Impulse Momentum Theorem: Definition, Derivation & Equation

The impulse-momentum theorem shows that the ​impulse​ an object experiences during a collision is equal to its ​change in momentum​ in that same time.

One of its most common uses is to solve for the average force an object will experience in different collisions, which is the basis for many real-world safety applications.

The impulse-momentum theorem can be expressed like this:

\(J = \Delta p\)

Where:

•​J​ is impulse in newton-seconds (Ns) or kgm/s, and 
•​p​ is linear momentum in kilogram-meters per second or kgm/s

Both are vector quantities. The impulse-momentum theorem can also be written out using the equations for impulse and momentum, like this:

\(J = m\Delta v = F\footnotesize avg \normalsize \Delta t\)

Where:

•​J​ is impulse in newton-seconds (Ns) or kgm/s,
•​m​ is mass in kilograms (kg), 
•​Δ v​ is final velocity minus initial velocity in meters per second (m/s), 
•​F​ is net force in Newtons (N), and 
•​t​ is time in seconds (s).

The impulse-momentum theorem can be derived from Newton's second law, ​F = ma​, and rewriting ​a​ (acceleration) as the change in velocity over time. Mathematically:

\(F = ma \newline
F = m (\Delta v / \Delta t) \newline
F \Delta t = m\Delta v\)

A major takeaway from the theorem is to explain how the force experienced by an object in a collision depends on the ​amount of time​ the collision takes.

For instance, a classic high school physics setup with impulse is the egg-drop challenge, where students must design a device to land an egg safely from a large drop. By adding padding to ​drag out​ the time when the egg is colliding with the ground and changing from its fastest velocity to a full stop, the forces the egg experiences must decrease. When the force is decreased enough, the egg will survive the fall without spilling its yolk.

This is the main principle behind an array of safety devices from everyday life, including airbags, seat belts and football helmets.

A 0.7 kg egg drops from the roof of a building and collides with the ground for 0.2 seconds before stopping. Just before hitting the ground, the egg was travelling at 15.8 m/s. If it takes approximately 25 N to break an egg, does this one survive?

\(m\Delta v = F\footnotesize avg \normalsize t \newline
m (v \footnotesize f \normalsize -v\footnotesize i \normalsize) = F\footnotesize avg \normalsize t \newline
0.7 kg (0-15.8m/s) = F\footnotesize avg \normalsize (0.2 s) \newline
-55.3 N = F\footnotesize avg\)

55.3 N is more than twice what it takes to crack the egg, so this one isn't making it back to the carton.

(Note that the negative sign on the answer indicates the force is in the opposite direction of the egg's velocity, which makes sense because it is the force from the ground acting upwards on the falling egg.)

Another physics student plans to drop an identical egg from the same roof. How long should she make sure the collision lasts thanks to her padding device, at a minimum, to save the egg?

\(m\Delta v = F\footnotesize avg \normalsize t \newline
m (v \footnotesize f \normalsize -v\footnotesize i \normalsize) = F\footnotesize avg \normalsize t \newline
0.7 kg (0-15.8m/s) = 25 N t \newline
0.44 s = t\)

Both collisions – where the egg breaks and where it doesn't – happen in less than half a second. But the impulse-momentum theorem makes it clear that even small increases in collision time can have a big impact on the outcome.