Billiards!

Find the distance between the ball and the pocket. Apply conservation of momentum conservation, and be careful with the velocities of each ball and their corresponding angles. 

Given \(L_1\) and \(L_2\), we can write the angle \(\theta\) as:

\begin{equation*}
\frac{L_2}{L_1} = \tan \theta,
\end{equation*}

Solving for \(\theta\):

\begin{equation*}
\theta = 34.99^\circ.
\end{equation*}

Conservation of momentum gives \(\vec{P}_{i}=\vec{P}_{f}\), or explicitly:

\begin{equation}
m_w \vec{v}_{iw}+m_r \vec{v}_{ir}= m_w \vec{v}_{fw}+m_r \vec{v}_{fr},
\end{equation}

where \(\vec{v}_{ir} = 0\) and the masses are the same. So, for both \({x}\) and \({y}\) components:

\begin{equation*}
\vec{v}_{iw} =  \vec{v}_{fw}+ \vec{v}_{fr}.
\end{equation*}

Two velocities in the \({y-}\)direction can be written in terms of the given variables and the unknown angle as \(v_{fw_y} = v \cos \theta\) and \(v_{fr_y} = v \cos \alpha\), respectively. Solving for the \({y-}\)component and the unknown angle (let’s call it \(\alpha\) ):

\begin{equation}
\frac{v_{iw_y} – v_{fw} \cos ( \theta)}{v_{fr}} = \cos \alpha.
\end{equation}

which, with numerical values, is:

\begin{equation*}
\alpha = 16.10 ^\circ.
\end{equation*}

For a more detailed explanation of any of these steps, click on “Detailed Solution”.

In order to determine if the red ball goes into the upper right pocket, we need to find the direction it moves after being hit by the white ball. More precisely, the red ball would only go into the pocket in question if it moves in the direction of angle \(\theta\) in figure 1.

 

Figure 1: On the left, we see the moment right before the collision where the red ball is static and the white ball as a vertical velocity. On the right, we see the wanted trajectory for the red ball in terms of the horizontal and vertical distances \(L_2\) and \(L_1\) and the angle \(\theta\) required for the red ball to go into the pocket.

 

Clearly, with \(L_1\) and \(L_2\) we can write the angle \(\theta\) as:

\begin{equation}
\frac{L_2}{L_1} = \tan \theta.
\end{equation}

If we use the numerical values for \(L_1\) and \(L_2\), we get:

\begin{equation}
\frac{(0.7 \, \text{m})}{(1 \, \text{m})} = \tan \theta.
\end{equation}

Hence, the angle is

\begin{equation}
\arctan{ \frac{0.7}{1} } = \theta,
\end{equation}

or

\begin{equation}
\theta = 34.99^\circ.
\end{equation}

So, it all comes down to finding whether the direction of motion for the red ball, after the collision, is given by an angle of \(34.99^\circ\) degrees.

Given that we already know the the initial and final speeds of the white ball, we will use conservation of linear momentum to find the the red ball’s direction of motion.

The linear momentum of a system is conserved if the total external force over the system is zero. Particular to our case, the linear momentum of a system along a certain direction is conserved if the total external force over the system in that direction is zero. It is convenient to treat the two balls as a single system because the total external force over the two balls is zero (we are assuming there is no friction). Notice that if we were to choose just one ball as the system, the linear momentum would not be conserved because the other ball would exert an external force during the collision. But taken together, the forces during the collision between the balls are internal to the system, and there is nothing to worry about (the linear momentum would be conserved).

The conservation of linear momentum states that

\begin{equation}
\label{first}
\vec{P}_{i}=\vec{P}_{f},
\end{equation}

where \( \vec{P}_{i}\) is the initial momentum of the system and \(\vec{P}_{f}\) is the final one. Using the definition of linear momentum, we have

\begin{equation}
m_w \vec{v}_{iw}+m_r \vec{v}_{ir}= m_w \vec{v}_{fw}+m_r \vec{v}_{fr},
\end{equation}

where we have used the subscript ‘w’ for the mass, initial velocity and final velocity of the white ball, and ‘r’ for the mass, initial velocity and final velocity of the red ball.

