Kevin S. Huang

2015: Problem 21

Topic: Kinematics
Concepts: 1D inelastic collision, Free fall

Solution:

If we launch a ball upward with velocity $v$ , then it takes time

v-gt=0

t=\frac{v}{g}

to reach the top. Coming down takes the same amount of time so the time for the first bounce is

T_1=2t=\frac{2v}{g}

Letting the coefficient of restitution be $r$ , the starting speed for the second bounce is $rv$ . Then the time taken is

T_2=\frac{2rv}{g}

For each subsequent bounce, the speed (and hence time) is reduced by a factor of $r$ . The total time is thus

T_{\text{tot}}=\sum_{i=1}^{\infty} T_i=\sum_{n=0}^{\infty} \frac{2r^nv}{g}=\frac{2v}{g}\sum_{n=0}^{\infty} r^n=\frac{2v}{g}\frac{1}{1-r}

where we used the formula for the sum of an infinite geometric series. Since $v=50\,\mathrm{m/s}$ and $r=0.9$ ,

T_{\text{tot}}=\frac{2(50\,\mathrm{m/s^2})}{(10\,\mathrm{m/s^2})}\frac{1}{1-0.9}=100\,\mathrm{s}

so the answer is B.