A cricket ball hits the ground with a speed of 5 ~m ~s ⁻¹ making an angle 30^ with the horizontal. If the coefficient of restitution is 0.2 , the speed of the reflected ball in m s ⁻¹ is

Step-by-step solution

We know that, coefficient of restitution, e= ( Relative velocity of separation Relative velocity of approach ) Now, given collision is as shown As impact is along normal to surface it does not change the horizontal component of ball. Now if v_ 2 y is final vertical component then v_ 2 y =e . v_ 1 y =0.2 5 60^ or v_ 2 y =0.2 5 1 2 =0.5 ~m / s So components of final velocity of ball are v_ 2 x =v_ 1 x =5 60^ =(5 3 ) / 2 and v_ 2 y =e v_ 1 y =1 / 2 Hence, final speed of ball, v_f= (v_ 2 x )^2+ (v_ 1 x )^2 aligned & =