Chapter 9 Problem 36

Problem

A 0.280-kg croquet ball makes an elastic head-on collision with a second ball initially at rest. The second ball moves off with half the original speed of the first ball.

(a) What is the mass of the second ball?

(b) What fraction of the original kinetic energy ($\Delta K/K$) gets transferred to the second ball?

Solution

This is a head-on 1D collision. Call the direction of the first ball the positive direction. Let A represent the first ball, and B represent the second ball. $v_{B}=0$ and $v_{B}^{\prime }={\frac {1}{2}}v_{A}$

(a)

conservation of momentum: $p_{i}=p_{f}$ $m_{A}v_{A}+m_{B}v_{B}=m_{A}v_{A}^{\prime }+m_{B}v_{B}^{\prime }$

$\rightarrow m_{A}\left(v_{A}-v_{A}^{\prime }\right)=m_{B}\left(v_{B}^{\prime }-v_{B}\right)$

conservation of energy: ${\frac {1}{2}}m_{A}V_{A}^{2}+{\frac {1}{2}}m_{B}V_{B}^{2}={\frac {1}{2}}m_{A}V_{A}^{\prime 2}+{\frac {1}{2}}m_{B}V_{B}^{\prime 2}$

$\rightarrow m_{A}\left(v_{A}^{2}-v_{A}^{\prime 2}\right)=m_{B}\left(v_{B}^{\prime 2}-v_{B}^{2}\right)$

$\left[m_{A}\left(v_{A}-v_{A}^{\prime }\right)\right]\left(v_{A}+v_{A}^{\prime }\right)=\left[m_{B}\left(v_{B}^{\prime }-v_{B}\right)\right]\left(v_{B}^{\prime }+v_{B}\right)$

$\rightarrow \left(v_{A}+v_{A}^{\prime }\right)=\left(v_{B}^{\prime }+v_{B}\right)$

$v_{A}-v_{B}=-\left(v_{A}^{\prime }-v_{B}^{\prime }\right)$ $\rightarrow v_{A}^{\prime }=-{\frac {1}{2}}v_{A}$

pluging this back to conservation of momentum equation:

$\rightarrow m_{A}v_{a}=-{\frac {1}{2}}m_{A}v_{A}+m_{B}{\frac {1}{2}}v_{A}$

$\rightarrow m_{B}=3m_{A}=3\left(0.280{\text{kg}}\right)=\left(0.840{\text{kg}}\right)$

(b)

${\frac {K_{B}^{\prime }}{K_{A}}}={\frac {{\frac {1}{2}}m_{B}v_{B}^{\prime 2}}{{\frac {1}{2}}m_{A}v_{A}^{2}}}$ $={\frac {3m_{A}\left({\frac {1}{2}}v_{A}\right)^{2}}{m_{A}v_{A}^{2}}}$ $=0.75$