Questions
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
detailed solution
Correct option is C
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/sTalk to our academic expert!
Similar Questions
Two similar springs P and Q have spring constants such that Kp > . They are stretched first by the same amount (case a), then by the same force (case b). The work done by the springs are related as, in case (a) and case (b), respectively.
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