Chapter 7, Problem 22P

To determine

The speed and direction of each ball after the collision.

Answer to Problem 22P

The speed of first ball after the collision $v_1=1.1\text{m}/\text{s}$ towards east direction.

The speed of second ball after the collision $v_2=4.4\text{m}/\text{s}$ towards east direction.

Explanation of Solution

Given:

The mass of first ball is $m_1=0.440\text{ kg}$ .

The initial velocity of the first ball is $u_1=3.3\text{m}/\text{s}$ .

The mass of second ball is $m_2=0.220\text{ kg}$ .

The initial velocity of the second ball is $u_2=0$ .

Formula used:

The expression for relative velocity is,

  $u_1-u_2=-\left(v_1-v_2\right)$

Here, $u_1$ is the initial velocity of first ball, $u_2$ is the initial velocity of the second ball, $v_1$ is the final velocity of first ball, and $v_2$ is the final velocity of the second ball.

The expression for law of conservation of momentum is,

  $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

Here, $m_1$ is the mass of the first ball, $m_2$ is the mass of second ball, $u_1$ is the initial velocity of first ball, $u_2$ is the initial velocity of the second ball, $v_1$ is the final velocity of first ball, and $v_2$ is the final velocity of the second ball.

Calculation:

The velocity of first ball after collision is,

  $\begin{array}{l}{u}_1-u_2=-\left(v_1-v_2\right)\\ \left(\text{3}\text{.3 }\text{m}/\text{s}\right)-\left(\text{0}\right)=v_2-v_1\\ \text{3}\text{.3}=v_2-v_1\\ v_1=v_2-\text{3}\text{.3}\end{array}$ …… (1)

The speed of ball using the relation of conservation of momentum is,

  $\begin{array}{l}{m}_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2\\ \left[\left(0.440\text{ kg}\right)\left(3.3\text{m}/\text{s}\right)+\left(0.220\text{ kg}\right)\left(0\right)\right]=\left(0.440\text{ kg}\right)v_1+\left(0.220\text{ kg}\right)v_2\\ 1.452=0.44\left(v_2-\text{3}\text{.3}\right)+0.22v_2\\ 0.66v_2=2.904\\ v_2=4.4\text{ m/s}\end{array}$

  $v_1=3-2v_2$ …… (1)

The velocity of second ball after collision is $v_2=4.4\text{ m/s}$ towards east direction.

The speed of first ball after the collision using equation (1) as follows:

  $\begin{array}{l}{v}_1=v_2-\text{3}\text{.3}\\ v_1=4.4-3.3\\ v_1=1.1\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the speed of first ball after the collision $v_1=1.1\text{m}/\text{s}$ towards east direction.

Thus, the speed of second ball after the collision $v_2=4.4\text{m}/\text{s}$ towards east direction.