To test the manufactured properties of `10N` steel balls, each ball is released from rest as shown and strikes a `45^@` inclined surface. If the coefficient of restitution is to be `e=0.8`, determine the distance s, where the ball must strike the horizontal plane at A. At what speed does the ball stike at A? (`g=9.8m//s^2`)

Answer

A, C

Explanation

`v_0=sqrt(2gh)=sqrt(2xx9.8xx1.5)=5.42m//s`
Component of velocity parallel and perpendicular to plane at the time of collision.


Component parallel to plane `(v_1)` remains unchanged, while component perpendicular to plane becomes `ev_2`, where
`ev_2=0.8xx3.83=3.0m//s`
`:.` Component of velocity in horizontal direction after collision
`v_x=((v_1+ev_2))/(sqrt2)=((3.83+3.0))/(sqrt2)=4.83m//s`
While component of velocity in vertical direction after collision.
`v_y=(v_1-ev_2)/(sqrt2)=(3.83-3.0)/(sqrt2)=0.59m//s`
Let t be the time, the particle takes from point C to A, then
`1.0=0.59t+1/2xx9.8xxt^2`
Solving this we get,
`t=0.4s` (Positive value)
`:. DA=v_xt=(4.83)(0.4)=1.93m`
`:. s=DA-DE`
`=1.93-1.0`
`s=0.93m`
`v_yA=v_(yc)+"gt"`
`=(0.59)+(9.8)(0.4)=4.51m//s`
`v_(xA)=v_(xC)=4.83m//s`
`:. v_A=sqrt((v_xA)^2+(v_yA)^2)`
`=6.6m//s`