Chapter 7, Problem 63P

(a)

To determine

The $x$ and $y$ component of momentum change of the ball of mass $m_1$.

Answer to Problem 63P

The $x$ and $y$ components of momentum change of the ball of mass $m_1$ are $\underset{\_}{\Delta p_{1x}=-1.00m_1{v}_{\text{i}}}$ and $\underset{\_}{\Delta p_{1y}=0.751m_1{v}_{\text{i}}}$ respectively.

Explanation of Solution

The momentum change of the ball of mass $m_1$ is equal but opposite to that of the ball of mass $m_2$.

$\Delta p_{1x}=-\Delta p_{2x}$ (I)

$\Delta p_{1y}=-\Delta p_{2y}$ (II)

Write the expression for the momentum change of the ball of mass $m_2$ in $x$ and $y$ directions.

$\Delta p_{2x}=m_2{v}_{2\text{f}x}-m_2{v}_{2\text{i}x}$ (III)

$\Delta p_{2y}=m_2{v}_{2\text{f}y}-m_2{v}_{2\text{i}y}$ (IV)

Here, $v_{2\text{i}}$ is the initial speed of the ball of mass $m_2$, and $v_{2\text{f}}$ is the final speed of the ball of mass $m_2$.

Use equation (III) in (I).

$\Delta p_{1x}=m_2{v}_{2\text{i}x}-m_2{v}_{2\text{f}x}$ (V)

Use equation (IV) in (II).

$\Delta p_{1y}=m_2{v}_{2\text{i}y}-m_2{v}_{2\text{f}y}$ (VI)

The initial speed of the ball of mass $m_2$ is zero in both $x$ and $y$ directions. Thus, equations (V) and (VI) can be modified as,

$\begin{array}{c}\Delta p_{1x}=m_2\left(0-v_{2\text{f}x}\right)\\ =-m_2{v}_{2\text{f}x}\end{array}$ (VII)

$\begin{array}{c}\Delta p_{1y}=m_2\left(0-v_{2\text{f}y}\right)\\ =-m_2{v}_{2\text{f}y}\end{array}$ (VIII)

Use $m_2=5m_1$ in equations (VII) and (VIII).

$\Delta p_{1x}=-5m_1{v}_{2\text{f}x}$ (IX)

$\Delta p_{1y}=-5m_1{v}_{2\text{f}y}$ (X)

Write the expression for the $x$ and $y$ components of the final velocity of the ball of mass $m_2$.

$v_{2\text{f}x}=\frac{1}{4}{v}_{\text{i}}\cos \left(-36.9°\right)$ (XI)

$v_{2\text{f}y}=\frac{1}{4}{v}_{\text{i}}\sin \left(-36.9°\right)$ (XII)

Use equation (XI) in equation (IX).

$\begin{array}{c}\Delta p_{1x}=-5m_1\left[\frac{1}{4}{v}_{\text{i}}\cos \left(-36.9°\right)\right]\\ =-1.00m_1{v}_{\text{i}}\end{array}$

$\begin{array}{c}\Delta p_{1y}=-5m_1\left[\frac{1}{4}{v}_{\text{i}}\sin \left(-36.9°\right)\right]\\ =0.751m_1{v}_{\text{i}}\end{array}$

Conclusion:

Therefore, the $x$ and $y$ components of momentum change of the ball of mass $m_1$ are $\underset{\_}{\Delta p_{1x}=-1.00m_1{v}_{\text{i}}}$ and $\underset{\_}{\Delta p_{1y}=0.751m_1{v}_{\text{i}}}$ respectively.

(b)

To determine

The $x$ and $y$ component of momentum change of the ball of mass $m_2$ and the relation between the momentum changes.

Answer to Problem 63P

The $x$ and $y$ components of momentum change of the ball of mass $m_2$ are $\underset{\_}{\Delta p_{2x}=+1.00m_1{v}_{\text{i}}}$ and $\underset{\_}{\Delta p_{2y}=-0.751m_1{v}_{\text{i}}}$ respectively. The momentum changes for each mass are equal in magnitude and opposite.

Explanation of Solution

The momentum change of the ball of mass $m_2$ is equal but opposite to that of the ball of mass $m_1$.

$\Delta p_{2x}=-\Delta p_{1x}$ (XIII)

$\Delta p_{2y}=-\Delta p_{1y}$ (XIV)

In part (a) it is obtained that the change in momentum of the ball of mass $m_1$ in $x$ and $y$ directions are as, $\Delta p_{1x}=-1.00m_1{v}_{\text{i}}$ and $\Delta p_{1y}=0.751m_1{v}_{\text{i}}$.

Thus, from equations (XIII) and (XIV) it can be deduced that,

$\Delta p_{2x}=+1.00m_1{v}_{\text{i}}$

$\Delta p_{2y}=-0.751m_1{v}_{\text{i}}$

The momentum changes for each mass are equal in magnitude and opposite.

Conclusion:

Therefore, the $x$ and $y$ components of momentum change of the ball of mass $m_2$ are $\underset{\_}{\Delta p_{2x}=+1.00m_1{v}_{\text{i}}}$ and $\underset{\_}{\Delta p_{2y}=-0.751m_1{v}_{\text{i}}}$ respectively. The momentum changes for each mass are equal in magnitude and opposite.