Calculating relative velocity -- Jumping between two moving trains

Click For Summary
Jumping between two moving trains is theoretically possible due to the conservation of velocity; when a person jumps, they maintain the same forward velocity as the train. However, air resistance poses a significant challenge, potentially hindering the jump and making it unsafe. The typical gap between trains in the U.S. is about 4.5 feet, which could allow for a jump but raises safety concerns. Additionally, obstacles like signal posts must be considered when attempting such a maneuver. Overall, while the physics may support the jump, practical execution is fraught with risks.
verkamp
Messages
1
Reaction score
0
Homework Statement
two trains running parallel, same velocity. what happens if I jump from one train to another?
Relevant Equations
Vba = Vb - Va
is it possible?

I've given it quite some thought, and my conclusion was that it would be possible, cuz right after I jump, I'll still have the same velocity in the same direction as the running train?
 
Physics news on Phys.org
verkamp said:
Homework Statement: two trains running parallel, same velocity. what happens if I jump from one train to another?
Relevant Equations: Vba = Vb - Va

is it possible?

I've given it quite some thought, and my conclusion was that it would be possible, cuz right after I jump, I'll still have the same velocity in the same direction as the running train?
You will have the same velocity but there will be a wind blowing you back at the same velocity. You cannot ignore air resistance in this case, so don't try it.
 
kuruman said:
You will have the same velocity but there will be a wind blowing you back at the same velocity. You cannot ignore air resistance in this case, so don't try it.
Track separation (center to center) in the U.S. is generally 14 feet minimum (source Michigan DOT department of rail). Passenger rail cars are a bit under 10 feet side to side (source various). So that is about 4.5 feet of train to train gap measured from toe of jumping foot to heel of landing foot. Doable, but not supremely safe.

Watch out for signal posts.
 
  • Like
Likes nasu and PeroK
Looks like this experiment has been carried out, at least once...
 
Last edited:
  • Haha
Likes jbriggs444 and kuruman
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 13 ·
Replies
13
Views
1K
Replies
24
Views
2K
  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
14
Views
452
  • · Replies 16 ·
Replies
16
Views
1K
  • · Replies 21 ·
Replies
21
Views
976
  • · Replies 16 ·
Replies
16
Views
1K