Question

A force in the negative direction of an x axis is applied for 34 ms to a 0.40 kg ball initially moving at 16 m/s in the positive direction of the axis. The force varies in magnitude, and the impulse has magnitude 32.4 N·s.(a) What is the ball's speed just after the force is applied?____m/s(b) What is the average magnitude of the force?___N

Answer 1

(a) 96 m/s

The impulse theorem states that the impulse exerted on the ball is equal to the change in momentum of the ball:

`I=\Delta p=m \Delta v = m(v-u)`

where

`I=32.4 Ns` is the impulse exerted on the ball

`m=0.40 kg` is the mass of the ball

`v` is the final speed of the ball

`u=16 m/s` is the initial speed of the ball

Solving the equation for v, we find

`v=(I)/(m)+u=(32.4 Ns)/(0.40 kg)+16 m/s=96 m/s`

(b) 952.9 N

The impulse is also equal to the product between the average force and the contact time:

`I=F\Delta t`

where here we have

`I=32.4 Ns` is the impulse

`F` is the average force

`\Delta t=34 ms=0.034 s` is the contact time

Solving the equation for F, we find the force:

`F=(I)/(\Delta t)=(32.4 Ns)/(0.034 s)=952.9 N`