A grenade moving at v = 400 m/s breaks into two equal parts.…

A grenade moving at $v = 400\ \mathrm{m/s}$ breaks into two equal parts. Each part moves along a straight line at $30^\circ$ to the original direction of motion (one on each side). Find the speeds of the parts.

Explanation of Method

A grenade splitting into two equal parts.

Step 1: Sideways momentum

Before the break there is no sideways momentum, so the sideways parts must cancel: $\tfrac{m}{2}v_1\sin 30^\circ - \tfrac{m}{2}v_2\sin 30^\circ = 0 \;\Longrightarrow\; v_1 = v_2$.

Step 2: Forward momentum

\[ m v = \tfrac{m}{2}v_1\cos 30^\circ + \tfrac{m}{2}v_2\cos 30^\circ = m\,v_1\cos 30^\circ. \]

\[ v_1 = \frac{v}{\cos 30^\circ} = \frac{400}{0.866}. \]

\[v_1 = v_2 \approx 462\ \mathrm{m/s}\]

$\vec p = m\vec v$ $\tfrac{m}{2}\vec v_1$ $\tfrac{m}{2}\vec v_2$ $30^\circ$ $30^\circ$

The two fragment momenta must add up to the grenade's momentum. Their sideways parts cancel, and each forward part is $\tfrac{m}{2}v_1\cos30^\circ$, so each must equal half of $mv$.

Result

\[\boxed{\,v_1 = v_2 \approx 462\ \mathrm{m/s}\,}\]

Remark

Check. The original result (461.9 m/s) is correct. The equal speeds do not need to be assumed: they follow from the sideways momentum balance. The kinetic energy grows by a factor $1/\cos^2 30^\circ = 4/3$; the extra energy comes from the explosion.