Chapter 7, Problem 26P

(a)

To determine

The velocity of the target ball after collision.

Answer to Problem 26P

The velocity of the target ball after collision is $\text{4}\text{.8 }\text{m}/\text{s}$ .

Explanation of Solution

Given:

The mass of the soft ball is $m_1=0.220\text{ kg}$ .

The initial velocity of the soft ball is $u_1=8.5\text{m}/\text{s}$ .

The initial velocity of the target ball is $u_2=\text{0}$ .

The final velocity of the soft ball is $v_1=-3.7\text{m}/\text{s}$ .

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 soft ball, $u_2$ is the initial velocity of the target ball, $v_1$ is the final velocity of soft ball, and $v_2$ is the final velocity of the target 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 soft ball, $m_2$ is the mass of target ball, $u_1$ is the initial velocity of soft ball, $u_2$ is the initial velocity of the target ball, $v_1$ is the final velocity of soft ball, and $v_2$ is the final velocity of the target ball.

Calculation:

The velocity of target ball after collision is,

  $\begin{array}{l}{u}_1-u_2=-\left(v_1-v_2\right)\\ \left(8.5\text{m}/\text{s}\right)-\left(0\right)=v_2-v_1\\ \left(8.5\text{m}/\text{s}\right)-\left(0\right)=v_2-\left(-\text{3}\text{.7 }\text{m}/\text{s}\right)\\ v_2=\text{4}\text{.8 }\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the velocity of target ball after collision is $v_2=\text{4}\text{.8 }\text{m}/\text{s}$ .

(b)

To determine

The mass of the target ball.

Answer to Problem 26P

The mass of the target ball is $0.560\text{ kg}$ .

Explanation of Solution

Given:

The mass of the soft ball is $m_1=0.220\text{ kg}$ .

The initial velocity of the soft ball is $u_1=8.5\text{m}/\text{s}$ .

The initial velocity of the target ball is $u_2=\text{0}$ .

The final velocity of the soft ball is $v_1=-3.7\text{m}/\text{s}$ .

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 soft ball, $u_2$ is the initial velocity of the target ball, $v_1$ is the final velocity of soft ball, and $v_2$ is the final velocity of the target 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 soft ball, $m_2$ is the mass of target ball, $u_1$ is the initial velocity of soft ball, $u_2$ is the initial velocity of the target ball, $v_1$ is the final velocity of soft ball, and $v_2$ is the final velocity of the target ball.

Calculation:

From part (a), the velocity of target ball after collision is $v_2=\text{4}\text{.8 }\text{m}/\text{s}$ .

The mass of target ball using the relation of conservation of momentum is,

  $\begin{array}{l}{m}_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2\\ \left(0.220\text{ kg}\right)\left(\text{8}\text{.5 }\text{m}/\text{s}\right)+m_2\left(0\right)=\left(0.220\text{ kg}\right)\left(-\text{3}\text{.7 }\text{m}/\text{s}\right)+m_2\left(\text{4}\text{.8 }\text{m}/\text{s}\right)\\ 4.8m_2=2.684\\ m_2=0.560\text{ kg}\end{array}$

Conclusion:

Hence, the mass of the target ball is $m_2=0.560\text{ kg}$ .