Answer to Question #116403 in Quantum Mechanics for Foibe Kambala

Question

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?

Answer 1

Potential energy of the ball before the throw

$W_1=mgh_1$

Potential energy of the ball after the rebound

$W_2=mgh_2$

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

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

$mgh_1=\frac{mv^2}{2}$

$gh_1=\frac{v^2}{2}$

b).The speed during the bounce is

$v=\sqrt{2gh_1}=\sqrt{2\cdot 9.81 \cdot2}=6.264m$

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