Consider a horizontal surface moving vertically upward with velocity `2ms^(-1)`. A small ball of mass 2 kg is moving with velocity `(2hati-2hatj)ms^(-2)` If the coefficient of restitution and coefficient of friction are `(1)/(2) and (1)/(3` respectively, find the horizontal velocity `("in "ms^(-1))` of the ball after the collision.

Answer

0

Explanation

The relative velocity of the particle with respect to the surface, along the normal, is the approach speed of the particle.
`u_("app") = 4 ms^(-1)`
The separation velocity will be
`u_(sep) = ev_(app) = (1)/(2) xx 4 = 2 ms^(-1)`
`:. vecv_(m,gr) = vecv_(m,s) + vecv_(s,gr) = 4 m//s`
`int `N dt = 4 (2) + 2(2) =12
Now `int mu N dt t = 2 ( v_(xf ) -2 ) `
`implies - (1)/(3) xx 12 =2 ( v_(xf ) -2) `
`implies - (1)/(3) xx 12 = 2 (v_(xf )-2)`
Therefore `v_(x ,f ) = (v_(x))_("final") =0`