Question:
a boom of mass 9 kg split into two part.one part is mass 3 kg moves with velocity16m/s.t hen the k.e of other part is .
Answer:
By the law of conservation of momentum,
The momentum associated with mass 3 kg= The momentum associated with mass 6 kg
m1v1 = m2v2
3×16 =6×v2
v2 = 8 m/sec
So, kinetic energy of mass 6 kg
=1/2(6)(8)2
= 192 J
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