A particle of mass m strikes a smooth floor with speed u at angle of incidence theta with the normal. The coefficient of resultant is e. Find the magnitude and direction of velocity with which the particle rebounds. Also, find the impulse and loss in K.E.

A particle of mass m strikes a smooth floor with speed u at angle of i

1.	A particle of mass m strikes a smooth floor with speed u at angle of incidence theta with the normal. The coefficient of resultant is e. Find the magnitude and direction of velocity with which the particle rebounds. Also, find the impulse and loss in K.E.
Answer» Solution :Before the collision: After the collision: Since there is no force ALONG the surface , hence the velocity component remains same. `V sin alpha = u sin theta`(i) Due to the normal force between the particle and the floor , the velocity component becomes `e TIMES` in opposite direction `v cos alpha = e u cos theta`(II) `v = sqrt((v sin alpha)^(2) + (v cos alpha)^(2)) = u sqrt(sin^(2) theta + e^(2) cos^(2) theta)` `tan alpha = ( v sin alpha)/(v cos alpha) = (1)/(e) tan theta` If the collision is elastic , `e = 1 , v = u , alpha = theta` `J_(x) = 0` `J_(y) = m(v cos alpha + u cos theta)` `= m ( e u cos theta + u cos theta) = mu cos theta ( 1 + e)` `J = mu cos theta ( 1 + e)` `Delta K = (1)/(2) mu^(2) - (1)/(2) m u^(2)= (1)/(2) mu^(2) cos^(2) theta ( 1 - e^(2))` `Delta` K: loss in `K.E.`