Chapter 14.2, Problem 14.48P

In the scattering experiment of Prob. 14.26, it is known that the alpha particle is projected from A₀(300, 0, 300) and that it collides with the oxygen nucleus C at Q(240, 200, 100), where all coordinates are expressed in millimeters. Determine the coordinates of point B₀ where the original path of nucleus B intersects the zx plane. (Hint: Express that the angular momentum of the three particles about Q is conserved.)

14.26 In a scattering experiment, an alpha particle A is projected with the velocity u₀ = −(600 m/s)i + (750 m/s)j − (800 m/s)k into a stream of oxygen nuclei moving with a common velocity v₀ = (600 m/s)j. After colliding successively with the nuclei B and C, particle A is observed to move along the path defined by the points A₁ (280, 240, 120) and A₂ (360, 320, 160), while nuclei B and C are observed to move along paths defined, respectively, by B₁ (147, 220, 130) and B₂ (114, 290, 120), and by C₁ (240, 232, 90) and C₂ (240, 280, 75). All paths are along straight lines and all coordinates are expressed in millimeters. Knowing that the mass of an oxygen nucleus is four times that of an alpha particle, determine the speed of each of the three particles after the collisions.

Fig. P14.26

To determine

The coordinates of point $B_0$ where the original path of nucleus B intersects the zx plane.

Answer to Problem 14.48P

The coordinates of point $B_0$ is $\underset{\_}{\left(181.7\text{mm},0\text{mm},139.4\text{mm}\right)}$.

Explanation of Solution

Given information:

The velocity of the alpha particle A is ${\text{u}}_0=-\left(600\text{m}/\text{s}\right)\text{i}+\left(750\text{m}/\text{s}\right)\text{j}-\left(800\text{m}/\text{s}\right)\text{k}$.

The common velocity of oxygen nuclei is ${\text{v}}_0=\left(600\text{m}/\text{s}\right)\text{j}$.

The alpha particle A projected from $A_0\left(300,0,300\right)$.

The position of the alpha particle A is $A_1\left(280,240,120\right)$ and $A_2\left(360,320,160\right)$.

The position of the nuclei B is $B_1\left(147,220,130\right)$ and $B_2\left(114,290,120\right)$.

The position of the nuclei C is $C_1\left(240,232,90\right)$ and $C_2\left(240,280,75\right)$.

The mass of an oxygen nucleus is four times that of an alpha particle.

The alpha particle collides with the oxygen nucleus C at $Q\left(240,200,100\right)$.

Calculation:

Provide the positions of each point in vector form as shown below.

$\begin{array}{l}{A}_1=280\text{i}+240\text{j}+120\text{k}\\ A_2=360\text{i}+320\text{j}+160\text{k}\\ B_1=147\text{i}+220\text{j}+130\text{k}\\ B_2=114\text{i}+290\text{j}+120\text{k}\end{array}$

$\begin{array}{l}{C}_1=240\text{i}+232\text{j}+90\text{k}\\ C_2=240\text{i}+280\text{j}+75\text{k}\\ A_0=300\text{i}+300\text{k}\\ Q=240\text{i}+200\text{j}+100\text{k}\end{array}$

Sketch the scattering of the alpha and nuclei particles as shown in Figure 1.

Refer to Figure 1.

Calculate the position vectors as shown below.

Calculate $\overrightarrow{A_1{A}_2}$ as shown below.

$\overrightarrow{A_1{A}_2}=A_2-A_1$

Substitute $280\text{i}+240\text{j}+120\text{k}$ for $A_1$ and $360\text{i}+320\text{j}+160\text{k}$ for $A_2$.

$\begin{array}{c}\overrightarrow{A_1{A}_2}=\left(360\text{i}+320\text{j}+160\text{k}\right)-\left(280\text{i}+240\text{j}+120\text{k}\right)\\ =80\text{i}+80\text{j}+40\text{k}\end{array}$

$\begin{array}{c}{A}_1{A}_2=\sqrt{{80}^2+{80}^2+{40}^2}\\ =\sqrt{14,400}\\ =120\text{mm}\end{array}$

Calculate $\overrightarrow{B_1{B}_2}$ as shown below.

$\overrightarrow{B_1{B}_2}=B_2-B_1$

Substitute $147\text{i}+220\text{j}+130\text{k}$ for $B_1$ and $114\text{i}+290\text{j}+120\text{k}$ for $B_2$.

