Question

A 2-kg bowling ball is accidentally dropped from a window and strikes the sidewalk at a speed of 30 m/s. It bounces back up with an initial speed of 20 m/s from the sidewalk. What is the magnitude of the impulse the sidewalk imparts to the bowling ball

Answer 1

`I=20\ kg.m.s^(-1)`

Step-by-step explanation:

Given:

mass of the ball,
`m=2\ kg`

initial speed of the ball before collision,
`u=30\ m.s^(-1)`

final speed of hte ball after collision,
`u=20\ m.s^(-1)`

We have mathematical relation for impulse as:

`I=F.dt` ......................(1)

From the Newton's second law of motion:

`F=(d)/(dt) (p)`

`F=m.(dv)/(dt)` ................................(2)

where:

`dt=` time duration

`dv=` change in velocity

`dp=` change in momentum

Now from (1) & (2)

`I=m.(dv)/(dt) .dt` (since this impulse is due to the impact force of the ball falling on the ground)

`I=m.dv`

putting the values

`I=2* (30-20)`

`I=20\ kg.m.s^(-1)` is the impulse transfer to the bowling by the sidewalk.