Chapter 7, Problem 67GP

To determine

The mass of the second ball

Answer to Problem 67GP

  $1.47m$

Explanation of Solution

Given:

Mass of the first ball $=m_1=m$

Mass of the second ball $=m_2$

Initial velocity of first ball before collision $=v_{1i}=v$

Initial velocity of second ball before collision $=v_{2i}=0\frac{\text{m}}{\text{s}}$

Final velocity of first ball after collision $=v_{1f}=\frac{-v}{4}$

Final velocity of second ball after collision $=v_{2f}$

Formula Used:

According conservation of momentum principle,

  $m_1{v}_{1i}+m_2{v}_{2i}=m_1{v}_{1f}+m_2{v}_{2f}$

Also, for elastic collision,

  $m_1{v}_{1i}^2+m_2{v}_{2i}^2=m_1{v}_{1f}^2+m_2{v}_{2f}^2$

Calculation:

According to conservation of momentum for the elastic collision between the two balls,

  $\begin{array}{l}{m}_1{v}_{1i}+m_2{v}_{2i}=m_1{v}_{1f}+m_2{v}_{2f}\\ mv+m_2(0)=m\left(\frac{-v}{4}\right)+m_2{v}_{2f}\\ m_2{v}_{2f}=m\left(\frac{5v}{4}\right)\\ v_{2f}=m\left(\frac{5v}{4m_2}\right)\left(\text{1}\right)\end{array}$

Using conservation of kinetic energy

  $\begin{array}{l}{m}_1{v}_{1i}^2+m_2{v}_{2i}^2=m_1{v}_{1f}^2+m_2{v}_{2f}^2\\ $ m{v}^2+m_2{(0)}^2=m{\left(\frac{-v}{4}\right)}^2+m_2{v}_{2f}^2$m_2{v}_{2f}^2=m\left(\frac{17v^2}{16}\right)\\ \text{Using equation }\left(\text{1}\right)\\ m_2{\left(\frac{5mv}{4m_2}\right)}^2=m\left(\frac{17v^2}{16}\right)\\ m_2\left(\frac{25m^2{v}^2}{16m^2{}_2}\right)=m\left(\frac{17v^2}{16}\right)\\ \left(\frac{25m}{m_2}\right)=17\\ m_2=\frac{25m}{17}=1.47m\end{array}$

Conclusion:

Therefore mass of the second ball is $1.47m$