Change in Momentum Formula: Δp = mv − mu with Examples

The change in momentum formula is $\Delta p = mv - mu = m(v - u)$ : subtract the starting momentum from the final momentum. A 0.5 kg ball that speeds up from 4 m/s to 10 m/s gains $0.5 \times 6 = 3\ \text{kg m/s}$ of momentum.

Indian books write u and v for the initial and final velocity; US books often write $v_i$ and $v_f$ . The same change also equals the impulse of the net force, $\Delta p = F\Delta t$ , which is why this formula links momentum to force in the laws of motion chapter.

Quick facts

Formula
$\Delta p = m(v - u)$
With force
$\Delta p = F\,\Delta t$
SI unit
kg m/s, equal to N s
Dimensional formula
[M¹ L¹ T⁻¹]
Type
Vector, it points along the net force
Graph meaning
Area under a force against time graph

Formula of change in momentum and its symbols

Symbols in Δp = m(v − u)
Symbol	Meaning	Unit
$\Delta p$	Change in linear momentum	kg m/s
$m$	Mass of the body, assumed fixed	kg
$u$	Velocity before the event	m/s
$v$	Velocity after the event	m/s

Momentum has a direction, so you must subtract with directions in mind. Along a straight line, give each velocity a plus or minus sign and the arithmetic takes care of it. The result $\Delta p$ is always in the direction of the net force that caused it, even when that is opposite to the motion.

Using the change in momentum formula step by step

Choose a positive direction and write it next to your sketch.
Write u and v with signs. A ball that bounces back has its velocity sign flipped.
Convert km/h to m/s and grams to kg.
Compute $\Delta p = m(v - u)$ . If the time of contact is given, divide by it to get the average force.

Solved examples for the formula for change in momentum

Example 1: speeding up

A 0.50 kg ball goes from 4.0 m/s to 10 m/s in the same direction. Find Δp.

$\Delta p = 0.50 \times (10 - 4.0) = 3.0\ \text{kg m/s}$

Example 2: a rebound from a garage door

A 58 g ball strikes a garage door at 18 m/s and comes straight back at 12 m/s. Find the change in momentum.

Call the direction toward the door positive. m = 0.058 kg, u = +18 m/s, v = −12 m/s.

$\Delta p = 0.058 \times (-12 - 18) = 0.058 \times (-30) = -1.74\ \text{kg m/s}$

The size is about 1.7 kg m/s, directed away from the door. Writing 12 − 18 instead would give only 0.35 kg m/s, a classic error.

Example 3: catching a ball, with km/h

A 150 g ball flying at 90 km/h is caught and brought to rest in 0.10 s. Find Δp and the average force.

Convert: 90 ÷ 3.6 = 25 m/s, and 150 g = 0.150 kg.

$\Delta p = 0.150 \times (0 - 25) = -3.75\ \text{kg m/s}$

$F = \frac{\Delta p}{\Delta t} = \frac{-3.75}{0.10} = -37.5\ \text{N}$

The hands push on the ball with about 38 N, against its motion.

Change in momentum and Newton's second law

Newton's second law in its original form says force is how fast momentum changes with time, $F = \frac{\Delta p}{\Delta t}$ . Multiply through by the time and you get $\Delta p = F\Delta t$ , the quantity called impulse. For a fixed mass, $\frac{\Delta p}{\Delta t} = m\frac{\Delta v}{\Delta t} = ma$ , which is the familiar force formula.

This tells you two ways to change momentum by the same amount: a big force for a short time or a small force for a longer time. Crumple zones, helmets and sand pits all stretch the time so the force stays small. The average force formula page puts numbers on this.

Change in momentum on graphs

Plot force on the vertical axis against time, and the area enclosed under the curve equals the change in momentum. A force that rises from 0 to 200 N and falls back to 0 over 0.03 s, in a triangle shape, gives an area of $\frac{1}{2} \times 0.03 \times 200 = 3.0\ \text{N s}$ . That is a 3.0 kg m/s change in momentum, whatever the mass of the body.

On a momentum against time graph, the slope is the net force. A straight rising line means a steady force.

Traps in change in momentum questions

Rebounds. When the direction reverses, add the speeds inside the bracket: Δp has size m(u + v) when u and v are the two speeds.
Change of direction at constant speed. A ball moving at 5 m/s that turns through 90° still changes momentum, by $m\sqrt{5^2 + 5^2}$ .
Variable mass. For a rocket, $\Delta p = m\Delta v$ is not enough. You need the thrust approach on the thrust formula page.
Missing units. State kg m/s or N s, never just a bare number.

Frequently asked questions

What is the formula of change in momentum?

Change in momentum is final momentum minus initial momentum, Δp = mv − mu = m(v − u). It also equals the net force times the time it acts, FΔt.

What is the unit of change in momentum?

It has the same unit as momentum, kg m/s. Because it equals force times time, N s is also correct.

Is change in momentum the same as impulse?

Yes. The impulse momentum theorem says the impulse of the net force equals the change in momentum, so the two are numerically equal and share units.

How do you find change in momentum when a ball bounces back?

Give the two velocities opposite signs. The bracket v − u then becomes the sum of the two speeds, so Δp is larger than either momentum alone.