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 magnitude of the change in momentum of the cannonball during firing is 800 kg·m/s.

Step-by-step explanation:

In order to answer this question, we need to use the equation:

change in momentum = average force x time

Given that the mass of the cannonball is 1.0 kilogram and the average force is 8.0 × 10^3 newtons, we can calculate the magnitude of the change in momentum as:

change in momentum = (8.0 × 10^3 N) x (1.0 × 10^(-1) s)

change in momentum = 800 kg·m/s

Therefore, the magnitude of the change in momentum of the cannonball during firing is 800 kg·m/s

Answer 2

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`