A shell moving horizontally at v_0 = 15 m/s explodes into…

A shell moving horizontally at $v_0 = 15\ \mathrm{m/s}$ explodes into two fragments of masses $m_1 = 0.5\ \mathrm{kg}$ and $m_2 = 1\ \mathrm{kg}$. The larger fragment moves at $v_2 = 30\ \mathrm{m/s}$ along the original direction of the shell. Find the velocity of the smaller fragment, which moves along the same line.

Explanation of Method

A shell splitting into two fragments.

Step 1: Momentum conservation

The explosion forces are internal, so the momentum just before equals the momentum just after: $(m_1 + m_2)\,v_0 = m_1 v_1 + m_2 v_2$.

\[ 1.5 \cdot 15 = 0.5\,v_1 + 1 \cdot 30. \]

\[ 0.5\,v_1 = 22.5 - 30 = -7.5. \]

\[v_1 = -15\ \mathrm{m/s} \ \ (15\ \mathrm{m/s} \text{ backward})\]

$\text{before}$ $1.5\ \mathrm{kg},\ 15\ \mathrm{m/s}$ $p = 22.5$ $\text{after}$ $1\ \mathrm{kg}:\ p = 30$ $0.5\ \mathrm{kg}:\ p = -7.5$ $v_1 = -15\ \mathrm{m/s}$

Momentum arrows to scale (in kg·m/s). The larger fragment carries more momentum (30) than the whole shell had (22.5), so the smaller one must carry −7.5, i.e. move backward.

Result

\[\boxed{\,v_1 = -15\ \mathrm{m/s} \ \ (15\ \mathrm{m/s} \text{ backward})\,}\]

Remark

Note. The original result is correct. The statement says both fragments keep the shell's direction, but that is impossible here: the larger fragment alone has more momentum (30) than the whole shell (22.5), so the smaller one must fly backward. The fragments move along the same line, not in the same direction.