37-5

In this problem we can to compare measurements made in the stationary lab frame with those 'made' by a moving particle. In the diagram below (fig.5), $x$ is the distance traveled by the particle in the lab frame before it disintegrates, and $t$ is the corresponding time (measured in the lab frame). Then $t'$ is the time interval until the disintegration as measured by the particle.

Figure 5: Space-time diagram for 37-5.

We can see that the problem is very similar to the last one. In fact, all the algebra is going to be the same and we will just quote the result that

$$ct'=\frac{ct}{\gamma}.$$

Now, we don't know $t$ this time, but we do know $x$ and $\beta$, the other quantities measured in the lab frame. We have

$$ct = \frac{x}{\beta}=\frac{1.05 {\;\rm mm}}{.992}=1.0585\;{\rm mm}.$$

We can also compute $\gamma$:

$$\gamma=\left(1-.992^2\right)^{-1/2}=7.9216.$$

Then we have

$$ct'=\frac{ct}{\gamma}=\frac{1.0585\;{\rm mm}}{7.9216}=.13362\;{\rm mm},$$

or $t'=4.45\times 10^{-13}$ s.