Question

A stationary shell breaks into three
fragments. The momentum of the
two fragments is P each and moves at 60° to each
other. The magnitude of momentum of the
third fragment is
(1) Р
(2) 2P
(3) P/root 3 (4)root 3 P​

Answer 1

Answer:

Option (4) $\sqrt{3}P$

Explanation:

The resultant of the two momentums will be just like the resultant of two forces making an angle $\theta$

$P'=\sqrt{P^2+P^2+2\times P\times P\cos\theta}$

$\implies P'=\sqrt{2P^2+2P^2\cos60^\circ}$

$\implies P'=\sqrt{2P^2+2P^2\cos60^\circ}$

$\implies P'=\sqrt{2P^2+2P^2\times\frac{1}{2}}$

$\implies P'=\sqrt{3P^2}$

$\implies P'=\sqrt{3}P$

From the principle of conservation of linear momentum, the momentum of the third fragment will be just opposite to the resultant momentum of the two fragment

Thus, the momentum of the third fracment = -$\sqrt{3}P$

The magnitude of the momentum = $\sqrt{3}P$

Hope this helps.