A small rubber ball strikes a rough horizontal surface with a velocity v, at an angle thetawith vertical. If the coefficient of restitution and coefficient of friction between the ball and horizontal surface are e and u respectively, find the velocity of the ball just after the impact.

A small rubber ball strikes a rough horizontal surface with a velocity

1.	A small rubber ball strikes a rough horizontal surface with a velocity v, at an angle thetawith vertical. If the coefficient of restitution and coefficient of friction between the ball and horizontal surface are e and u respectively, find the velocity of the ball just after the impact.
Answer» Solution :As the particle collides with the rough horizontal impulse of `int `N dt and horizontal impulse be `int F dt` where friction `F = muN ….(i)` Writing the impulse momentum equation for the particle in x-direction, we have, `mv _(0) sin theta - int F dt = mv _(x) .....(ii)` Similarly, the impulse momentum equation in y-direction gives `- mv _(0) cos theta + int N dt = mv _(y)....(iii)` The coefficient of RESTITUTION along y-direction (normal) gives `v _(y) =- e (- v _(0) cos thea)` Then`v_(y) = ev _(0) cos theta ""...(iv)` We have four equations for four UNKNOWN quantities , i.e., `int dt , int N dt, v _(x) and v _(y).` Substituting `F = mu N ` in eq. (ii) and eliminating `int N dt` from eqs. (ii) and (iii), we have `v _(x) = v _(0) [ sin theta - mu (1 + e ) cos theta ] ......(v)` From eqs. `(iv) and (v),` we have, `[ v _(0) sin theta - mu (1 + e ) cos theta ] hati + ev _(0) cos theta hatj.`