$\begin{array}{c}\overrightarrow{B_1{B}_2}=\left(114\text{i}+290\text{j}+120\text{k}\right)-\left(147\text{i}+220\text{j}+130\text{k}\right)\\ =-33\text{i}+70\text{j}-10\text{k}\end{array}$

$\begin{array}{c}{B}_1{B}_2=\sqrt{{\left(-33\right)}^2+{70}^2+{\left(-10\right)}^2}\\ =\sqrt{6,089}\\ =78.032\text{mm}\end{array}$

Calculate $\overrightarrow{C_1{C}_2}$ as shown below.

$\overrightarrow{C_1{C}_2}=C_2-C_1$

Substitute $240\text{i}+232\text{j}+90\text{k}$ for $C_1$ and $240\text{i}+280\text{j}+75\text{k}$ for $C_2$.

$\begin{array}{c}\overrightarrow{C_1{C}_2}=\left(240\text{i}+280\text{j}+75\text{k}\right)-\left(240\text{i}+232\text{j}+90\text{k}\right)\\ =48\text{j}-15\text{k}\end{array}$

$\begin{array}{c}{C}_1{C}_2=\sqrt{{48}^2+{\left(-15\right)}^2}\\ =\sqrt{2,529}\\ =50.289\text{mm}\end{array}$

Calculate $\overrightarrow{Q{A}_0}$ as shown below.

$\overrightarrow{Q{A}_0}=A_0-Q$

Substitute $300\text{i}+300\text{k}$ for $A_0$ and $240\text{i}+200\text{j}+100\text{k}$ for $Q$.

$\begin{array}{c}\overrightarrow{Q{A}_0}=\left(300\text{i}+300\text{k}\right)-\left(240\text{i}+200\text{j}+100\text{k}\right)\\ =60\text{i}-200\text{j}+200\text{k}\end{array}$

Calculate $\overrightarrow{Q{B}_1}$ as shown below.

$\overrightarrow{Q{B}_1}=B_1-Q$

Substitute $147\text{i}+220\text{j}+130\text{k}$ for $B_1$ and $240\text{i}+200\text{j}+100\text{k}$ for $Q$.

$\begin{array}{c}\overrightarrow{Q{B}_1}=\left(147\text{i}+220\text{j}+130\text{k}\right)-\left(240\text{i}+200\text{j}+100\right)\text{k}\\ =-93\text{i}+20\text{j}+30\text{k}\end{array}$

Calculate $\overrightarrow{Q{B}_0}$ as shown below.

$\overrightarrow{Q{B}_0}=\left(\Delta x\right)\text{i}+\left(\Delta y\right)\text{j}+\left(\Delta z\right)\text{k}$

Calculate the unit vector $\left({\lambda }_A\right)$ as shown below.

${\lambda }_A=\frac{\overrightarrow{A_1{A}_2}}{A_1{A}_2}$

Substitute $80\text{i}+80\text{j}+40\text{k}$ for $\overrightarrow{A_1{A}_2}$ and $120\text{mm}$ for $A_1{A}_2$.

$\begin{array}{c}{\lambda }_A=\frac{80\text{i}+80\text{j}+40\text{k}}{120}\\ =0.66667\text{i}+0.66667\text{j}+0.33333\text{k}\end{array}$

Calculate the unit vectors $\left({\lambda }_B\right)$ as shown below.

${\lambda }_B=\frac{\overrightarrow{B_1{B}_2}}{B_1{B}_2}$

Substitute $-33\text{i}+70\text{j}-10\text{k}$ for $\overrightarrow{B_1{B}_2}$ and $78.032\text{mm}$ for $B_1{B}_2$.

$\begin{array}{c}{\lambda }_B=\frac{-33\text{i}+70\text{j}-10\text{k}}{78.032}\\ =-0.42290\text{i}+0.89707\text{j}-0.12815\text{k}\end{array}$

Calculate the unit vectors $\left({\lambda }_C\right)$ as shown below.

${\lambda }_C=\frac{\overrightarrow{C_1{C}_2}}{C_1{C}_2}$

Substitute $48\text{j}-15\text{k}$ for $\overrightarrow{C_1{C}_2}$ and $50.289\text{mm}$ for $C_1{C}_2$.

