Chapter 7, Problem 24P

To determine

The speeds of billiard balls after collision

Answer to Problem 24P

The speed of two balls will interchange.

  $3\frac{\text{m}}{\text{s}}\text{ and 2}\frac{\text{m}}{\text{s}}$

Explanation of Solution

Given:

Mass of the each ball $=m_1=m_2=m$

Initial velocity of one ball before collision $=v_{1i}=2\frac{\text{m}}{\text{s}}$

Initial velocity of other ball before collision $=v_{2i}=-3\frac{\text{m}}{\text{s}}$

Final velocity of one ball after collision $=v_{1f}$

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

Formula Used:

Momentum of an object is given as

  $\begin{array}{l}p=mv\\ \text{where }p=\text{momentum}\\ m=\text{mass}\\ v=\text{velocity}\end{array}$

According conservation of momentum principle, for perfectly elastic collision,

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

Also, for perfectly 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,

  $\begin{array}{l}{m}_1{v}_{1i}+m_2{v}_{2i}=m_1{v}_{1f}+m_2{v}_{2f}\\ m(2)+m(-3)=m{v}_{1f}+m{v}_{2f}\\ v_{1f}+v_{2f}=-1\\ v_{2f}=-1-v_{1f}\left(1\right)\end{array}$

Using conservation of kinetic energy

  $\begin{array}{l}{m}_1{v}_{1i}^2+m_2{v}_{2i}^2=m_1{v}_{1f}^2+m_2{v}_{2f}^2\\ m{(}^2+m{(}^{-}=m{v}_{1f}^2+m{v}_{2f}^2\\ v_{1f}^2+v_{2f}^2=13\\ \text{Using equation }\left(1\right)\\ v_{1f}^2+{(}^{-}=13\\ v_{1f}=-3\frac{\text{m}}{\text{s}}\end{array}$

From equation (1),

  $\begin{array}{l}{v}_{2f}=-1-v_{1f}\\ v_{2f}=-1-(-3)\\ v_{2f}=2\frac{\text{m}}{\text{s}}\end{array}$

Conclusion:

Therefore, speed of balls is $3\frac{\text{m}}{\text{s}}\text{ and 2}\frac{\text{m}}{\text{s}}$ opposite to each other.