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Q.

A block of mass 2 kg slides along a frictionless table with a speed of 10 m/sec. A block of mass 5 kg moving at 3 m/sec directly in front of 2 kg mass. A massless spring of spring constant k = 1120 N/m is attached to the back side of 5 kg mass as shown in figure. When the blocks collide the maximum compression in the spring (if the spring does not bend) will be:

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a

1.12 m

b

0.33 m

c

0.25 m

d

0.4 m

answer is A.

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Detailed Solution

At maximum compression, both the block will have common velocity and it is given by:
$v = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}} = \frac{2 (10) + 5 (3)}{2 + 5} = 5 m / s$
Now ,
$\frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2} = \frac{1}{2} {Kx}^{2} + \frac{1}{2} (m_{1} + m_{2}) v^{2}$
$\Rightarrow 1120 x^{2} = 70 \Rightarrow x^{2} = \frac{1}{16} \Rightarrow x = \frac{1}{4} = 0 .25 m$

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