First slide

Third law

Question

A body of mass M at rest explodes into three pieces, two of which of mass $\frac{M}{4}$ 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

1.5 m/s
B

2.0 m/s
C

2.5 m/s
D

3.0 m/s

Solution

Momentum of one piece = $\frac{M}{4} \times 3$
Momentum of the other piece = $\frac{M}{4} \times 4$
Resultant momentum = $\sqrt{\frac{9 M^{2}}{16} + M^{2}} = \frac{5 M}{4}$
The third piece should also have the same momentum.
Let its velocity be v, then
$\frac{5 M}{4} = \frac{M}{2} \times v or v = \frac{5}{2} = 2.5 m / \sec$

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