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`)

Explanation

`v_(0)=sqrt(2gh)=sqrt(2x9.8xx1.5)=5.42` m/s
Component of velocity parallel and perpendicular to plane at the time of collision.
`v_(1)=v_(2)=(v_(0))/(sqrt2)=3.83` m/sec.

Component parallel to plane `(v_(1))` remains unchanged, while component perpendicular to plane becomes `ev_(2)`, where
`ev_(2)=0.8xx3.83=3.0` m/s
`therefore` Component of velocity in horizontal direction after collision
`v_(x)=((v_(1)+ev_(2)))/(sqrt2)=((3.83+3.0))/(sqrt2)=4.83` m/s
While component of velocity in vertical direction after collision.
`v_(y)=(v_(1)-ev_(2))/(sqrt2)=(3.83-3.0)/(sqrt2)=0.59` m/s
Let t be the time, the particle takes from point C to A, then
`1.0=0.59t+1/2xx9.8xxt^(2)` t=0.4 sec
Solving this we get,
`thereforeDA=v_(x)t=(4.83)(0.4)=1.93m`
`thereforeS=DA-DE=1.93-1.0`
S=0.93
`v_(yA)=v_(yc)+gt=(0.59)+(9.8)(0.4)=4.51` m/s
`v_(sA)=v_(xC)=4.83` m/s
`thereforev_(A)=sqrt((v_(xA))^(2)+(v_(yA))^(2))=6.6` m/s