Chapter 7, Problem 57QAP

To determine

The velocities of the balls after the collision

Answer to Problem 57QAP

The velocity of first ball after collision is 1 m/s towards left and second ball is 2 m/s towards right.

Explanation of Solution

Given info:

A 2.00 kg ball moves at 3.00 m/s towards the right. It collides elastically with a 4.00 kg ball that is initially at rest.

Formula used:

Using the formula, conservation of linear momentum

  $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$ (1)

Here, $m_1\text{and }{m}_2$ are the masses of first and second balls respectively and $u_1\text{and }{u}_2$ are the velocities of the both balls before collision and $v_1\text{and }{v}_2$ are the velocities after collision.

Also, for elastic collision

  $e=\frac{v_2-v_1}{u_1-u_2}$ (2)

Calculation:

We have, $m_1=2.00\text{kg, }{m}_2=4.00\text{kg and }{u}_1=3.00\text{m/s and }{u}_2=0$

Substituting the given values in equ.(1), we get

  $\begin{array}{l}\text{2}\text{kg}\text{×}\text{3}\text{m/s+4}\text{kg×0}=2\text{kg}\times v_1+4\text{kg}\times v_2\\ \Rightarrow v_1+2v_2=3(3)\end{array}$

Now from equation (2)

  $\begin{array}{l}1=\frac{v_2-v_1}{3-0}(e=1,forelasticcollision)\\ \Rightarrow v_2-v_1=3(4)\end{array}$

Adding equation (3) and (4), we get

  $\begin{array}{l}(v_1+2v_2)+\left(v_2-v_1\right)=3+3\\ \Rightarrow 3v_2=6\\ \Rightarrow v_2=2\text{m/s (towards right)}\end{array}$

Substituting in equation (3), we get

  $\begin{array}{l}(v_1+2\times 2)=3\\ \Rightarrow v_1=3-4=-1(towardsleft)\end{array}$

Conclusion:

Thus, the velocity of first ball after collision is 1 m/s towards left and second ball is 2 m/s towards right.