Chapter 7, Problem 42P

(a)

To determine

The equation expressing the conservation of the momentum for the components in the x and y direction.

Answer to Problem 42P

The the expression for the conservation of momentum along x direction is $m_1{u}_1=m_1{v}_1\cos {\theta }_A+m_2{v}_2\cos {\theta }_B$ .

The expression for the conservation of momentum along y direction is $m_1{v}_1\sin {\theta }_A=m_2{v}_2\sin {\theta }_B$ .

Explanation of Solution

Given:

The mass of ball A is $m_1=0.400\text{ kg}$ .

The initial velocity of the ball A is $u_1=1.8\text{m}/\text{s}$ .

The mass of ball B is $m_2=0.500\text{ kg}$ .

The final velocity of the ball A is $v_1=1.10\text{m}/\text{s}$ .

The deflected angle of ball A is ${\theta }_{\text{A}}=30°$ .

Formula used:

The expression for law of conservation of momentum is,

  $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

Here, $m_1$ is the mass of the ball A, $m_2$ is the mass of ball B, $u_1$ is the initial velocity of ball A, $u_2$ is the initial velocity of the ball B, $v_1$ is the final velocity of ball A, and $v_2$ is the final velocity of the ball B.

Calculation:

Sketch the representation of the billiard ball before and after collision as shown in Figure 1.

  

Apply the conservation of momentum along x direction.

  $m_1{u}_1=m_1{v}_1\cos {\theta }_A+m_2{v}_2\cos {\theta }_B$ …… (1)

Apply the conservation of momentum along y direction.

  $\begin{array}{l}0=m_1{v}_1\sin {\theta }_A-m_2{v}_2\sin {\theta }_B\\ m_1{v}_1\sin {\theta }_A=m_2{v}_2\sin {\theta }_B\end{array}$ …… (2)

Conclusion:

Thus, the expression for the conservation of momentum along x direction is $m_1{u}_1=m_1{v}_1\cos {\theta }_A+m_2{v}_2\cos {\theta }_B$ .

Thus, the expression for the conservation of momentum along y direction is $m_1{v}_1\sin {\theta }_A=m_2{v}_2\sin {\theta }_B$ .

(b)

To determine

The speed and deflected angle of the ball B.

Answer to Problem 42P

The deflected angle of ball B is ${\theta }_B=26.57°$ .

The speed of ball B is $v_2=1.61\text{ m/s}$ .

Explanation of Solution

Given:

The mass of ball A is $m_1=0.400\text{ kg}$ .

The initial velocity of the ball A is $u_1=1.8\text{m}/\text{s}$ .

The mass of ball B is $m_2=0.500\text{ kg}$ .

The final velocity of the ball A is $v_1=1.10\text{m}/\text{s}$ .

The deflected angle of ball A is ${\theta }_{\text{A}}=30°$ .

Formula used:

The expression for law of conservation of momentum is,

  $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

Here, $m_1$ is the mass of the ball A, $m_2$ is the mass of ball B, $u_1$ is the initial velocity of ball A, $u_2$ is the initial velocity of the ball B, $v_1$ is the final velocity of ball A, and $v_2$ is the final velocity of the ball B.

Calculation:

Refer to part (a).

The expression for the conservation of momentum along x direction is $m_1{u}_1=m_1{v}_1\cos {\theta }_A+m_2{v}_2\cos {\theta }_B$ .

The expression for the conservation of momentum along y direction is $m_1{v}_1\sin {\theta }_A=m_2{v}_2\sin {\theta }_B$ .

The speed and deflected angle of ball B using the relation of conservation of momentum along x direction is,

  $\begin{array}{l}{m}_1{u}_1=m_1{v}_1\cos {\theta }_A+m_2{v}_2\cos {\theta }_B\\ \left(0.400\text{kg}\right)\left(1.8\text{ m/s}\right)=\left(0.400\text{kg}\right)\left(0\right)\cos 30°+\left(0.500\text{ kg}\right)v_2\cos {\theta }_B\\ 0.5v_2\cos {\theta }_B=0.72\\ v_2\cos {\theta }_B=1.44\end{array}$ …… (3)

The speed and deflected angle of ball B using the relation of conservation of momentum along y direction is,

  $\begin{array}{l}{m}_1{v}_1\sin {\theta }_A=m_2{v}_2\sin {\theta }_B\\ \left(0.400\text{kg}\right)\left(1.8\text{ m/s}\right)\sin 30°=\left(0.500\text{ kg}\right)v_2\sin {\theta }_B\\ 0.5v_2\sin {\theta }_B=0.36\\ v_2\sin {\theta }_B=0.72\end{array}$ …… (4)

Find the deflected angle of ball B using equations (3) and (4).

  $\begin{array}{l}\frac{v_2\sin {\theta }_B}{v_2\cos {\theta }_B}=\frac{0.72}{1.44}\\ \tan {\theta }_B=0.5\\ {\theta }_B=26.57°\end{array}$

The speed of the ball B using equation (3) is,

  $\begin{array}{l}{v}_2\mathrm{co}{\theta }_B=1.44\\ v_2\cos 26.57°=1.44\\ v_2=1.61\text{ m/s}\end{array}$

Conclusion:

Thus, the deflected angle of ball B is ${\theta }_B=26.57°$ and the speed of ball B is $v_2=1.61\text{ m/s}$ .