$\begin{array}{c}{\lambda }_C=\frac{48\text{j}-15\text{k}}{50.289}\\ =0.95448\text{j}-0.29828\text{k}\end{array}$

Provide the velocity vectors after the collisions as shown below.

$\begin{array}{l}{\text{v}}_A={\lambda }_A{v}_A\\ {\text{v}}_B={\lambda }_B{v}_B\\ {\text{v}}_C={\lambda }_C{v}_C\end{array}$

Apply the conservation of momentum as shown below.

$\begin{array}{l}m{\text{u}}_0+4m{\text{v}}_0+4m{\text{v}}_0=m{\text{v}}_A+4m{\text{v}}_B+4m{\text{v}}_C\\ {\text{u}}_0+8{\text{v}}_0={\text{v}}_A+4{\text{v}}_B+4{\text{v}}_C\end{array}$

Substitute ${\lambda }_A{v}_A$ for ${\text{v}}_A$, ${\lambda }_B{v}_B$ for ${\text{v}}_B$, and ${\lambda }_C{v}_C$ for ${\text{v}}_C$.

${\text{u}}_0+8{\text{v}}_0={\lambda }_A{v}_A+4{\lambda }_B{v}_B+4{\lambda }_C{v}_C$

Substitute $-600\text{i}+750\text{j}-800\text{k}\left(\text{m}/\text{s}\right)$ for ${\text{u}}_0$, $600\text{j}\left(\text{m}/\text{s}\right)$ for ${\text{v}}_0$, $0.66667\text{i}+0.66667\text{j}+0.33333\text{k}$ for ${\lambda }_A$, $-0.42290\text{i}+0.89707\text{j}-0.12815\text{k}$ for ${\lambda }_B$, and $0.95448\text{j}-0.29828\text{k}$ for ${\lambda }_C$.

$\begin{array}{l}-600\text{i}+750\text{j}-800\text{k}+8\left(600\text{j}\right)=\left[\begin{array}{l}\left(0.66667\text{i}+0.66667\text{j}+0.33333\text{k}\right)v_A\\ +4\left(-0.42290\text{i}+0.89707\text{j}-0.12815\text{k}\right)v_B\\ +4\left(0.95448\text{j}-0.29828\text{k}\right)v_C\end{array}\right]\\ -600\text{i}+5,550\text{j}-800\text{k}=\left[\begin{array}{l}\left(0.66667\text{i}+0.66667\text{j}+0.33333\text{k}\right)v_A\\ +\left(-1.6916\text{i}+3.58828\text{j}-0.5126\text{k}\right)v_B\\ +\left(3.81792\text{j}-1.19312\text{k}\right)v_C\end{array}\right]\end{array}$

Equating the components of $\text{i},\text{j},$ and $\text{k}$ as shown below.

$\begin{array}{l}0.66667v_A-1.6916v_B=-600\\ 0.66667v_A+3.58828v_B+3.81792v_C=5,550\\ 0.33333v_A-0.5126v_B-1.19312v_C=-800\end{array}$

Solve the Equations to get the speed of the particles.

$\begin{array}{l}{v}_A=919\text{m}/\text{s}\\ v_B=717\text{m}/\text{s}\\ v_C=619\text{m}/\text{s}\end{array}$

Calculate the velocity vector $\left({\text{v}}_B\right)$ as shown below.

${\text{v}}_B={\lambda }_B{v}_B$

Substitute $717\text{m}/\text{s}$ for $v_B$ and $-0.42290\text{i}+0.89707\text{j}-0.12815\text{k}$ for ${\lambda }_B$.

$\begin{array}{c}{\text{v}}_B=717\times \left(-0.42290\text{i}+0.89707\text{j}-0.12815\text{k}\right)\\ =-303.2193\text{i}+643.19919\text{j}-91.88355\text{k}\end{array}$

Apply the conservation of momentum about Q as shown below.

