Chapter 8, Problem 50TP

To determine

(a)

The final velocity of mass B after the collision.

Answer to Problem 50TP

The final velocity of mass B after collision is $5\text{m}/\text{s}$.

Explanation of Solution

Given:

The mass A is moving in $+x$ direction with an initial velocity, $v_1=15\text{m}/\text{s}$.

The mass of B is, $m_2=2m_1$.

The mass B is moving in $-x$ direction with an initial velocity, $v_2=10\text{m}/\text{s}$.

The mass A is moving in $-x$ direction after collision with a final velocity, ${v^{\prime }}_1=15\text{m}/\text{s}$.

Formula used:

Write the expression for conservation of momentum.

  $m_1{v}_1+m_2{v}_2=m_1{v^{\prime }}_1+m_2{v^{\prime }}_2$

Here, $m_1$ is the mass of A, $m_2$ is the mass of B, $v_1$ is the initial velocity of mass A, $v_2$ is the initial velocity of mass B, ${v^{\prime }}_1$ is the final velocity of the mass A and ${v^{\prime }}_2$ the final velocity of mass B.

Calculation:

The final velocity of mass B is calculated as

  $\begin{array}{c}{m}_1{v}_1+m_2{v}_2=m_1{v^{\prime }}_1+m_2{v^{\prime }}_2\\ m_1\left(15\text{m}/\text{s}\right)+\text{2}{m}_1\left(-10\text{m}/\text{s}\right)=m_1\left(-15\text{m}/\text{s}\right)+\text{2}{m}_1{v^{\prime }}_2\\ 2m_1{v^{\prime }}_2=10m_1\\ {v^{\prime }}_2=5\text{m}/\text{s}\end{array}$

Conclusion:

The velocity of mass B after collision is $5\text{m}/\text{s}$.

To determine

(b)

The change in kinetic energy after the collision.

Answer to Problem 50TP

The change in kinetic energy after collision is $-375\text{J}$.

Explanation of Solution

Given:

The mass of A is, $m_1=5\text{kg}$.

Formula used:

The initial kinetic energy ${\text{KE}}_{\text{i}}$ of the system is given by

  ${\text{KE}}_{\text{i}}=\frac{1}{2}\left(m_1{v}_1^2+m_2{v}_2^2\right)$

The final kinetic energy ${\text{KE}}_{\text{f}}$ of the system is given by

  ${\text{KE}}_{\text{f}}=\frac{1}{2}\left[m_1{\left({v^{\prime }}_1\right)}^2+m_2{\left({v^{\prime }}_2\right)}^2\right]$

The formula for change in kinetic energy $\Delta \text{KE}$ is given by

  $\Delta \text{KE}={\text{KE}}_{\text{f}}-{\text{KE}}_{\text{i}}$

Calculation:

The mass of B is calculated as

  $\begin{array}{c}{m}_2=2m_1\\ =2\left(5\text{kg}\right)\\ =\text{10}\text{kg}\end{array}$

The initial kinetic energy ${\text{KE}}_{\text{i}}$ of the system is calculated as

  $\begin{array}{c}{\text{KE}}_{\text{i}}=\frac{1}{2}\left[\left(5\text{kg}\right){\left(15\text{m}/\text{s}\right)}^2+\left(10\text{kg}\right){\left(-10\text{m}/\text{s}\right)}^2\right]\\ =1062.5\text{J}\end{array}$

The final kinetic energy ${\text{KE}}_{\text{f}}$ of the system is calculated as,

  $\begin{array}{c}{\text{KE}}_{\text{f}}=\frac{1}{2}\left[\left(5\text{kg}\right){\left(-15\text{m}/\text{s}\right)}^2+\left(10\text{kg}\right){\left(5\text{m}/\text{s}\right)}^2\right]\\ =687.5\text{J}\end{array}$

The change in kinetic energy is calculated as

  $\begin{array}{c}\Delta \text{KE}=687.\text{5}\text{J}-1062.5\text{J}\\ =-375\text{J}\end{array}$

Conclusion:

The change in kinetic energy after collision is $-375\text{J}$.