Solution:
As the bomb was initially at rest and no external force acts on it total momentum of the bomb should remain constant.
so, (m/4) 3 i +(m/4) 4 j + (m/2) v1 = 0
v1 = 3/2 i + 4/2 j
|V1|= 2.5 m/s
A bomb of mass M at rest explodes into three pieces, two of which of mass M/4 each, are thrown off in perpendicular directions with speeds of 3 m/s and 4 m/s. The third piece is thrown off with a speed
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