Q.

A spring is held compressed. Its stored energy is 2. 4 joule. Its ends are in contact with masses of 1 gm and 48 gm placed on a smooth horizontal surface. When the spring is released, the mass will acquire a velocity of

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a

2¯⋅40149m/s

b

24×4948m/s

c

107m/s

d

1047m/s

answer is C.

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Detailed Solution

12×11000×v12+12×481000×v22=2⋅4  or  v12+48v22=2⋅4×2000=4800 Applying the law of conservation of momentum 11000v1=481000v2  or  v1=48v2 Frorn eq. [1), we have 48v22+48v22=4800  Solving, we get v2=(10/7)m/s
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