Chapter 8, Problem 62P

(a)

To determine

The final speed of the particles in terms of $v_{\text{i}}$.

Answer to Problem 62P

The final speed of the particles of mass $m$ and $3m$ are $\sqrt{2}{v}_{\text{i}}$ and $\sqrt{\frac{2}{3}}{v}_{\text{i}}$ respectively.

Explanation of Solution

Given information: The masses of the particle are $m$ and $3m$. The initial speed of both the particle is $v_{\text{i}}$.

Write the expression to calculate the initial momentum of both particle.

$\begin{array}{c}{P}_{\text{i}}=-m{v}_{\text{i}}\widehat{\text{i}}+3m{v}_{\text{i}}\widehat{\text{i}}\\ =2m{v}_{\text{i}}\widehat{\text{i}}\end{array}$

Here,

$P_{\text{i}}$ is the initial momentum of both particle.

Write the expression to calculate the initial kinetic energy of both particle.

$\begin{array}{c}K{E}_{\text{i}}=\frac{1}{2}m\times v_{\text{i}}^{\text{2}}+\frac{1}{2}3m\times v_{\text{i}}^{\text{2}}\\ =2m{v}_{\text{i}}^{\text{2}}\end{array}$

Here,

$K{E}_{\text{i}}$ is the initial kinetic energy of both particle.

Write the expression of final momentum of the particle of mass m.

$P_{\text{mf}}=-m{v}_{\text{mf}}\widehat{\text{j}}$

Here,

$v_{\text{mf}}$ is the final velocity of the particle of mass m.

Write the expression of conservation of momentum equation.

$\begin{array}{c}{P}_{\text{i}}=P_{\text{mf}}+P_{3\text{mf}}\\ P_{3\text{mf}}=P_{\text{i}}-P_{\text{mf}}\end{array}$ (1)

Here,

$P_{\text{mf}}$ is the final momentum of the particle of mass $m$.

$P_{\text{3mf}}$ is the final momentum of the particle of mass $3m$.

Substitute $2m{v}_{\text{i}}\widehat{\text{i}}$ for $P_{\text{i}}$ and $-m{v}_{\text{mf}}\widehat{\text{j}}$ for $P_{\text{mf}}$ in equation (1) to find $P_{\text{3mf}}$.

$P_{3\text{mf}}=2m{v}_{\text{i}}\widehat{\text{i}}+m{v}_{\text{mf}}\widehat{\text{j}}$

Write the expression to calculate the final velocity of the particle of mass $3m$.

$v_{3\text{mf}}=\frac{P_{3\text{mf}}}{3m}$ (2)

Here,

$v_{3\text{mf}}$ is the final velocity of the particle of mass $3m$.

Substitute $2m{v}_{\text{i}}\widehat{\text{i}}+m{v}_{\text{mf}}\widehat{\text{j}}$ for $P_{\text{3mf}}$ in equation (2) to find $v_{3\text{mf}}$.

$\begin{array}{c}{v}_{3\text{mf}}=\frac{2m{v}_{\text{i}}\widehat{\text{i}}+m{v}_{\text{mf}}\widehat{\text{j}}}{3m}\\ =\frac{2}{3}{v}_{\text{i}}\widehat{\text{i}}+\frac{1}{3}{v}_{\text{mf}}\widehat{\text{j}}\end{array}$

Write the expression of conservation of energy.

$\begin{array}{c}K{E}_{\text{mf}}+K{E}_{3\text{mf}}=K{E}_{\text{i}}\\ \frac{1}{2}m{v}_{\text{mf}}^{\text{2}}+\frac{1}{2}3m{v}_{\text{3mf}}^{\text{2}}=K{E}_{\text{i}}\end{array}$ (3)

Here,

$K{E}_{\text{mf}}$ is the final kinetic energy of the particle of mass $m$.

$K{E}_{\text{3mf}}$ is the final kinetic energy of the particle of mass $3m$.

Substitute $2m{v}_{\text{i}}^{\text{2}}$ for $K{E}_{\text{i}}$ and $\frac{2}{3}{v}_{\text{i}}\widehat{\text{i}}+\frac{1}{3}{v}_{\text{mf}}\widehat{\text{j}}$ for $v_{3\text{mf}}$ in equation (3).

