Length Contraction

Question

A rod of proper length l₀ is at rest in a frame of reference S travelling at a speed v relative to the reference frame S' of a brick wall. The rod collides end-on with the wall, which brings that end of the rod instantly to rest relative to the wall. A shock wave travels down the length of the rod at a speed v_s > v as measured in S. Behind the shock wave, the crumpled material of the rod is left at rest relative to the wall. Show that in S, the fully crumpled length of the rod will be l₀(1-v/v_s). Derive the length of the crumpled rod in S', the rest frame of the wall and hence show that the rod must end up being compressed to a proper length less than or equal to l₀sqrt[(c-v)/(c+v)]

Answer 1

Lena,

 

Here is the answer to the first part.

 

 

The shock wave travels along the length of the rod, l₀, with speed, v_s.  The time of propagation for the shock wave along the entire length of the rod is t_s = l₀/v_s.

 

In this time, the trailing end of the rod has traveled vt_s = v(l₀/v_s).

 

Therefore, the difference in length is Δl = l₀ - v(l₀/v_s) = l₀(1 - v/v_s).