Physics, asked by moriarty42, 1 year ago

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.

Answers

Answered by PoojaBurra
2

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.

 

 

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