In the system shown in fig. block of mass M is placed on a smooth horizontal surface. There is a mass less rigid support attached to the block. Block of mass m is placed on the first block and it is connected to the support with a spring of force constant K. There is no friction between the blocks. A bullet of mass $m_{0}$ , moving with speed u hits the block of mass M and gets embedded into it. The collision is instantaneous. Assuming that m always stays over M, calculate the maximum extension in the spring ( in cm ) caused during the subsequent motion.

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answer is B.

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Detailed Solution

After collision velocity of (M + $m_{0}$ ) system is given by momentum conservation.
$(M + m_{0}) V = m_{0} u 10. V = 0 .14 \times 400 V = 5.6 m / s$
Now the system is equivalent to that shown in figure below

Extension is maximum when both blocks have same velocity
$(4 + 10) V_{0} = 10 \times 5 .6 \Rightarrow V_{0} = 4 m / s$
Energy conservation
$\frac{1}{2} {Kx}^{2} + \frac{1}{2} (14) V_{0}^{2} = \frac{1}{2} \times 10 \times 5 {.6}^{2}$
Solving x = 0.1 m