Question

A neutral π-meson is a particle that can be created by accelerator beams. If one such particle lives 1.40×10^(-16) s as measured in the laboratory, and 0.840×10^(-16) s when at rest relative to an observer, what is its velocity relative to the laboratory?

a) 0.40c
b) 0.60c
c) 0.80c
d) 0.90c

Answer 1

To find the velocity of the neutral π-meson relative to the laboratory, we can use the concept of time dilation. By using the given time intervals, we can calculate the Lorentz factor and then use it to find the velocity relative to the laboratory.

Step-by-step explanation:

To find the velocity of the neutral π-meson relative to the laboratory, we can use the concept of time dilation. Time dilation occurs when the observed time interval is longer for an object moving relative to an observer than when the object is at rest relative to the observer.

We can use the time dilation equation Δt = γΔt', where Δt is the observed time interval in the laboratory, Δt' is the time interval at rest, and γ is the Lorentz factor. By rearranging the equation, we can solve for γ: γ = Δt/Δt'.

Plugging in the given values, γ = 1.40 × 10-16 s / 0.840 × 10-16 s = 1.667.

Now, we can use the formula v = c(1 - 1/γ2)0.5 to find the velocity relative to the laboratory, where c is the speed of light. Plugging in the value of γ, we can calculate the velocity: v = c(1 - 1/1.6672)0.5 ≈ 0.40c.

Therefore, the velocity of the neutral π-meson relative to the laboratory is approximately 0.40c.