A partical of mass `m` and velocity `v_(1)` in positive y direction is projected on to a belt that is moving with uniform velocity `v_(2)` in x- direction as shown in figure. Coefficient of friction between partical and belt is `mu` . Assuming that the partical first touches the belt at the origin of fixed `x-y` coordinate system and remains on the belt , find co-ordinates `(x-y)` of the point where slidding stop.

Explanation

`v_(r) = sqrt(v_(1)^(2) + v_(2)^(2))`
Reterdation `a = mug`
`:.` Time when slipping stop is `t = (v_(r))/(a)`
or `t = (sqrt(v_(1)^(2) + v_(2)^(2)))/(mug)`
`s_(r) = (v_(r)^(2))/(2 a) = (v_(1)^(2) + v_(2)^(2))/(2 mu g)`
`x_(r) = - s_(r), cos theta = - ((v_(1)^(2) + v_(2)^(2))/(2 mu g)) ((v_(2))/(sqrt (v_(1)^(2) + v_(2)^(2))))`
`(-v_(2)sqrt(v_(1)^(2) + v_(2)^(2)))/(2mug)`
`y_(r) = s_(r) sin theta = ((v_(1)^(2) + v_(2)^(2))/(2 mu g)) ((v_(1))/(sqrt (v_(1)^(2) + v_(2)^(2))))
`= (v_(1)sqrt (v_(1)^(2) + v_(2)^(2)))/(2 mu g)`

In time `t` belt will move a distance `s = v_(2) t`
or `v_(2)sqrt (v_(1)^(2)) + v_(2)^(2))/( mu g)` in x- direction
Hence , coordinate of particle,
`x = x_(r) + s = v_(2)sqrt (v_(1)^(2) + v_(2)^(2))/( 2 mu g)`
and `y = y_(r) = v_(1)sqrt (v_(1)^(2)) + v_(2)^(2))/( 2 mu g)`