Impulse and force-momentumEdexcel A-Level Physics: Revision notes

The momentum of an object is $p = mv$, a vector quantity measured in kg m s⁻¹ with the same direction as the velocity.

Newton's second law in its general form says that the resultant force on an object is equal to the rate of change of momentum:

$F = \dfrac{\Delta p}{\Delta t}$

For constant mass this gives the familiar $F = ma$, since $\Delta p = m\Delta v$ and $a = \Delta v / \Delta t$. The momentum form is more useful when forces act for short times, as in collisions.

The impulse of a force is the force multiplied by the time for which it acts:

$\text{impulse} = F\Delta t = \Delta p$

The impulse on an object equals its change in momentum. It is measured in newton seconds (N s), and 1 N s = 1 kg m s⁻¹.

When the force varies during a collision, the impulse is the area under the force against time graph, and the average force is the impulse divided by the contact time. The maximum force is larger than the average because the force rises and falls during the contact.

Momentum is a vector, so choose a positive direction and give velocities in the opposite direction a negative sign. When an object rebounds, the change in momentum is larger than its initial momentum.

Worked example. A 0.058 kg ball arrives at 20 m s⁻¹ and rebounds at 30 m s⁻¹ in the opposite direction. Taking the initial direction as positive, $\Delta p = 0.058 \times (-30) - 0.058 \times 20 = -2.9$ kg m s⁻¹. If the contact lasts 5.0 ms, the average force is $2.9 \div 5.0 \times 10^{-3} = 5.8 \times 10^2$ N, directed opposite to the original velocity.

For a given change in momentum, a longer contact time means a smaller average force: $F = \Delta p / \Delta t$. This is the principle behind many safety features:

• Crumple zones and airbags lengthen the time taken to stop, reducing the force on the occupants.
• A cricketer moves their hands back when catching a fast ball.
• Landing with bent knees, or on a soft surface, increases the stopping time.

In each case the impulse is the same, but the force is smaller because it acts for longer.

A trolley is released down a track so that it hits a force sensor connected to a data logger. A light gate and an interrupt card measure the speed before and after the collision, using speed = card width ÷ time.

1. Measure the mass of the trolley with a balance.
2. Record the force against time during the collision, and find the impulse from the area under the data.
3. Calculate $\Delta p = m(v - u)$, adding the speeds if the trolley rebounds.
4. Repeat for different initial speeds and plot impulse against $\Delta p$. A straight line through the origin with gradient 1 supports $F\Delta t = \Delta p$.

Sources of error include friction on the track, uncertainty in the card width and a sampling rate that is too low for the short collision. Repeating and averaging improves reliability.