Question

If a bomb at rest explodes and splits into two equal fragments, the fragments will move in:

(a) One direction and with equal velocities.
(b) Opposite directions with different velocities.
(c) One direction with different velocities.
(d) Opposite directions with equal velocities.

[1 mark]

Answer 1

Solution:

Correct Option: Option (d)
Let the mass of the bomb be M.
$\therefore$The masses of two equal fragments of the bomb is $\frac{M}{2}$ each.
Let $U$ be the initial velocity of the bomb and $v_1$ and $v_2$ be the velocity of the fragments after explosion.
As there is no external force applied in this case,
$\therefore$ momentum before explosion = momentum after explosion.
i.e, $MU=\frac{M}{2}{v}_1+\frac{M}{2}{v}_2$
As bomb was at rest initially, $MU=0$
we have, $\frac{M}{2}{v}_1+\frac{M}{2}{v}_2$ will also be equal to zero.
$\therefore$ $\frac{M}{2}{v}_1+\frac{M}{2}{v}_2=0$
$\frac{M}{2}{v}_1=-\frac{M}{2}{v}_2$
$\Rightarrow$ $v_1=-v_2$
This means the fragments will move in opposite directions with equal velocity.