In the arrangement of Fig. 9-21, billiard ball 1 moving at a speed of 2.2 m/s undergoes a glancing collision with identical billiard ball 2 that is at rest. After the collision, ball 2 moves at speed 1.1 m/s , at an angle of${\mathit{\theta }}_{\mathbf{2}}\mathbf{=}\mathbf{60}\mathbf{°}$. What are (a) the magnitude and (b) the direction of the velocity of ball 1 after the collision? (c) Do the given data suggest the collision is elastic or inelastic?

1. Masses of both balls, $m_1=m_2=m|$.
2. The initial speed of the first ball,$v_{1i}=2.2m/s$.
3. The initial speed of the second ball,$v_{2i}=0.0m/s$.
4. Speed of the ball 2 after the collision,$v_{2f}=1.1m/s$.
5. Ball 2 makes an angle after the collision, ${\theta }_2=60°$.

Using the law of conservation of momentum, you can write the equation for the x and y components of the momentum. From these components, you can find the direction and magnitude of the velocity of ball 1 after the collision.

Using the formula of kinetic energy, you can find the kinetic energy before and after the collision, and determine whether the collision is elastic or not.

Let us assume that $v_{1f}$is the speed of ball 1 after the collision.

According to the law of conservation of momentum, momentum will be conserved in both the direction, so for y direction, we can write,

$$\begin{gathered} m_1{v}_{1i}+m_2{v}_{2i}=m_2{v}_{2f}\sin {\theta }_1+m_2{v}_{2f}\sin {\theta }_2 \\ 0=m{v}_{1f}\sin {\theta }_1+m{v}_{2f}\sin {\theta }_2 \\ 0=v_{1f}\sin {\theta }_1+v_{2f}\sin {\theta }_2 \\ {v}_{1f}\sin {\theta }_1=-v_{2f}\sin {\theta }_2 \end{gathered}$$

Substitute the values in the above expression, and we get,

$$\begin{gathered} v_{1f}\sin {\theta }_1=-1.1\times \sin 60° \\ {v}_{1f}\sin {\theta }_1=-0.9526 \end{gathered}$$ (1)

For x direction, we can write,

$$\begin{gathered} m_1{v}_{1i}+m_2{v}_{2i}=m_2{v}_{2f}\cos {\theta }_1+m_2{v}_{2f}\cos {\theta }_2 \\ {v}_{1i}=m{v}_{1f}\cos {\theta }_1+m{v}_{2f}\cos {\theta }_2 \\ {v}_{1f}\cos {\theta }_1=v_{1f}-v_{2f}\cos {\theta }_2 \end{gathered}$$

Substitute the values in the above expression, and we get,

$\begin{array}{ll}{v}_{1f}& \cos {\theta }_1=2.2-1.1\times (\cos 60°)\\ v_{1f}& \cos {\theta }_1=1.65\end{array}$ (2)

Divide equation 1 by equation 2, and we get,

$$\begin{gathered} \frac{v_{1f}\sin {\theta }_1}{v_{1f}\cos {\theta }_1}=\frac{0.9526}{1.65} \\ \tan {\theta }_1=-0.05773 \\ {\theta }_1={\tan }^{-1}\left(-0.05773\right) \\ {\theta }_1=-29.{999}^0 \\ =-{30}^0 \end{gathered}$$

Substitute the value in equation 1, and we get,

$$\begin{gathered} v_{1f}\cos \left(-{30}^0\right)=1.65 \\ {v}_{1f}=\frac{1.65}{0.866} \\ {v}_{1f}=1.90\frac{m}{s} \end{gathered}$$

Therefore, the magnitude of the velocity of ball 1 after the collision is $1.90\frac{m}{s}$.

We found,

$$\begin{gathered} {\theta }_1=29.999° \\ \approx -30° \end{gathered}$$

Thus, the direction of the veclocity of ball 1 after the collision is $-30°$ counterclockwise from the x-axis.

We can calculate the initial kinetic energy as,

$K_i=\frac{1}{2}m{v}_{1i}^2$

Substitute the values in the above expression, and we get,

$$\begin{gathered} K_i=\frac{1}{2}m{\left(2.2\frac{m}{s}\right)}^2 \\ =2.42mJ \end{gathered}$$

We can calculate the final kinetic energy as,

$K_f=\frac{1}{2}m\left(v_{1f}^2+v_{2f}^2\right)$

Substitute the values in the above expression, and we get,

role="math" localid="1661314935819" $$\begin{gathered} K_f=\frac{1}{2}m\left({\left(1.1\right)}^2+{\left(1.9\right)}^2\right) \\ =2.41mJ \\ \approx K_i \end{gathered}$$

We can see that initial kinetic energy and final kinetic energy are the same. It means that total energy is conserved, and that implies that the collision is elastic.

Thus, the collision is elastic.