How Do You Solve Momentum Conservation Problems for Explosions?

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5. Substitute masses and signed velocities, then solve for the unknown.
6. Check that the directions and units are physically reasonable.

Momentum is conserved because forces between the fragments are internal forces. They are equal and opposite, so they do not change the total momentum of the system. Conservation applies when the resultant external impulse during the short explosion is negligible.

Worked Example

A stationary object of mass $5.0\,\text{kg}$ explodes into two fragments. A $2.0\,\text{kg}$ fragment moves right at $6.0\,\text{m s}^{-1}$. Determine the velocity of the $3.0\,\text{kg}$ fragment.

Take right as positive. Initially, the object is stationary, so:

After the explosion:

The negative sign means the second fragment moves left at $4.0\,\text{m s}^{-1}$.

A common misconception is that kinetic energy must also be conserved. In an explosion, total momentum is conserved, but kinetic energy usually increases because internal energy is transferred into kinetic energy.

An additional check is to compare kinetic energies. Here, the final kinetic energy is $60\,\text{J}$, whereas the initial kinetic energy is zero. The increase came from stored internal energy. This does not violate conservation of energy, because total energy includes forms other than kinetic energy.

Exam Technique

For a Calculate question, show the conservation equation, substitution, and final answer with direction. Never omit signs: writing only speed can lose the vector meaning that the question is testing.