Q.

A block of mass m moving with a velocity v0 collides with a stationary block of mass M at the back of which a spring of spring constant k is attached, as shown in the figure. Select the correct alternative(s)

Moderate

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By Expert Faculty of Sri Chaitanya

The velocity of centre of mass is mm+Mv0

The initial kinetic energy of the system in the centre of mass frame is 14mMM+mv02

The maximum compression in the spring is v0mMm+M1k

When the spring is in the state of maximum compression the kinetic energy in the centre of mass frame is zero

answer is 1,3,4.

(Detailed Solution Below)

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Detailed Solution

Since, vcm=mvo+M(0)m+M=mvom+MSo, OPTION (A) is correct.At the time of maximum compression both the blocks will be moving with same common velocity. Applying Law of Conservation of Linear Momentum and Conservation of Mechanical Energy, we getmvo=(M+m)v .......(1)12mvo2=12(M+m)v2+12kxmax2 ......(2) Solving (1) and (2), we getxmax=vomM(m+M)1kSo, OPTION (C) is also correct.In the state of maximum compression, both the blocks are moving with the same velocity. Therefore, velocity of centre of mass in the CM frame is zero.Hence OPTION (D) is also correct.

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