Two masses $m_{1} = 2 k g$ and $m_{2} = 5 k g$ are moving on a frictionless surface with velocities 10 m/s and 3 m/s respectively. $m_{2}$ is ahead of $m_{1}$ . An ideal spring of spring constant $k = 1120 N / m$ is attached on the back side of $m_{2}$ .The maximum compression of the spring will be

0.51 m

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0.062 m

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0.37 m

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0.72 m

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Solution

The correct option is A $0.37 m$
At the time of maximum compression of spring velocity of both blocks will same.
By momentum conservation,
$m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) u$
$2 \times 10 + 5 \times 3 = (2 + 5) u$
$25 = 7 u$
$u = \frac{25}{7} m / s = 3.57 m / s$
By energy conservation,
$\frac{1}{2} m_{1} v_{1}^{2}$ + $\frac{1}{2} m_{2} v_{2}^{2}$ = $\frac{1}{2} m_{1} u^{2}$ + $\frac{1}{2} m_{2} u^{2}$ + $\frac{1}{2} k x^{2}$
$\frac{1}{2} \times 2 \times 100 + \frac{1}{2} \times 5 \times 9 = \frac{1}{2} \times 2 \times (3.57)^{2} + \frac{1}{2} \times 5 \times (3.57)^{2} + \frac{1}{2} k x^{2}$
$122.5 = 44.6 + \frac{1}{2} k x^{2}$
$\frac{1}{2} k x^{2} = 77.9$
$x^{2} = \frac{2 (77.9)}{k} = \frac{77.9}{1120} \times 2$
$x = 0.37$

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