Question

A vessel at rest explodes into three pieces. Two pieces having equal masses fly off perpendicular to one another with the same velocity of 30 m/s. The third piece has three times mass of each of the other piece. The magnitude and direction of the velocity of the third piece would be

• A. $10\sqrt{2}m/secondand{135}^{\circ }fromeither$
• B. $10\sqrt{2}m/secondand{45}^{\circ }fromeither$
• C. $\frac{10}{\sqrt{2}}m/secondand{135}^{\circ }fromeither$
• D. $\frac{10}{\sqrt{2}}m/secondand{45}^{\circ }fromeither$

Answer 1

The correct option is A

$10\sqrt{2}m/secondand{135}^{\circ }fromeither$

Let's say two pieces are having equal mass 'm' and third piece have a mass of '3m'.

According to law of conservation of linear momentum. Since the initial momentum of the system was zero, therefore final momentum of the system must be zero i.e. the resultant of momentum of two pieces must be equal to the momentum of third piece. We know that if two particle possess same momentum and angle in between them is ${90}^{\circ }$ then resultant will be given by $P\sqrt{2}=mv\sqrt{2}=m30\sqrt{2}$

Let the velocity of mass 3m is V. So $3mV=30m\sqrt{2}$

$\therefore V=10\sqrt{2}andangle{135}^{\circ }$ from either.