Inelastic Collision and average force between 2 trains

[Sam Fred]

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

 

[Doc Al]

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.

 

[Sam Fred]

Ok...
then :
-m2v2 + F(avg) t = (m1+m2) v3
F(avg) = 28125 N

 

[Doc Al]

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?

 

[Sam Fred]

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

 

[Doc Al]

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!

 

[Sam Fred]

Oh yes ... it was a typo
Thanks very much for helping me through the problem