Question #116403

8. You drop a ball from a height of 2.0 m and it bounce back to a height of 1.5 m

a) What fraction of its initial energy is lost during the bounce?

b) What is the ball’s speed just as it leaves the ground after the bounce?

c) Where did the energy go?

Expert's answer

Potential energy of the ball before the throw

W1=mgh1W_1=mgh_1

Potential energy of the ball after the rebound

W2=mgh2W_2=mgh_2

a). Lost during the bounce δ=W1W2=mgh1mgh2=h1h2=21.5=1.333\delta=\frac{W_1}{W_2}=\frac{mgh_1}{mgh_2}=\frac{h_1}{h_2}=\frac{2}{1.5}=1.333 parts of energy.

We write down the law of conservation of energy for the moment the ball bounces

mgh1=mv22mgh_1=\frac{mv^2}{2}

gh1=v22gh_1=\frac{v^2}{2}

b).The speed during the bounce is

v=2gh1=29.812=6.264mv=\sqrt{2gh_1}=\sqrt{2\cdot 9.81 \cdot2}=6.264m

c).The energy went into heating the ball during deformation when bouncing.


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