Physics Problems & Solutions: Classical Mechanics

A ball is dropped from a height h. As it bounces off the floor, its speed is 80% of what it was just before it hit the floor. The ball will then rise to a height of most nearly

A. 0.94 h
B. 0.80 h
C. 0.75 h
D. 0.64 h
E. 0.50 h

(GR9277 #45)

Solution:

Conservation of energy before and when the ball hits the ground:

$\begin{aligned} \frac{1}{2}m \cdot 0+mgh&=\frac{1}{2}mv^2+mg\cdot 0 \\mh&=\frac{1}{2}v^2 \\v^2 &= 2gh \end{aligned}$

Conservation of energy when and after the ball hits the ground

$\begin{aligned} \frac{1}{2}mv'^2+mg\cdot 0&=\frac{1}{2}m\cdot 0+mgh' \\\frac{1}{2}mv'^2&=mgh' \\v'^2 &= 2gh' \\(0.8v)^2&=0.64v^2 = 0.64\times 2gh = 2gh' \\h'&=0.64h \end{aligned}$

Answer: D

Physics Problems & Solutions

Classical Mechanics - Energy

No comments :