Chapter 8, Problem 32TP

To determine

(a)

The velocity of object B after collision.

Answer to Problem 32TP

The velocity of object B after collision using conservation of momentum is $4\text{m}/\text{s}$.

Explanation of Solution

Given:

Velocity of object A before collision, $u_{\text{a}}=4\text{m}/\text{s}$.

Velocity of object B before collision, $u_{\text{b}}=-8\text{m}/\text{s}$.

Velocity of object A after collision, $v_{\text{a}}=-8\text{m}/\text{s}$.

Mass of object A = Mass of object B = $m$

Formula used:

Momentum is conserved.

The velocity of object B after collision will be calculated as

  $m_{\text{a}}{u}_{\text{a}}+m_{\text{b}}{u}_{\text{b}}=m_{\text{a}}{v}_{\text{a}}+m_{\text{b}}{v}_{\text{b}}$

Calculation:

The velocity of object B after collision will be calculated as

  $\begin{array}{c}{m}_{\text{a}}{u}_{\text{a}}+m_{\text{b}}{u}_{\text{b}}=m_{\text{a}}{v}_{\text{a}}+m_{\text{b}}{v}_{\text{b}}\\ m\left(4\text{m}/\text{s}\right)+m\left(-8\text{m}/\text{s}\right)=m\left(-8\text{m}/\text{s}\right)+m{v}_{\text{b}}\\ -m\left(4\text{m}/\text{s}\right)+m\left(8\text{m}/\text{s}\right)=v_{\text{b}}\\ v_{\text{b}}=4\text{m}/\text{s}\end{array}$

Conclusion:

The velocity of object B after collision, using conservation of momentum is $4\text{m}/\text{s}$.

To determine

(b)

The velocity of object B after collision.

Answer to Problem 32TP

The velocity of object B after collision using the conservation of kinetic energy principle is $4\text{m}/\text{s}$.

Explanation of Solution

Formula Used:

The total kinetic energy is conserved in elastic collision. It is given by

  $\frac{1}{2}{m}_{\text{a}}{\left(u_{\text{a}}\right)}^2+\frac{1}{2}{m}_{\text{b}}{\left(u_{\text{b}}\right)}^2=\frac{1}{2}{m}_{\text{a}}{\left(v_{\text{a}}\right)}^2+\frac{1}{2}{m}_{\text{b}}{\left(v_{\text{b}}\right)}^2$

Calculation:

The total kinetic energy is conserved in elastic collision. It is given by

  $\begin{array}{c}\frac{1}{2}{m}_{\text{a}}{\left(u_{\text{a}}\right)}^2+\frac{1}{2}{m}_{\text{b}}{\left(u_{\text{b}}\right)}^2=\frac{1}{2}{m}_{\text{a}}{\left(v_{\text{a}}\right)}^2+\frac{1}{2}{m}_{\text{b}}{\left(v_{\text{b}}\right)}^2\\ \frac{1}{2}m{\left(4\text{m}/\text{s}\right)}^2+\frac{1}{2}m{\left(-8\text{m}/\text{s}\right)}^2=\frac{1}{2}m{\left(-8\text{m}/\text{s}\right)}^2+\frac{1}{2}m{\left(v_{\text{b}}\right)}^2\\ 16{\text{m}}^2/{\text{s}}^2+64{\text{m}}^2/{\text{s}}^2-64{\text{m}}^2/{\text{s}}^2={\left(v_{\text{b}}\right)}^2\\ v_{\text{b}}=4\text{m}/\text{s}\end{array}$

Conclusion:

The velocity of object B after collision, using conservation of kinetic energy principle is $4\text{m}/\text{s}$.