The three balls in the overhead view of Fig. 9-76 are identical. Balls 2 and 3 touch each other and are aligned perpendicular to the path of ball 1 . The velocity of ball 1 has magnitude ${\mathbf{v}}_{\mathbf{0}}\mathbf{=}\mathbf{10}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$ and is directed at the contact point of balls 1 and 2. After the collision, what are the (a) speed and (b) direction of the velocity of ball 2, the (c) speed and (d) direction of the velocity of ball 3, and the (e) speed and (f) direction of the velocity of ball 1? (Hint: With friction absent, each impulse is directed along the line connecting the centers of the colliding balls, normal to the colliding surfaces.)

1. Masses of both balls,$m_1=m_2=m_3=m$.
2. The initial speed of the first ball,$v_{1i}=10m/s$.
3. Angle, $\theta =30°$.

The incident ball exerts an impulse of the same magnitude on each ball along the line, which joins the centers of the target balls and incident ball. After the collision, the incident ball leaves along the x-axis, whereas the target balls make an equilateral triangle with the incident ball.

Using the law of conservation of momentum, you can write an equation for the x component of the momentum. From this, you can find the direction and magnitude of the velocity of ball 1, ball 2, and ball 3 after the collision.

1. X-component:$m{v}_{1i}=m{v}_{1f}+mv\cos \theta$
2. $K=\frac{1}{2}m{v}^2$

The incident ball exerts an impulse of the same magnitude on each ball along the line, which joins the centers of the target balls and the incident ball. After the collision, the incident ball leaves along the x-axis, whereas target balls make an equilateral triangle with the incident ball. Therefore, it makes an angle of $\theta =30°$ with the x-axis.

Let$v_{2f}=v_{3f}=v$ be the velocity of target balls after the collision.

According to the law of conservation of momentum along x component can be written as,

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

Squaring both sides of the equation, we get,

$$\begin{gathered} {\left(v_{1f}\right)}^2={\left(v_{1i}-2v\cos \theta \right)}^2 \\ {v}_{1f}^2=v_{1i}^2-4v_{1j}v\cos \theta +4v^2\theta \end{gathered}$$ (1)

Now, using the conservation of kinetic energy as,

$$\begin{gathered} \frac{1}{2}{m}_{1i}^2=\frac{1}{2}{m}_{1f}^2+2\left(\frac{1}{2}{m}^2\right) \\ {v}_{1i}^2=v_{1f}^2+2N_2^2 \end{gathered}$$ (2)

Therefore, substituting values from equation 1 into equation 2 as,

$$\begin{gathered} v_{1i}^2=v_{1i}^2-4v_{1i}v\cos \theta +4v^2\theta +2v^2 \\ 4v_{1i}v\cos \theta =4v^2\theta +2v^2 \\ v=\frac{\left(4v^2\theta +2v^2\right)}{4v_{1i}\cos \theta } \end{gathered}$$

Solving further as,

$$\begin{gathered} v=\frac{2v^2\left(1+2{\cos }^2\theta \right)}{4v_{1i}\cos \theta } \\ v=\frac{v^2\left(1+2{\cos }^2\theta \right)}{2v_{1i}\cos \theta } \\ v=\frac{2v_{1i}\cos \theta }{\left(1+2{\cos }^2\theta \right)} \end{gathered}$$

Substitute the values in the above expression, and we get

$$\begin{gathered} v=\frac{2\left(10{\displaystyle \frac{m}{s}}\right)\cos 30°}{\left(1+2{\cos }^{2}30°\right)} \\ =6.93\frac{m}{s} \\ \approx 6.9m/s \end{gathered}$$

Therefore, the speed of ball 2 after the collision is 6.9 m/s .

As after the collision, the incident ball leaves along the x-axis, whereas target balls make an angle of $\theta =30°$with an x-axis.

As ball 2 moves above the x-axis, the angle is$30°$counterclockwise from the x-axis.

Thus, the direction of the velocity of ball 2 after the collision is $30°$.

After the collision, the incident ball leaves along the x-axis, whereas target balls make an angle of $\theta =30°$with the x-axis.

As the speed of ball 2 after the collision is 6.9 m/s , the speed of ball 3 after the collision is 6.9 m/s .

Therefore, the speed of the ball 3 after the collision is 6.9 m/s .

After the collision, the incident ball leaves along the x-axis, whereas target balls make an angle of $\theta =30°$with the x-axis.

As ball 3 moves below the x-axis, the angle is$-30°$counterclockwise from the x-axis.

Thus, The direction of the velocity of ball 3 after the collision is$-30°$ counterclockwise from the x-axis.

The magnitude of ball 1 after the collision can be calculated as,

$v_{1f}=v_{1i}-2v\cos \theta$

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

$$\begin{gathered} v_{1f}=10\frac{m}{s}-2\left(6.93\frac{m}{s}\right)\cos 30° \\ =2.0\frac{m}{s} \end{gathered}$$

Therefore, the magnitude of the speed of ball 1 after the collision is 2.0 m/s.

After the collision, the incident ball leaves along the x-axis, the direction of the velocity of ball 1 after the collision is $-180°$.

Therefore, the direction of the velocity of ball 1 after the collision is $-180°$.