Question

A cannonball with a mass of 1.0 kilogram is fired horizontally from a 500.-kilogram cannon, initially at rest, on a horizontal, frictionless surface. The cannonball is acted on by an average force of 8.0 × 10 3 newtons for 1.0 × 10 −1 second. What is the magnitude of the change in momentum of the cannonball during firing?

Answer 1

The impulse J is equal to the magnitude of the force applied to the cannonball times the time it is applied:
$J=F \Delta t$
But the impulse is also equal to the change in momentum of the cannonball:
$J=\Delta p$
If we put the two equations together, we find
$F \Delta t= \Delta p$
And since we know the magnitude of the average force and the time, we can calculate the change in momentum:
$\Delta p= F \Delta t=(8.0 \cdot 10^3 N)(1.0 \cdot 10^{-1} s)=800 kg m/s$