Question

A `70 g` ball B dropped from a height `h_(0) = 9m` reaches a height `h_(2) = 0.25 m` after bouncing twice from identical `210 g` plates. Plate A rests directly on hard ground, while plate C rests on a foam-rubber mat. Determine (i) the coefficient of restituation between the ball and the plates, (ii) the height `h_(1)` of the ball's first bounce.

Answer 1

Given :

Mass of the ball = 70g

Height at which ball is released = 9m

Height of the ball after rebouncing twice from the ground = 0.25m

Mass of the plates = 210g

To Find :

The coefficient of restitution between the ball and the plates

The height of the ball after first bounce

Solution :

• After first collision of the ball with the plates final velocity of the ball

               $v_{1} =ev_{0}$

               $v_{1} =e\sqrt{2gh_{0} }$

• For the second collision

               $mv=4mv^{1}$    

               $v^{1}=\frac{v}{4}$

               $e\sqrt{\frac{2gh_{0} }{4} } =\sqrt{2gh_{2} }$

               $e=\frac{2}{3}$

The coefficient of restitution between the ball and the plates is 2/3.

• Height attained after first collision

              $h_{1}=e^{2} h_{0}$

              $h_{1} =\frac{2}{3} ^{2} \times9$

              $h_{1}=4m$

The height of the ball after first bounce is 4m.