Conservation of Momentum: Explosion

When an object explodes at rest, the pieces fly apart in different directions. Even though new kinetic energy appears (from chemical or stored energy), the total momentum is still conserved — because no external force acts during the explosion. The pieces must fly apart in such a way that their momenta add to the original (zero) momentum.

The Conservation Law

Before the explosion: $p_i = M\,v_M = 0$ (object at rest).

After the explosion: $p_f = m_1 v_1 + m_2 v_2$ (two pieces moving).

Setting $p_i = p_f$:

$0 = m_1 v_1 + m_2 v_2$

Solving for $v_2$:

$v_2 = -\dfrac{m_1 v_1}{m_2}$

The negative sign tells us that $v_2$ points opposite to $v_1$ — the two pieces fly apart.

Step-by-Step Solution

Given: Original mass $M = 10\;\text{kg}$ at rest splits into $m_1 = 6\;\text{kg}$ (moving right at $v_1 = +5\;\text{m/s}$) and $m_2 = 4\;\text{kg}$ (moving with unknown velocity $v_2$).

Find: $v_2$ and the kinetic energy released by the explosion.

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Step 1 — Verify the masses add up.

$m_1 + m_2 = 6 + 4 = 10 = M \;\;\checkmark$

Mass is conserved (otherwise we'd have a different problem).

Step 2 — Apply conservation of momentum.

Initial momentum: $p_i = (10)(0) = 0$.

Final momentum:

$p_f = m_1 v_1 + m_2 v_2 = 6(5) + 4 v_2 = 30 + 4 v_2$

Set $p_i = p_f$:

$0 = 30 + 4 v_2$

Step 3 — Solve for $v_2$.

$4 v_2 = -30$

$v_2 = -7.5\;\text{m/s}$

The negative sign means the 4 kg piece moves to the left (opposite direction from the 6 kg piece), at 7.5 m/s.

Step 4 — Verify by computing the total final momentum.

$p_f = (6)(5) + (4)(-7.5) = 30 - 30 = 0 \;\;\checkmark$

Conservation holds.

Step 5 — Compute the kinetic energy released.

Initial KE = 0 (object at rest).

Final KE:

$\text{KE}_f = \tfrac{1}{2}m_1 v_1^{2} + \tfrac{1}{2}m_2 v_2^{2}$

$= \tfrac{1}{2}(6)(25) + \tfrac{1}{2}(4)(56.25)$

$= 75 + 112.5$

$= 187.5\;\text{J}$

So the explosion released 187.5 J of energy from chemical bonds (or stored elastic energy, etc.) — kinetic energy is not conserved in an explosion. Momentum is.

Step 6 — Notice the lighter piece moves faster.

The 4 kg piece moves at 7.5 m/s while the 6 kg piece moves at only 5 m/s. The lighter object always recoils faster — this is exactly the same physics as a gun's recoil. A heavy bullet leaves the muzzle at high speed, and the much heavier rifle pushes back at a low speed.

The ratio of speeds is the inverse of the ratio of masses:

$\dfrac{|v_2|}{|v_1|} = \dfrac{m_1}{m_2} = \dfrac{6}{4} = 1.5 \;\;\checkmark$

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Answer:

$\boxed{\;v_2 = -\dfrac{m_1 v_1}{m_2} = -7.5\;\text{m/s}\;}$

The 4 kg piece moves to the left at 7.5 m/s. The total momentum after the explosion is zero (just like before), and the explosion released 187.5 J of kinetic energy.

Try It

• Adjust the mass split and the velocity of piece 1 with the sliders.
• Watch the two pieces fly apart in opposite directions.
• The HUD verifies that $\sum p = 0$ at all times after the explosion.
• Try equal masses — both pieces move with the same speed in opposite directions.
• Try a very heavy piece 1 — the lighter piece 2 shoots away much faster (like a bullet leaving a gun).