Momentum conservation and collisions

Learning objectives

Outcome
Use the impulse–momentum relation $\vec F\,\Delta t = \Delta \vec p$ to decide whether the total momentum of two bodies can be treated as conserved during a collision.
Outcome
Calculate an unknown post-collision velocity from $m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'$ in one dimension, assigning projection signs correctly.
Outcome
Classify a collision as elastic, inelastic, or perfectly inelastic from given before-and-after data, and compute the change in kinetic energy.

Understand the idea

Momentum conservation in collisions: from bullets to cars

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The momentum of a body is $\vec p = m\vec v$ , and Newton's second law in impulse form $\vec F\,\Delta t = \Delta \vec p$ links force to the change in momentum.

During a collision the internal impact forces are large, while external forces (gravity, friction) deliver only a small impulse over the short time $\Delta t$ , so along the line of impact the system may be treated as closed and $m_1\vec v_1 + m_2\vec v_2 = m_1\vec v_1' + m_2\vec v_2'$ applies.

This vector equality is easiest to use in projections onto a chosen axis, taking careful account of the signs of the velocities.

In a perfectly elastic collision kinetic energy is conserved; in a perfectly inelastic collision the bodies move together and the kinetic-energy loss is $\Delta K = -\dfrac{1}{2}\,\dfrac{m_1 m_2}{m_1+m_2}\,(v_1-v_2)^2$ .

This is why, when a light car meets an oncoming truck, the total momentum of the system is conserved while the light car changes its velocity far more than the heavy one.

Worked examples

Direct

Problem

A bullet of mass $m = 0{,}010$ kg leaves a rifle of mass $M = 4{,}0$ kg with speed $v = 500$ m/s. Find the recoil velocity $V$ of the rifle.

Solution

Let the $x$ -axis point along the bullet's direction. Before the shot the system is at rest, so $p_x = 0$ . After the shot

m v + M V = 0 \quad\Rightarrow\quad V = -\frac{m v}{M}.

Substituting the numbers:

V = -\frac{0{,}010 \cdot 500}{4{,}0} = -1{,}25\ \text{m/s}.

The minus sign means the rifle moves opposite to the bullet; the recoil speed is $1{,}25$ m/s. Note that the bullet's and the rifle's momenta are equal in magnitude and opposite in direction, so their sum stays zero.

Application

Problem

A cart of mass $m_1 = 2{,}0$ kg moving at $v_1 = 3{,}0$ m/s catches up with a cart of mass $m_2 = 4{,}0$ kg moving in the same direction at $v_2 = 1{,}0$ m/s. After the collision the carts lock together and move as one. Find their common velocity and the kinetic energy lost.

Solution

The collision is perfectly inelastic, so momentum is conserved while kinetic energy is not:

v' = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{2{,}0 \cdot 3{,}0 + 4{,}0 \cdot 1{,}0}{2{,}0 + 4{,}0} = \frac{10}{6{,}0} \approx 1{,}67\ \text{m/s}.

Kinetic energy before the collision:

K_i = \tfrac{1}{2} m_1 v_1^2 + \tfrac{1}{2} m_2 v_2^2 = \tfrac{1}{2}\cdot 2{,}0\cdot 9{,}0 + \tfrac{1}{2}\cdot 4{,}0\cdot 1{,}0 = 9{,}0 + 2{,}0 = 11\ \text{J}.

After the collision:

K_f = \tfrac{1}{2}(m_1+m_2)v'^2 = \tfrac{1}{2}\cdot 6{,}0\cdot (1{,}67)^2 \approx 8{,}3\ \text{J}.

The loss is $\Delta K = K_f - K_i \approx -2{,}7$ J. The same result follows at once from $\Delta K = -\tfrac{1}{2}\mu (v_1-v_2)^2$ with the reduced mass $\mu = \frac{m_1 m_2}{m_1+m_2} = \frac{8{,}0}{6{,}0} \approx 1{,}33$ kg and relative speed $(v_1-v_2)=2{,}0$ m/s.

Transfer

Problem

Ballistic pendulum: a bullet of mass $m = 0{,}010$ kg embeds itself in a block of mass $M = 2{,}0$ kg hanging on a long string, and the block with the bullet rises to a height $h = 0{,}20$ m. Find the bullet's speed before impact and the fraction of mechanical energy lost.

Solution

The problem is solved in two stages: the impact (momentum conserved, energy lost) and the subsequent rise (mechanical energy conserved).

Stage 1 — the rise. The speed $v'$ just after impact follows from energy conservation:

\tfrac{1}{2}(M+m)v'^2 = (M+m)gh \quad\Rightarrow\quad v' = \sqrt{2gh} = \sqrt{2\cdot 9{,}8\cdot 0{,}20} \approx 1{,}98\ \text{m/s}.

Stage 2 — the impact. Momentum is conserved:

m v = (M+m)v' \quad\Rightarrow\quad v = \frac{M+m}{m}\,v' = \frac{2{,}01}{0{,}010}\cdot 1{,}98 \approx 4{,}0\cdot 10^{2}\ \text{m/s}.

Energy loss. Before impact $K_i = \tfrac{1}{2} m v^2 \approx \tfrac{1}{2}\cdot 0{,}010\cdot (398)^2 \approx 7{,}9\cdot 10^{2}$ J; after impact $K_f = \tfrac{1}{2}(M+m)v'^2 \approx 3{,}9$ J. The lost fraction is $1 - K_f/K_i \approx 0{,}995$ , i.e. about $99{,}5\,\%$ of the bullet's kinetic energy becomes internal energy (heating and deformation). The pendulum acts as a speed detector: it measures $v$ through the easily observed height $h$ .

