Inelastic Collision and average force between 2 trains

AI Thread Summary
The discussion revolves around solving an inelastic collision problem involving two trains. Train A, weighing 15,000 kg and moving at 1.5 m/s, collides with Train B, weighing 12,000 kg and moving at 0.75 m/s in the opposite direction. The correct speed of both trains after coupling is calculated to be 0.5 m/s to the right. For the average force during the collision, initial calculations were incorrect, but after clarifying the approach, the average force was determined to be 18,750 N to the left. The conversation highlights the importance of consistency in applying momentum principles to solve collision problems accurately.
Sam Fred
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I solved the question , but i need to check it with you guys ... I believe there is something wrong with part b .

Homework Statement



The 15000kg train A is running at 1.5 m/s on the horizontal tack (to the right) when it encounters a 12000kg train B running at 0.75m/s toward it (to the left) . If the trains meet and
couple together, determine (a) the speed of both cars just after the couple, and (b) the
average force between them if the coupling takes place in 0.8sec.

Homework Equations


m1v1 + m2v2 = (m1+m2)v3
momentum 1 + F(avg)t = momentum 2

The Attempt at a Solution


a-
v3 = (m1v1 + m2v2)/m1+m2
v3 = 0.5 m/s to the right

b-
momentum 1 + F (avg) t = momentum 2
(m1+m2)v3 + F(avg) 0.8 = 0
F(avg) = -16875 N
 
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Sam Fred said:

The Attempt at a Solution


a-
v3 = (m1v1 + m2v2)/m1+m2
v3 = 0.5 m/s to the right
Good.

b-
momentum 1 + F (avg) t = momentum 2
(m1+m2)v3 + F(avg) 0.8 = 0
F(avg) = -16875 N
To find the average force, just look at one of the cars. It doesn't matter which one, since the force on either will be the same magnitude.
 
Ok...
then :
-m2v2 + F(avg) t = (m1+m2) v3
F(avg) = 28125 N
 
Sam Fred said:
Ok...
then :
-m2v2 + F(avg) t = (m1+m2) v3
F(avg) = 28125 N
Still not right. Stick to a single mass, all the way through. How does the velocity, and thus the momentum, of mass m2 change?
 
Aha ... I guess I was confused ...
for train a which is going to the right
m1v1 - F(avg from b) t = m2v3
F(avg from b) = (m1v1 - m2v3 ) / t
F(avg from b) = 18750 N to the left
 
Sam Fred said:
Aha ... I guess I was confused ...
for train a which is going to the right
m1v1 - F(avg from b) t = m2v3
F(avg from b) = (m1v1 - m2v3 ) / t
I believe you have a typo. The only mass that should appear is m1.

F(avg from b) = 18750 N to the left
Good!
 
Oh yes ... it was a typo
Thanks very much for helping me through the problem
 
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