Perfectly Inelastic Collision

In a perfectly inelastic collision, the two objects stick together and move as a single combined mass. Momentum is conserved, but kinetic energy is not — some of it gets converted to heat, sound, or deformation. This is the simplest collision type after elastic.

The Formula

For two objects of masses $m_1, m_2$ with initial velocities $v_1, v_2$, that stick together after impact:

$m_1 v_1 + m_2 v_2 = (m_1 + m_2)\,v'$

Solving for the final velocity:

$v' = \dfrac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$

This is just the center-of-mass velocity of the system — the entire combined object continues moving at the velocity that the center of mass had all along.

Step-by-Step Solution

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

Find: The final velocity $v'$ of the combined object, and how much kinetic energy was lost.

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Step 1 — Apply conservation of momentum.

$m_1 v_1 + m_2 v_2 = (m_1 + m_2)\,v'$

$(1)(5) + (2)(0) = (1 + 2)\,v'$

$5 = 3\,v'$

$v' = \dfrac{5}{3} \approx 1.667\;\text{m/s}$

Step 2 — Compute kinetic energy before the collision.

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

$= \tfrac{1}{2}(1)(25) + 0 = 12.5\;\text{J}$

Step 3 — Compute kinetic energy after the collision.

$\text{KE}_{\text{after}} = \tfrac{1}{2}(m_1 + m_2)\,v'^{2}$

$= \tfrac{1}{2}(3)(5/3)^{2} = \tfrac{1}{2}(3)\cdot\dfrac{25}{9}$

$= \dfrac{75}{18} = \dfrac{25}{6} \approx 4.167\;\text{J}$

Step 4 — Compute the energy lost.

$\Delta\text{KE} = \text{KE}_{\text{before}} - \text{KE}_{\text{after}} = 12.5 - 4.167 = 8.333\;\text{J}$

Express as a fraction:

$\dfrac{\Delta\text{KE}}{\text{KE}_{\text{before}}} = \dfrac{8.333}{12.5} \approx 0.667 = 66.7\%$

So two-thirds of the kinetic energy disappears into heat and deformation when these two specific masses collide inelastically.

Step 5 — A general formula for the energy loss.

For a perfectly inelastic collision with one mass at rest, the fraction of KE lost is:

$\dfrac{\Delta\text{KE}}{\text{KE}_{\text{initial}}} = \dfrac{m_2}{m_1 + m_2}$

For our case: $2/(1 + 2) = 2/3 \approx 66.7\%$. ✓

Notice the larger the second mass relative to the first, the more KE is lost. If $m_2 \gg m_1$, almost all the energy disappears (think of throwing a ball at a wall).

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Answer:

$\boxed{\;v' = \dfrac{5}{3} \approx 1.667\;\text{m/s},\quad \Delta\text{KE} = \dfrac{25}{3} \approx 8.333\;\text{J lost}\;(66.7\%)\;}$

Momentum is conserved (5 kg·m/s before, 5 kg·m/s after). But KE is not — about 67% of it converts to non-mechanical forms (heat, sound, deformation). This is what distinguishes inelastic collisions from elastic ones.

Try It

• Adjust the masses and initial velocity with the sliders.
• Watch the two balls combine into one larger object after impact.
• The HUD shows the energy lost in real time.
• Try $m_2 = m_1$ — exactly 50% of KE is lost. Try $m_2 = 9 m_1$ — 90% is lost.