We can now use the fact that the initial velocity of the red ball is zero, and we can also use the fact that the masses are the same. So, we can cancel them:

\begin{equation}
\vec{v}_{iw}= \vec{v}_{fw}+\vec{v}_{fr}.
\end{equation}

Now, since this is a two-dimensional problem (the balls can move in different directions over the table), it is important to distinguish between the momentum in X and the momentum in Y. In both cases, it is conserved, and so

\begin{equation}
\vec{v}_{iw_x}= \vec{v}_{fw_x}+\vec{v}_{fr_x},
\end{equation}

and

\begin{equation}
\vec{v}_{iw_y}= \vec{v}_{fw_y}+\vec{v}_{fr_y}.
\end{equation}

In order to use these equations, let’s insert a coordinate system with the Y axis pointing in the direction that the white ball is initially moving, as is shown in figure 2.

 

Figure 2: We place the coordinate system on the table such that the Y axis is vertical and the X axis is horizontal. The initial velocity of the white ball is along the Y axis.

 

According to this system, the final velocities of the white ball and the red ball will have an X component and an Y component, as illustrated in figure 3.

 

Figure 3: Final velocities for both balls and their components. In the left, we see the velocities of the white ball, and in the right the velocities of the red ball.

 

Notice that in the coordinate system used, the initial velocity of the white ball is positive in Y, and the Y component of its final velocity is also positive. Furthermore, the red ball will start moving with some positive velocity in Y and some (positive or negative) velocity in X. So, for the linear momentum in Y, we have

\begin{equation}
\label{Billiard_conservacionX}
v_{iw_y} \, \hat{\textbf{i}} =v_{fw_y} \, \hat{\textbf{j}} +v_{fr_y} \, \hat{\textbf{j}}.
\end{equation}

Now, from the diagram above, we see that the Y component of the final velocities is given by

\begin{equation}
\label{Billiard_vf_wx}
v_{fw_y} = v_{fw} \cos (30 ^\circ),
\end{equation}

and

\begin{equation}
\label{Billiard_vf_rx}
v_{fr_y} = v_{fr} \cos \alpha,
\end{equation}

where \(v_{fw}\) is the final speed of the white ball and \(v_{fr}\) the final speed of the red ball. Also, we have called \(\alpha\) the angle that the red ball forms with the Y axis (this is the angle we want to compare with \(\theta =34.99^\circ\)).

 

If we use equations \eqref{Billiard_vf_wx} and \eqref{Billiard_vf_rx} into the equation \eqref{Billiard_conservacionX} and focus only on magnitudes, we get

\begin{equation}
v_{iw_y} = v_{fw} \cos (30 ^\circ) + v_{fr} \cos \alpha .
\end{equation}

 

Let’s then move the \(v_{fw} \cos 30\) term to the left side to get

\begin{equation}
v_{iw_y} – v_{fw} \cos (30 ^\circ) = v_{fr} \cos \alpha.
\end{equation}

If we divide by \(v_{fr}\), we get

\begin{equation}
\frac{v_{iw_y} – v_{fw} \cos (30 ^\circ)}{v_{fr}} = \cos \alpha.
\end{equation}

If we insert the numerical values, we get

\begin{equation}
\frac{(1 \,\text{m/s}) – (0.6 \,\text{m/s}) \cos (30 ^\circ)}{(0.5 \,\text{m/s})} = \cos \alpha.
\end{equation}

That is,

\begin{equation}
\arccos ( 0.96 ) = \alpha,
\end{equation}

to finally get

\begin{equation}
\alpha = 16.10 ^\circ.
\end{equation}

Notice that \(\alpha\) is not the same as \(\theta =34.99^\circ\), and so we can infer that the red ball does not go into the pocket!

 

You need to be registered and logged in to take this quiz. Log in or Register