distance equal to the radius RR.

Therefore, the displacement of mass mm relative to the block is: Δxm/B=R\Delta x_{m/B} = R

Using the relation for relative displacement, Δxm=ΔxB+Δxm/B=ΔxB+R\Delta x_m = \Delta x_B + \Delta x_{m/B} = \Delta x_B + R: MΔxB+m(ΔxB+R)=0M \Delta x_B + m (\Delta x_B + R) = 0 (M+m)ΔxB=−mR(M + m) \Delta x_B = -m R ΔxB=−mRM+m\Delta x_B = -\frac{m R}{M + m}

Thus, the xx-component of displacement of the center of mass of the block MM is −mRM+m-\frac{m R}{M + m}. This makes Option C correct.

Now, we calculate the final position xx of the point mass mm: x=xinitial+Δxm=−R+(ΔxB+R)=ΔxB=−mRM+mx = x_{\text{initial}} + \Delta x_m = -R + \left(\Delta x_B + R\right) = \Delta x_B = -\frac{m R}{M + m}

Thus, Option A is incorrect (it incorrectly includes a factor of 2\sqrt{2}).

2. Analysis of Velocities at the Instant of Separation

Let vv be the horizontal speed of the point mass mm and VV be the velocity of the block MM at the moment mass mm leaves the track.

By conservation of linear momentum in the horizontal direction: mv+MV=0  ⟹  V=−mMvm v + M V = 0 \implies V = -\frac{m}{M} v

Since the system is frictionless, mechanical energy is conserved: ΔU+ΔK=0\Delta U + \Delta K = 0 mgR=12mv2+12MV2m g R = \frac{1}{2} m v^2 + \frac{1}{2} M V^2

Substituting V=−mMvV = -\frac{m}{M} v into the energy equation: mgR=12mv2+12M(−mMv)2m g R = \frac{1}{2} m v^2 + \frac{1}{2} M \left(-\frac{m}{M} v\right)^2 mgR=12mv2(1+mM)m g R = \frac{1}{2} m v^2 \left(1 + \frac{m}{M}\right) v2=2gR1+mMv^2 = \frac{2 g R}{1 + \frac{m}{M}} v=2gR1+mMv = \sqrt{\frac{2 g R}{1 + \frac{m}{M}}}

Thus, Option B is correct.

Substituting vv back into the expression for VV: V=−mM2gR1+mMV = -\frac{m}{M} \sqrt{\frac{2 g R}{1 + \frac{m}{M}}}

Thus, Option D is incorrect.

Conclusion

The correct options are B and C.

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