$\begin{array}{l}\left(\begin{array}{l}\overrightarrow{Q{A}_0}\times m{\text{u}}_0+\overrightarrow{Q{B}_0}\times \left(4m{\text{v}}_0\right)\\ +\overrightarrow{Q{C}_0}\times \left(4m{\text{v}}_0\right)\end{array}\right)=\overrightarrow{Q{A}_1}\times m{\text{v}}_A+\overrightarrow{Q{B}_1}\times \left(4m{\text{v}}_B\right)+\overrightarrow{Q{C}_1}\times \left(4m{\text{v}}_C\right)\\ \overrightarrow{Q{A}_0}\times m{\text{u}}_0+\overrightarrow{Q{B}_0}\times \left(4m{\text{v}}_0\right)+0=0+\overrightarrow{Q{B}_1}\times \left(4m{\text{v}}_B\right)+0\\ \overrightarrow{Q{A}_0}\times {\text{u}}_0+\overrightarrow{Q{B}_0}\times \left(4{\text{v}}_0\right)=\overrightarrow{Q{B}_1}\times \left(4{\text{v}}_B\right)\end{array}$

Substitute $60\text{i}-200\text{j}+200\text{k}$ for $\overrightarrow{Q{A}_0}$, $-600\text{i}+750\text{j}-800\text{k}\left(\text{m}/\text{s}\right)$ for ${\text{u}}_0$, $\left(\Delta x\right)\text{i}+\left(\Delta y\right)\text{j}+\left(\Delta z\right)\text{k}$ for $\overrightarrow{Q{B}_0}$, $600\text{j}\left(\text{m}/\text{s}\right)$ for ${\text{v}}_0$, $-93\text{i}+20\text{j}+30\text{k}$ for $\overrightarrow{Q{B}_1}$, and $-303.2193\text{i}+643.19919\text{j}-91.88355\text{k}$ for ${\text{v}}_B$.

$\begin{array}{l}\left[\begin{array}{l}\left(60\text{i}-200\text{j}+200\text{k}\right)\times \\ \left(-600\text{i}+750\text{j}-800\text{k}\right)\\ +\left(\left(\Delta x\right)\text{i}+\left(\Delta y\right)\text{j}+\left(\Delta z\right)\text{k}\right)\times \left(4\times 600\text{j}\right)\end{array}\right]=\left[\begin{array}{l}\left(-93\text{i}+20\text{j}+30\text{k}\right)\\ \times 4\left(\begin{array}{l}-303.2193\text{i}\\ +643.19919\text{j}\\ -91.88355\text{k}\end{array}\right)\end{array}\right]\\ \left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 60& -200& 200\\ -600& 750& -800\end{array}\right|+\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ \Delta x& \Delta y& \Delta z\\ 0& 2,400& 0\end{array}\right|=\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ -93& 20& 30\\ -1,212.877& 2,572.797& -367.534\end{array}\right|\\ \left[\begin{array}{l}\left(10,000\text{i}-72,000\text{j}-75,000\text{k}\right)\\ +\left(-2,400\Delta z\text{i}+2,400\Delta x\text{k}\right)\end{array}\right]=\left(-84,534.59\text{i}-70,566.972\text{j}-215,012.581\text{k}\right)\end{array}$

Equating the components of $\text{i},\text{j},$ and $\text{k}$ as shown below.

$\begin{array}{l}10,000-2,400\Delta z=-84,534.59\\ -2,400\Delta z=-94,534.59\\ \Delta z=39.389\text{mm}\end{array}$

$\begin{array}{l}-75,000+2,400\Delta x=-215,012.581\\ 2,400\Delta x=-140,0012.581\\ \Delta x=-58.339\text{mm}\end{array}$

Refer to Figure 1.

Calculate the coordinates of $B_0$ as shown below.

Coordinate of along x direction.

$x_{B_0}=x_Q+\Delta x$

Substitute $240\text{mm}$ for $x_Q$ and $-58.339\text{mm}$ for $\Delta x$.

$\begin{array}{c}{x}_{B_0}=240-58.339\\ =181.66\text{mm}\\ \approx 181.7\text{mm}\end{array}$

Coordinate of along y direction.

$y_{B_0}=0$

Coordinate of along z direction.

$z_{B_0}=z_Q+\Delta z$

Substitute $100\text{mm}$ for $x_Q$ and $39.389\text{mm}$ for $\Delta x$.

$\begin{array}{c}{z}_{B_0}=100+39.389\\ =139.389\text{mm}\\ \approx 139.4\text{mm}\end{array}$

Therefore, the coordinates of point $B_0$ is $\underset{\_}{\left(181.7\text{mm},0\text{mm},139.4\text{mm}\right)}$.