A body of mass M at rest explodes into three pieces, two of which of mass M/4 each are thrown off in perpendicular directions with velocities of 3m/s and 4m/s respectively . Find the velocity of third piece.

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Solution

Initial mass = M
initial velocity = 0

After explosion,
m1 = M/4
v1 = 3m/s i (i means along x-direction)
m2 = M/4
v2 = 4m/s j (j means along y-direction)
m3 = M-(M/4 + M/4) = M/2
v3 = V

As there is no external force,
momentum before explosion = momentum after explosion
⇒M(0) = m1v1 + m2v2 + m3v3
⇒ 0 = M/4 (3)i + M/4 (4)j + M/2(V)
⇒ 0 = M [ (3/4)i + 1j + V/2 ]
⇒ 0 = (3/4)i + 1j + V/2
⇒ V/2 = -[(3/4)i + 1j]
⇒ V = 2×-[(3/4)i + 1j]
⇒ V = -[ (3/2)i +(2)j ]
⇒ V = -(3/2)i -2j
|V| = √[(3/2)² + 2²] = 2.5m/s

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