$\begin{array}{c}\frac{1}{2}m{v}_{\text{mf}}^{\text{2}}+\frac{1}{2}3m{\left(\frac{2}{3}{v}_{\text{i}}\widehat{\text{i}}+\frac{1}{3}{v}_{\text{mf}}\widehat{\text{j}}\right)}^2=2m{v}_{\text{i}}^{\text{2}}\\ \frac{1}{2}m{v}_{\text{mf}}^{\text{2}}+\frac{1}{2}3m\left(\frac{4}{9}{v}_{\text{i}}^{\text{2}}+\frac{1}{9}{v}_{\text{mf}}^{\text{2}}\right)=2m{v}_{\text{i}}^{\text{2}}\\ v_{\text{mf}}=\sqrt{\frac{3}{2}\left(2m{v}_{\text{i}}^{\text{2}}-\frac{2}{3}m{v}_{\text{i}}^{\text{2}}\right)}\\ =\sqrt{2}{v}_{\text{i}}\end{array}$

Thus, the final speed of the particle of mass $m$ is $\sqrt{2}{v}_{\text{i}}$.

The final velocity of the particle of mass $3m$ is,

$v_{3\text{mf}}=\frac{2}{3}{v}_{\text{i}}\widehat{\text{i}}+\frac{1}{3}{v}_{\text{mf}}\widehat{\text{j}}$ (4)

The final speed of the particle of mass $m$ is $\sqrt{2}{v}_{\text{i}}$.

Substitute $\sqrt{2}{v}_{\text{i}}$ for $v_{\text{mf}}$ in equation (4) to find $v_{3\text{mf}}$.

$v_{3\text{mf}}=\frac{2}{3}{v}_{\text{i}}\widehat{\text{i}}+\frac{1}{3}\sqrt{2}{v}_{\text{i}}\widehat{\text{j}}$

The magnitude of the velocity $v_{3\text{mf}}$ is,

$\begin{array}{c}\left|v_{3\text{mf}}\right|=\sqrt{{\left(\frac{2}{3}{v}_{\text{i}}\widehat{\text{i}}\right)}^2+{\left(\frac{1}{3}\sqrt{2}{v}_{\text{i}}\widehat{\text{j}}\right)}^2}\\ =\sqrt{\frac{2}{3}}{v}_{\text{i}}\end{array}$

Thus, the final speed of the particle of mass $3m$ is $\sqrt{\frac{2}{3}}{v}_{\text{i}}$.

Conclusion:

Therefore, the final speed of the particles of mass $m$ and $3m$ are $\sqrt{2}{v}_{\text{i}}$ and $\sqrt{\frac{2}{3}}{v}_{\text{i}}$ respectively.

(b)

To determine

The angle $\theta$ at which the particle $3m$ is scattered.

Answer to Problem 62P

The angle $\theta$ at which the particle $3m$ is scattered is $35.3°$.

Explanation of Solution

Given information: The masses of the particle are $m$ and $3m$. The initial speed of both the particle is $v_{\text{i}}$.

Write the expression to calculate the angle $\theta$ at which the particle $3m$ is scattered.

$\theta ={\tan }^{-1}\left(\frac{v_{3\text{mfy}}}{v_{3\text{mfx}}}\right)$

Here,

$v_{3\text{mfy}}$ is the final velocity of the particle of mass $3m$ in y direction.

$v_{3\text{mfx}}$ is the final velocity of the particle of mass $3m$ in x direction.

The final velocity of the particle of mass $3m$ is,

$v_{3\text{mf}}=\frac{2}{3}{v}_{\text{i}}\widehat{\text{i}}+\frac{1}{3}\sqrt{2}{v}_{\text{i}}\widehat{\text{j}}$ . (5)

The x and y component of $v_{3\text{mf}}$ is,

$v_{3\text{mfx}}=\frac{2}{3}{v}_{\text{i}}$ and $v_{3\text{mfy}}=\frac{1}{3}\sqrt{2}{v}_{\text{i}}$

Substitute $\frac{2}{3}{v}_{\text{i}}$ for $v_{3\text{mfx}}$ and $\frac{1}{3}\sqrt{2}{v}_{\text{i}}$ for $v_{3\text{mfy}}$ in equation (5) to find $\theta$.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{\frac{1}{3}\sqrt{2}{v}_{\text{i}}}{\frac{2}{3}{v}_{\text{i}}}\right)\\ =35.3°\end{array}$

Thus, the angle $\theta$ at which the particle $3m$ is scattered is $35.3°$.

Conclusion:

Therefore, the angle $\theta$ at which the particle $3m$ is scattered is $35.3°$.