In the previous problem , if the collision is inelastic and `e = 1//2`

Options

A. After collision , the ball moves with `5 sqrt(13) m//s` at angle `tan^(-1) (2 sqrt(3))` with horizontal
B. Force by wall on the ball is `375 N` to right
C. Change in momentum of ball is `0.75 kg m//s`
D. All options are correct

Answer

D

Explanation

In this case , the velocity component , perpendicular to the wall becomes `e` times and reverses its direction
`v' = sqrt((10 sqrt(3))^(2) + (5)^(2)) = sqrt(325) = 5 sqrt(13) m//s`
`tan alpha = (10 sqrt(3))/(5) = 2 sqrt(3) rArr alpha = tan^(-1) (2 sqrt(3))`
`Delta p_(x) = m ev cos 60^(@) - (-v cos 60^(@))]`
`= mv cos 60^(@) ( 1 + e)`
`= 50 xx 10^(-3) xx 20 xx (1)/(2) (1 + (1)/(2)) = 0.75 kg m//s`
`F_(x) = (Delta p_(x))/(Delta t) = (0.75)/( 2 xx 10^(-3)) = 375 N`, to the right