Question

A force in the negative direction of an x axis is applied for 17 ms to a 0.36 kg ball initially moving at 11 m/s in the positive direction of the axis. The force varies in magnitude, and the impulse has magnitude 21.9 N s. (a) What is the ball's velocity (including sign for direction) just after the force is applied? (b) What is the average magnitude of the force on the ball?

Answer 1

Answer:891.11 N

Step-by-step explanation:

Initial velocity
`\left ( u\right )=11 m/s`

momentum magnitude
`\left ( P\right )=21.9 Ns`

final velocity
`\left ( v\right )`

mv-mu=21.9

`0.36* v-0.36* 11=-21.9`

v=-49.833 m/s

`\left ( b\right )`

t=27 ms

v=u+at

`-49.833=17+a\left ( 27* 10^(-3)\right )`

`a=2475.308 m/s^2`

F=ma=0.36\times 2475.308=891.11 N