If ball hits floor with velocity v10ms at angle 45 degrees w

If ball hits floor with velocity v=10[m/s] at angle =45 degrees with respect to normal, and rebounds with velocity v’=8.5[m/s/] at angle ’=40 degrees with respect to normal, what is the coefficient of restitution.

Solution

As, Line of Impact in this case will be normal

and, as impulse (or force) is acting along this line only, so MOMENTUM will change only in this direction and will be constant for other directions

>> Now, U = Initial Velocity = 10 m/s , at angle 45° with normal

So, Initial Component of Velocity along normal, U\" = 10*cos45 = 7.07 m/s

>> Also, After Rebounding, Velocity = 8.5 m/s, at an angle 40° with normal

So, Final Component of Velocity along normal, V\" = 8.5*cos40 = 6.51 m/s

>> As, Coefficient of Restitution, e = Relative Velocity of separation/Relative Velocity of approach

>> As, surface will remain at rest

=> e = (V\" - 0)/(U\" - 0)

=> e = 6.51/7.07

=> Coefficient of Restitution , e = 0.921 ..........ANSWER......