First slide

Second law

Question

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 3 m,/s and 4 m/s respectively, The third piece will be thrown off with a velocity of

Moderate

A

1.5m/s
B

2.0 m/s
C

2.5 m/s
D

3.0 m/s

Solution

See fig.11
$Momentum of A = \frac{M}{4} \times 3 Momentum of B = \frac{M}{4} \times 4 = M Resultant momentum = \sqrt{\{{(\frac{3 M}{4})}^{2} + M^{2}\}} = \frac{5}{4} M Now C must have a momentum 5 / 4 . M along OC ∴ \frac{M}{2} \times V^{'} = \frac{5}{4} M or V^{'} = 2 .5 m / s$

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