Equations 11.31 assumed that ball 2 was at rest prior to the collision. Suppose, however, you need to analyze the perfectly elastic collision that is just about to take place in FIGURE 11.21. What are the direction and speed of each ball after the collision? You could solve the simultaneous momentum and energy equations, but the mathematics becomes quite messy when both balls have an initial velocity. Fortunately, there’s an easier way.

FIGURE 11.21 A perfectly elastic collision in which both balls have an initial velocity.

You already know the answer—Equations 11.31 —when ball 2 is initially at rest. And in Chapter 4 you learned the Galilean transformation of velocity. This transformation relates an object’s velocity as measured in one reference frame to its velocity in a different reference frame that moves with respect to the first. The Galilean transformation provides an elegant and straightforward way to analyze the collision of Figure 11.21.

FIGURE 11.22a shows the situation, just before the collision, in the lab frame L. Ball 1 has initial velocity $(v_{\text{i}x}){}_{1\text{L}}=2.0\text{m/s}$. Recall from Chapter 4 that the subscript notation means “velocity of ball 1 relative to the lab frame L.” Because ball 2 is moving to the left, it has $(v_{\text{i}x}){}_{2\text{L}}=-3.0\text{m/s}$. We would like to observe the collision from a reference frame in which ball 2 is at rest. That will be true if we choose a moving reference frame M that travels alongside ball 2 with the same velocity: ${\left(v_x\right)}_{\text{ML}}=-3.0\text{m/s}$.

FIGURE 11.22 The collision seen in two reference frames: the lab frame L and a moving frame M in which ball 2 is initially at rest.

Full Alternative Text

We first need to transform the balls’ velocities from the lab frame to the moving reference frame. From Chapter 4, theThe Galilean transformation of velocity for an object O is

That is, O’s velocity in reference frame M is its velocity in reference frame L plus the velocity of frame L relative to frame M. Because reference frame M is moving to the left relative to L with ${\left(v_x\right)}_{\text{ML}}=-3.0\text{m}\text{/}\text{s}$, reference frame L is moving to the right relative to M with ${\left(v_x\right)}_{\text{LM}}=+3.0\text{m/s}$. Applying the transformation to the two initial velocities gives

$(v_{\text{i}x}){}_{2\text{M}}=0\text{m}\text{/}\text{s}$, as expected, because we chose a moving reference frame in which ball 2 would be at rest.

FIGURE 11.22b now shows a situation—with ball 2 initially at rest—in which we can use Equations 11.31 to find the post-collision velocities in frame M:

Reference frame M hasn’t changed—it’s still moving to the left in the lab frame at $3.0\text{m}\text{/}\text{s}$—but the collision has changed both balls’ velocities in frame M.

To finish, we need to transform the post-collision velocities in frame M back to the lab frame L. We can do so with another application of the Galilean transformation:

FIGURE 11.23 shows the outcome of the collision in the lab frame. It’s not hard to confirm that these final velocities do, indeed, conserve both momentum and energy.

FIGURE 11.23 The post-collision velocities in the lab frame.