Rod colliding with the ground (Special relativity question)

[LCSphysicist]

Homework Statement

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Relevant Equations

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Ok, so basically: There is a frase with a particle moving horizontally with velocity v and a rod parallel to the ground, with rest length L, falling with speed u vertical constant. The rod bounces off on the ground. Describe what happens in the particle frame and find the angle the rod makes with the horizontal when it collides in this frame.

So i tried this:

The event A, the left corner of the rod hits the ground.
The event B, the right corner hits the ground.

In the ground frame, Ta=Tb=0. $\Delta x = L/\gamma_u$

So, LT to the particle frame, we have $$|\Delta t'| = |\gamma_v \beta_v \Delta x|, |\Delta x'| = | \gamma_v \Delta x|$$
That is, the rod don't hit the ground simultaneously, the right corner will hit it with a difference of time $|\Delta t'|$

So, when the left corner touches the ground, the time the right corner is elevated $y' = u \Delta t' = u' \gamma_v \beta_v \Delta x$

Since u is perpendicular to v, u', the vertical velocity of the rod in the particle frame, is $u' = u/\gamma_v$.

$y' = u '\Delta t' = (u/\gamma_v) \gamma_v \beta_v \Delta x = u\beta_v \Delta x$

$tan\theta' = y'/\Delta x' = u\beta_v \Delta x/\Delta x' = u\beta_v / \gamma_v$

Ignore the units, i put c=1.
This is just wrong. Could you help me to solve the question?

 

[anuttarasammyak]

According to IFRs, the rod is horizontal, right up or left up in touching the ground. I do not find anything wrong with it.

 

[TSny]

In the ground frame, Ta=Tb=0. $\Delta x = L/\gamma_u$

In the ground frame, the rod is moving perpendicularly to its length. Should the rod be contracted in the ground frame?

So, LT to the particle frame, we have $$|\Delta t'| = |\gamma_v \beta_v \Delta x|, |\Delta x'| = | \gamma_v \Delta x|$$
That is, the rod don't hit the ground simultaneously, the right corner will hit it with a difference of time $|\Delta t'|$

So, when the left corner touches the ground, the time the right corner is elevated $y' = u \Delta t' = u' \gamma_v \beta_v \Delta x$

Ok. In the middle expression, $u$ should have a prime. But you have it correct in the final expression.

Since u is perpendicular to v, u', the vertical velocity of the rod in the particle frame, is $u' = u/\gamma_v$.

$y' = u '\Delta t' = (u/\gamma_v) \gamma_v \beta_v \Delta x = u\beta_v \Delta x$

Ok

$tan\theta' = y'/\Delta x' = u\beta_v \Delta x/\Delta x' = u\beta_v / \gamma_v$

Here, $\Delta x'$ should be the (simultaneous) horizontal distance between the left and right ends of the rod as seen in the primed frame. This is not the same as the horizontal distance between events A and B as seen in the primed frame. In the primed frame, A and B do not occur simultaneously. So, the rod moves horizontally during the time between the events in the primed frame.

 

[LCSphysicist]

In the ground frame, the rod is moving perpendicularly to its length. Should the rod be contracted in the ground frame?Ok. In the middle expression, $u$ should have a prime. But you have it correct in the final expression.OkHere, $\Delta x'$ should be the (simultaneous) horizontal distance between the left and right ends of the rod as seen in the primed frame. This is not the same as the horizontal distance between events A and B as seen in the primed frame. In the primed frame, A and B do not occur simultaneously. So, the rod moves horizontally during the time between the events in the primed frame.

Thank you by the reply. So considering the last paragraph, i think we could say that:
The horizontal distance between the ends of the rod will be equal to $$\Delta S = \Delta X' - v\Delta t' = \gamma_v \Delta X - v \gamma_v \beta_v \Delta X$$

So that $$tan = u\beta_v \Delta X/(\Delta X (\gamma_v - v \gamma_v \beta_v)) = u v / (\gamma_v - v^2 \gamma_v ) = uv \gamma_v$$

So, is that right? I was expecting something involving $gamma_u$ too. At least to me, it is quite amazing it does not appear here.

 

[TSny]

The horizontal distance between the ends of the rod will be equal to $$\Delta S = \Delta X' - v\Delta t' = \gamma_v \Delta X - v \gamma_v \beta_v \Delta X$$

So that $$tan = u\beta_v \Delta X/(\Delta X (\gamma_v - v \gamma_v \beta_v)) = u v / (\gamma_v - v^2 \gamma_v ) = uv \gamma_v$$

That looks right.

It's interesting to consider the shape of the rod in the primed frame at times between the two events B and A for the case where the rod bounces elastically off the ground in the ground frame.