Practice

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Self-study plan (safe, tabletop).

Open an interactive collision simulation (for example, Collision Lab) and choose the one-dimensional mode. Set $m_1 = 1$ kg, $v_1 = 2$ m/s, $m_2 = 1$ kg, $v_2 = 0$ ; set the elasticity slider first to $1{,}0$ (elastic collision), then to $0$ (perfectly inelastic).
For each case record the before and after velocities in a table and compute $p_1+p_2$ and $K_1+K_2$ . Check that the momentum is the same before and after, while kinetic energy decreases only in the second case.
Repeat with $m_2 = 3$ kg and confirm that the lighter body changes its velocity more strongly.
Tabletop experiment (optional). On a smooth tabletop with a very slight tilt, send one coin into another, keeping away from the table edge and using nothing hot or sharp. Film the collision with a phone in slow-motion mode, mark the coin positions frame by frame, and estimate the speed ratio. Compare the vector sum of momenta before and after; friction makes conservation only approximate — explain this as the action of an external force.

Study all high-speed and bullet-like cases only in the simulation; never perform real shots, falls from height, or impacts with heavy objects.

Visual reference

Purpose

Show that the vector sum of momenta is the same before and after a collision, while kinetic energy decreases only in an inelastic collision.

Representation

A two-panel diagram: on the left, "before the collision" — two bodies with masses $m_1$ , $m_2$ , velocity arrows $\vec v_1$ , $\vec v_2$ , and momentum vectors $m_1\vec v_1$ , $m_2\vec v_2$ ; on the right, "after the collision" — the same bodies with velocities $\vec v_1'$ , $\vec v_2'$ (or one stuck-together body for the perfectly inelastic case). Below the panels: a table of $p_x$ projections with the same total on both sides, and two bar charts of kinetic energy for the elastic and perfectly inelastic cases.

Real-world examples

Ice hockey: two players collide

Scenario

Two hockey players meet on the ice. Player A has a mass of $85\ \text{kg}$ and glides at $6.0\ \text{m/s}$ ; player B, mass $75\ \text{kg}$ , stands still. They collide and hold on to each other, so after the collision they move as one body. The total momentum before the impact is $p = 85 \cdot 6.0 = 510\ \text{kg}\cdot\text{m/s}$ (player B contributes $0$ ), so afterwards the $160\ \text{kg}$ pair must still carry $510\ \text{kg}\cdot\text{m/s}$ : $v = 510/160 = 3.2\ \text{m/s}$ .

Connection

The collision is treated as perfectly inelastic: momentum is conserved because the pushes the skaters exert on each other are equal in magnitude, opposite in direction and last the same short time, so their impulses cancel pairwise. Kinetic energy is not conserved here: $\frac{1}{2}\cdot 85\cdot 6.0^2 = 1530\ \text{J}$ before against $\frac{1}{2}\cdot 160\cdot 3.2^2 \approx 819\ \text{J}$ after — the missing $711\ \text{J}$ goes into deformed kit, heating and sound.

Car crash: a rear-end impact

Scenario

On a city street a $1200\ \text{kg}$ car travelling at $15\ \text{m/s}$ ( $54\ \text{km/h}$ ) runs into a $1400\ \text{kg}$ car standing at a red light. The bumpers lock together. Before the impact the momentum is $p = 1200\cdot 15 = 1.8\times 10^4\ \text{kg}\cdot\text{m/s}$ ; the stationary car contributes nothing. After the crash the two cars, $2600\ \text{kg}$ in total, slide as one body at $v = 1.8\times10^4/2600 \approx 6.9\ \text{m/s}$ ( $25\ \text{km/h}$ ).

Connection

Momentum is conserved along the direction of motion even though the collision is strongly inelastic; the crumpled metal, heat and noise account for the drop in kinetic energy from $\frac{1}{2}\cdot1200\cdot15^2 = 1.35\times10^5\ \text{J}$ to $\frac{1}{2}\cdot2600\cdot 6.9^2 \approx 6.2\times10^4\ \text{J}$ . This is exactly the calculation used to reconstruct vehicle speeds from skid marks and damage.

A basketball bouncing off the court

Scenario

A basketball of mass $0.62\ \text{kg}$ lands on a court floor moving downwards at $5.0\ \text{m/s}$ and leaves the floor upwards at $4.0\ \text{m/s}$ . Taking downwards as negative, the ball's momentum changes from $-3.1\ \text{kg}\cdot\text{m/s}$ to $+2.5\ \text{kg}\cdot\text{m/s}$ , a change of $+5.6\ \text{kg}\cdot\text{m/s}$ .

Connection

The ball and the Earth form a closed system: the floor's push changes the ball's momentum by $+5.6\ \text{kg}\cdot\text{m/s}$ , so the Earth receives an equal downward momentum and recoils at $\Delta v = 5.6/6.0\times10^{24} \approx 9\times10^{-25}\ \text{m/s}$ — impossible to measure, yet the momentum balance is exact. The bounce is not perfectly elastic either: kinetic energy falls from $\frac{1}{2}\cdot0.62\cdot5.0^2 = 7.8\ \text{J}$ to $\frac{1}{2}\cdot0.62\cdot4.0^2 = 5.0\ \text{J}$ .