Question

Billiard balls A and B, of equal mass, move at right angles and meet at the origin of an xy coordinate system as shown in Fig. 7–36. Initially ball A is moving along the y axis at \({\bf{ + 2}}{\bf{.0}}\;{\bf{m/s}}\), and ball B is moving to the right along the x axis with speed \({\bf{ + 3}}{\bf{.7}}\;{\bf{m/s}}\).After the collision (assumed elastic), ball B is moving along the positive y axis (Fig. 7–36) with velocity What is the final direction of ball A, and what are the speeds of the two balls?

FIGURE 7-36

Problem 46. (Ball A after the Collison is not shown)

Short answer

The final direction of ball A is along the positive x-axis.

The final speed of ball A is \(3.7\;{\rm{m/s}}\)and the final speed of ball B is \(2.0\;{\rm{m/s}}\).

Step-by-step solution

Given data

The velocity of mass interchanges for an elastic collision between two similar masses.

The initial speed of ball A is \({v_{\rm{A}}} = 2.0\;{\rm{m/s}}\) and the initial speed of ball B is \({v_{\rm{B}}} = 3.7\;{\rm{m/s}}\).

Let m be the mass of the balls.

After collision, the speed of ball A is \({v'_{\rm{A}}}\) and the speed of ball B is \({v'_{\rm{B}}}\).

Also, let \({v'_{{\rm{Ax}}}}\) be the component of the speed of ball A along the x-axis and \({v'_{{\rm{Ay}}}}\) be the component of the speed of ball A along the y-axis.

Use of momentum conservation

For the speed of ball A after the collision, you can write:

\({\left( {{{v'}_A}} \right)^2} = {\left( {{{v'}_{{\rm{Ax}}}}} \right)^2} + {\left( {{{v'}_{{\rm{Ay}}}}} \right)^2}\) ... (i)

Using the momentum conservation along the x-axis, you get:

\(\begin{array}{c}m{v_{\rm{B}}} = m{{v'}_{{\rm{Ax}}}}\\{v_{\rm{B}}} = {{v'}_{{\rm{Ax}}}}\end{array}\) … (ii)

Using the momentum conservation along the y-axis, you get:

\(\begin{array}{c}m{v_{\rm{A}}} = m{{v'}_{{\rm{Ay}}}} + m{{v'}_{\rm{B}}}\\{v_{\rm{A}}} = {{v'}_{{\rm{Ay}}}} + {{v'}_{\rm{B}}}\end{array}\) … (iii)

Use kinetic energy conservation

Also, for the kinetic energy conserved in elastic collision, you get:

\(\begin{array}{l}\frac{1}{2}mv_{\rm{A}}^2 + \frac{1}{2}mv_{\rm{B}}^2 = \frac{1}{2}m{\left( {{{v'}_{\rm{A}}}} \right)^2} + \frac{1}{2}m{\left( {{{v'}_{\rm{B}}}} \right)^2}\\v_{\rm{A}}^2 + v_{\rm{B}}^2 = {\left( {{{v'}_{\rm{A}}}} \right)^2} + {\left( {{{v'}_{\rm{B}}}} \right)^2}\end{array}\)

Now, after further calculation of the above equation using the values from the previous equations, you get:

\(\begin{array}{c}{\left( {{{v'}_{{\rm{Ay}}}} + {{v'}_{\rm{B}}}} \right)^2} + {\left( {{{v'}_{{\rm{Ax}}}}} \right)^2} = {\left( {{{v'}_{\rm{A}}}} \right)^2} + {\left( {{{v'}_{\rm{B}}}} \right)^2}\\{\left( {{{v'}_{{\rm{Ax}}}}} \right)^2} + 2{{v'}_{{\rm{Ay}}}}{{v'}_{\rm{B}}} + {\left( {{{v'}_{\rm{B}}}} \right)^2} + {\left( {{{v'}_{{\rm{Ay}}}}} \right)^2} = {\left( {{{v'}_{\rm{A}}}} \right)^2} + {\left( {{{v'}_{\rm{B}}}} \right)^2}\\{\left( {{{v'}_{\rm{A}}}} \right)^2} + {\left( {{{v'}_{\rm{B}}}} \right)^2} + 2{{v'}_{{\rm{Ay}}}}{{v'}_{\rm{B}}} = {\left( {{{v'}_{\rm{A}}}} \right)^2} + {\left( {{{v'}_{\rm{B}}}} \right)^2}\\2{{v'}_{{\rm{Ay}}}}{{v'}_{\rm{B}}} = 0\end{array}\)

As \({v'_{\rm{B}}} \ne 0\), from the above equation, you get:

\({v'_{{\rm{Ay}}}} = 0\) … (iv)

Therefore, from equations (i) and (iv), you get:

\(\begin{array}{c}{\left( {{{v'}_A}} \right)^2} = {\left( {{{v'}_{{\rm{Ax}}}}} \right)^2} + {\left( 0 \right)^2}\\{{v'}_A} = {{v'}_{{\rm{Ax}}}}\end{array}\) … (v)

Hence, the final direction of ball A is along the positive x-axis.

Now, comparing equations (ii) and (v), you get:

\(\begin{array}{c}{{v'}_A} = {v_{\rm{B}}}\\ = 3.7\;{\rm{m/s}}\end{array}\)

Now, from equations (iii) and (iv), you get:

\(\begin{array}{c}{{v'}_{\rm{B}}} = {v_{\rm{A}}}\\ = 2.0\;{\rm{m/s}}\end{array}\)

Hence, the final speed of ball A is \(3.7\;{\rm{m/s}}\), and the final speed of ball B is \(2.0\;{\rm{m/s}}\).