Elastic Collision (1D)

In a 1D elastic collision, both momentum and kinetic energy are conserved. These two equations give you a unique solution for the two unknown final velocities. The closed-form answer is so famous it's worth memorizing:

$v_1' = \dfrac{m_1 - m_2}{m_1 + m_2}\,v_1 + \dfrac{2 m_2}{m_1 + m_2}\,v_2$

$v_2' = \dfrac{2 m_1}{m_1 + m_2}\,v_1 + \dfrac{m_2 - m_1}{m_1 + m_2}\,v_2$

When the second ball starts at rest ($v_2 = 0$), these simplify dramatically.

Step-by-Step Solution

Given: $m_1 = 1\;\text{kg}$, $v_1 = 5\;\text{m/s}$, $m_2 = 2\;\text{kg}$, $v_2 = 0$ (at rest).

Find: The final velocities $v_1'$ and $v_2'$ after the elastic collision.

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Step 1 — Apply conservation of momentum.

$p_{\text{before}} = p_{\text{after}}$

$m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'$

$(1)(5) + (2)(0) = (1)v_1' + (2)v_2'$

$5 = v_1' + 2 v_2' \quad\text{...(eq. 1)}$

Step 2 — Apply conservation of kinetic energy.

$\tfrac{1}{2}m_1 v_1^{2} + \tfrac{1}{2}m_2 v_2^{2} = \tfrac{1}{2}m_1 v_1'^{2} + \tfrac{1}{2}m_2 v_2'^{2}$

$\tfrac{1}{2}(1)(25) + 0 = \tfrac{1}{2}(1)(v_1'^{2}) + \tfrac{1}{2}(2)(v_2'^{2})$

$12.5 = 0.5\,v_1'^{2} + v_2'^{2} \quad\text{...(eq. 2)}$

Step 3 — Use the closed-form formula (which encodes both equations).

For $v_2 = 0$:

$v_1' = \dfrac{m_1 - m_2}{m_1 + m_2}\,v_1 = \dfrac{1 - 2}{1 + 2}(5) = \dfrac{-1}{3}(5) \approx -1.667\;\text{m/s}$

$v_2' = \dfrac{2 m_1}{m_1 + m_2}\,v_1 = \dfrac{2(1)}{3}(5) = \dfrac{10}{3} \approx 3.333\;\text{m/s}$

Step 4 — Verify both conservation laws.

Momentum:

$p' = (1)(-1.667) + (2)(3.333) = -1.667 + 6.667 = 5.000 \;\;\checkmark$

Kinetic energy:

$\text{KE}' = \tfrac{1}{2}(1)(2.778) + \tfrac{1}{2}(2)(11.111) = 1.389 + 11.111 = 12.500 \;\;\checkmark$

Both check out — this is a true elastic collision.

Step 5 — Interpret the negative sign.

$v_1' < 0$ means ball 1 bounces back in the opposite direction. That makes sense: it hit a heavier object and got rebuffed. The heavier ball goes forward at $\approx 3.33\;\text{m/s}$.

Step 6 — Equal masses special case (compare).

If $m_1 = m_2$, the formulas give $v_1' = 0$ and $v_2' = v_1$ — the moving ball stops dead and the stationary ball takes off with the original speed. This is the classic Newton's cradle behavior, only because the masses are equal.

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Answer:

$\boxed{\;v_1' = -\dfrac{5}{3} \approx -1.667\;\text{m/s},\quad v_2' = \dfrac{10}{3} \approx 3.333\;\text{m/s}\;}$

The lighter ball bounces backward; the heavier ball moves forward. Total momentum and total kinetic energy are both conserved.

Try It

• Adjust the masses and initial velocities with the sliders.
• Watch the animation: balls approach, collide, and separate at the new velocities.
• The HUD verifies both conservation laws are satisfied at every collision.
• Try equal masses with the second ball at rest — the first ball stops and the second one takes off (Newton's cradle).
• Try a very heavy ball 2 — the lighter ball 1 bounces back at almost its original speed.