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Three blocks are placed on smooth horizontal surface and lie on same horizontal straight line. Block 1 and block 3 have mass m each and block 2 has mass M (M > m). Block 2 and block 3 are initially stationary, while block 1 is initially moving towards block 2 with speed v as shown. Assume that all collisions are head on and perfectly elastic. What value of M/m ensures that block 1 and block 3 have the same final speed?

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$5 + \sqrt{2}$

$5 - \sqrt{2}$

$2 + \sqrt{5}$

$3 + \sqrt{5}$

answer is C.

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Detailed Solution

Let v₁, v₂ and v₃ be velocities of blocks 1, 2 and 3 after suffering collision each.

$mv = {mv}_{1} + {Mv}_{2} and v_{1} - v_{2} = - v$

Solving, we get $v_{1} = \frac{m - M}{m + M} < 0$

$∴ m < M$

$∴ |v_{1}| = \frac{M - m}{m + M} v$ …(i)

and $v_{3} = \frac{2 m}{m + M} \times v_{2} = \frac{4 Mmv}{(m + M)^{2}}$ …(ii)

$∴ \frac{M - m}{M + m} v = \frac{4 Mmv}{(M + m)^{2}}$

$M^{2} - m^{2} = 4 Mm \Rightarrow \frac{M}{m} = 2 + \sqrt{5}$

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