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Q.

Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass. Spring Constant of spring is K = 140 N/M and blocks are placed on a frictionless horizontal surface. An impulse gives a speed of 14 m/s to 10 kg block in the direction of the lighter block. Then, the maximum compression in the spring will be

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a

1 m

b

2 m

c

3 m

d

4 m

answer is B.

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Detailed Solution

12m1m3m1+m2v1-v22=12kx212×4014(14)2=12×140x24=x2X = 2m
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Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass. Spring Constant of spring is K = 140 N/M and blocks are placed on a frictionless horizontal surface. An impulse gives a speed of 14 m/s to 10 kg block in the direction of the lighter block. Then, the maximum compression in the spring will be