Suppose a newly found elementary particle has an average lifetime, T=30 ns, when at rest. In order for this new elementary particle to travel 15m in one lifetime, how fast must it be moving relative to the velocity of light? In other words, find v/c.
To solve this problem, we can use the concept of time dilation, specifically the equation for time dilation given by:
T_0 = T / sqrt(1 - (v/c)^2),
where T_0 is the proper time (lifetime at rest), T is the observed lifetime, v is the velocity of the particle relative to an observer, and c is the speed of light.
We are given T = 30 ns, and we need to find v/c such that the particle can travel a distance of 15m in one lifetime.
Let's first convert 30 ns to seconds by dividing it by 10^9:
T = 30 ns = 30 × 10^(-9) s.
Now, we can rearrange the time dilation equation to solve for v/c:
(v/c)^2 = 1 - (T_0 / T)^2.
Substituting the given values, we have:
(v/c)^2 = 1 - (15 / (30 × 10^(-9)))^2.
Simplifying, we get:
(v/c)^2 = 1 - (500 × 10^(-9))^2.
(v/c)^2 = 1 - 250000 × 10^(-18).
(v/c)^2 = 1 - 2.5 × 10^(-2).
(v/c)^2 = 0.975.
To find v/c, we can take the square root of both sides:
v/c = sqrt(0.975).
v/c ≈ 0.987.
Therefore, the particle must be moving at approximately 0.987 times the speed of light relative to an observer in order to travel a distance of 15m in one lifetime.