Math Is Fun Forum

fusilli_jerry89

Member

Registered: 2006-06-23

Posts: 86

Elastic Trains

A man on a small flatcar on a frictionless rail sees 2 trains. A and B approach him from opposite directions, with each train travelling at 60 km/hr. The flatcar, initially at rest, first recoils from a collision with A, then with B, then A again, etc. Assume that the collisions are all perfectly elastic, and that both trains are infinitely more massive than the flatcar.

What is the speed of the flat car after the first collision? the second?

Can some1 explain to me how to do this? bcuz wuldnt an infinitely larger mass equal infinite momentum, therefore wuldnt the flat car go infinite speed after one hit?

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Elastic Trains

That would only be the case if the collision caused the train to stop.

You can't actually work out the speeds analytically, because equating momentum before and after gives you that 60m_1 = v_1m_1 + v_2m_2.

m_1>>>m_2, and so m_2 can be approximated as 0. Then you have 60m_1 = v_1m_1 + 0v_2.
So the train retains its speed of 60m/s, but you then have that 0 x v_2 = 0, and so according to that, v_2 can be whatever it wants.

Resorting to intuition though, the flatcar will assume the train's speed of 60m/s.
Then I'd guess that its speed after the second collision would become 120m/s.

If you really want an analytic answer, a possible method would be to set m_1 as 100 (say) and m_2 as some very small, but non-zero, number. The answer you get out of it will be wrong, but it will exist and be fairly close to the actual answer.

The answer that you truly want is the limit as m_2 --> 0.

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Why did the vector cross the road?
It wanted to be normal.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Elastic Trains

The trick is to consider the collisions from the point of view of the trains. In the first collision, using the reference frame of train A, the flatcar is hurtling towards the train at 60 kph; it makes an elastic collision and rebounds away at the same speed (relative to A) in the opposite direction. (In an elastic collision, kinetic energy is conserved.) So the speed of the flatcar after the first collision is actually 120 kph. For the second collision, train B sees the flatcar approaching at 120 + 60 = 180 kph; after the elastic collision, by the foregooing reasoning, the flatcar's speed is moving away from B at 180 kph from B’s point of view. Thus, after the second collision, the flatcar’s speed is 240 kph. Hence the flatcar's speed is 120n kph after the nth collision, increasing by 120 kph with each collision until it gets smashed between the two trains.

Last edited by JaneFairfax (2007-10-17 22:10:06)