Chapter 7, Problem 93P

To determine

The speed of the target after collision if the collision is perfectly elastic.

Answer to Problem 93P

The speed of the target after collision if the collision is perfectly elastic is $\underset{\_}{2.8\text{m/s}}$.

Explanation of Solution

Given that the mass of the object is $2.0\text{kg}$, the object is approaching the target with speed $8.0\text{m/s}$, the angle of deflection is $90.0°$, and speed of the object after collision is $6.0\text{m/s}$.

Since the collision is perfectly elastic, the kinetic energy and momentum of the system is conserved.

Write the expression indicating the momentum conservation along $x$ direction, to the given system.

$m_1{v}_{1\text{i}x}+m_2{v}_{\text{2i}x}=m_1{v}_{1\text{f}x}+m_2{v}_{2\text{f}x}$ (I)

Here, $m_1$ is the mass of the object, $m_2$ is the mass of the target, $v_{1\text{i}x}$ is the initial speed of the object in $x$ direction, $v_{2\text{i}x}$ is the initial speed of the target in $x$ direction, $v_{1\text{f}x}$ is the final speed of the object in $x$ direction, and $v_{2\text{f}x}$ is the final speed of the target in $x$ direction.

Write the expression indicating the momentum conservation along $y$ direction, to the given system.

$m_1{v}_{1\text{i}y}+m_2{v}_{\text{2i}y}=m_1{v}_{1\text{f}y}+m_2{v}_{2\text{f}y}$ (II)

Here, $v_{1\text{i}y}$ is the initial speed of the object in $y$ direction, $v_{2\text{i}y}$ is the initial speed of the target in $y$ direction, $v_{1\text{f}y}$ is the final speed of the object in $y$ direction, and $v_{2\text{f}y}$ is the final speed of the target in $y$ direction.

Since the collision is elastic, $v_{1\text{i}x}=v_{1\text{i}}$, and $v_{1\text{f}y}=v_{1\text{f}}$.

The initial speed of the target in both $x$ and $y$ directions are zero. The final speed of the object in $x$ direction is zero. Moreover, the initial speed of the object in $y$ direction is zero. Thus, equation (I) and (II) can be modified as,

$m_1{v}_{1\text{i}x}=m_2{v}_{2\text{f}x}$ (III)

$m_1{v}_{1\text{f}y}=-m_2{v}_{2\text{f}y}$ (IV)

Square and add equation (III) and (IV).

$\begin{array}{c}{\left(m_1{v}_{1\text{i}x}\right)}^2+{\left(m_1{v}_{1\text{f}y}\right)}^2={\left(m_2{v}_{2\text{f}x}\right)}^2+{\left(-m_2{v}_{2\text{f}y}\right)}^2\\ m_2^2{v}_{2\text{f}}^2=m_1^2{v}_{1\text{i}}^2+m_1^2{v}_{1\text{f}}^2\\ m_2{v}_{2\text{f}}^2=\frac{m_1^2}{m_2}\left(v_{1\text{i}}^2+v_{1\text{f}}^2\right)\end{array}$ (V)

Write the expression indicating the conservation of kinetic energy of the system.

$\begin{array}{c}\frac{1}{2}{m}_1{v}_{1\text{i}}^2+\frac{1}{2}{m}_2{v}_{2\text{i}}^2=\frac{1}{2}{m}_1{v}_{1\text{f}}^2+\frac{1}{2}{m}_2{v}_{2\text{f}}^2\\ \frac{1}{2}{m}_1{v}_{1\text{i}}^2+0=\frac{1}{2}{m}_1{v}_{1\text{f}}^2+\frac{1}{2}{m}_2{v}_{2\text{f}}^2\\ m_2{v}_{2\text{f}}^2=m_1{v}_{1\text{i}}^2-m_1{v}_{1\text{f}}^2\\ v_{2\text{f}}=\sqrt{\frac{m_1}{m_2}\left(v_{1\text{i}}^2-v_{1\text{f}}^2\right)}\end{array}$ (VI)

Use equation (VI) in (V) and solve for $m_2$.

$\begin{array}{c}{m}_2\left(\frac{m_1}{m_2}\left(v_{1\text{i}}^2-v_{1\text{f}}^2\right)\right)=\frac{m_1^2}{m_2}\left(v_{1\text{i}}^2+v_{1\text{f}}^2\right)\\ m_2=m_1\left(\frac{v_{1\text{i}}^2+v_{1\text{f}}^2}{v_{1\text{i}}^2-v_{1\text{f}}^2}\right)\end{array}$ (VII)

Use equation (VII) in (VI).

$\begin{array}{c}{v}_{2\text{f}}=\sqrt{\frac{m_1\left(v_{1\text{i}}^2-v_{1\text{f}}^2\right)}{m_1\left(\frac{v_{1\text{i}}^2+v_{1\text{f}}^2}{v_{1\text{i}}^2-v_{1\text{f}}^2}\right)}}\\ =\sqrt{\frac{{\left(v_{1\text{i}}^2-v_{1\text{f}}^2\right)}^2}{v_{1\text{i}}^2+v_{1\text{f}}^2}}\\ =\frac{v_{1\text{i}}^2-v_{1\text{f}}^2}{\sqrt{v_{1\text{i}}^2+v_{1\text{f}}^2}}\end{array}$ (VIII)

Conclusion:

Substitute $8.0\text{m/s}$ for $v_{1\text{i}}$, and $6.0\text{m/s}$ for $v_{1\text{f}}$ in equation (VIII) to find $v_{2\text{f}}$.

$\begin{array}{c}{v}_{2\text{f}}=\frac{{\left(8.0\text{m/s}\right)}^2-{\left(6.0\text{m/s}\right)}^2}{\sqrt{{\left(8.0\text{m/s}\right)}^2+{\left(6.0\text{m/s}\right)}^2}}\\ =2.8\text{m/s}\end{array}$

Therefore, the speed of the target after collision if the collision is perfectly elastic is $\underset{\_}{2.8\text{m/s}}$.