A particle moving on a smooth horizontal surface strikes a stationary wall. The angle of strike is equal to the angle of rebound & is equal to `37^(@)` and the coefficient of restitution with wall is `e = (1)/(5)`. Find the friction coefficient between wall and the particle form `(X)/(10)` and fill value of X :

Answer

5

Explanation

Since `e = (1)/(5)`
`:.` Final normal component of velocity
`= (v cos 37^(@))/(5)`
As the angle of rebound is equal to the angle before impact.
Therefore, both normal & tangential components of velocities must change by the same factor.
`:.` Tangential velocity after impact becomes `(v sin 37^(@))/(5)`
Let the time of impact be `Delta t`
`N=(m(v cos 37^(@)+(v cos 37^(@))/(5)))/(Delta t)=(6mv cos 37^(2))/(5 Deltat)`
wher N is the normal force imparted on the ball by the wall.
Frictional force `= mu N = (6)/(5) (mu mv cos 37^(@))/(Delta t)`
Also frictional force
`=(m[v sin 37^(2)-(v sin 37^(@))/(5)])/(Delta t)`
`rArr(m[v sin 37^(@)-(v sin 37^(@))/(5)])/(Deltat)`
`=(6)/(5)(mu mv cos 37^(@))/(Delta t)`
`rArr mu=(2)/(3)tan 37^(@)`
`rArr mu =(2)/(3).(3)/(4)=(1)/(2)`.