Question

a ball having a mass of 500ghits wall with a 10m/s velocity. wall applies 4000 n force to the ball and it turns back with 8m/s velocity . what it the ball- wall contact ?

Answer 1

As per Newton's II law we know that

`F = (\Delta P)/(\Delta t)`

now here we will have

`F = (m(v_f - v_i))/(\Delta t)`

we know that

m = 500 g = 0.5 kg

`v_i = -10 m/s`

`v_f = +8 m/s`

F = 4000 N

now from above equation we will have

`4000 = (0.5(8 - (-10)))/(\Delta t)`

`\Delta t = (9)/(4000) = 2.25 ms`