What is the final velocity of the five-car train after a series of collisions?

  • Thread starter Thread starter grewas8
  • Start date Start date
  • Tags Tags
    Momentum Train
Click For Summary
The discussion revolves around calculating the final velocity of a five-car train after a series of collisions, using the principle of conservation of momentum. Initially, three train cars are moving at 3.97 m/s, while a fourth car approaches at 4.59 m/s, and a fifth car is at rest. Participants emphasize that the total momentum before the collisions must equal the total momentum after, despite the change in mass. It is clarified that while the masses of the cars can be considered equal, the final velocity will differ due to the increase in mass after the collisions. The importance of correctly applying momentum conservation principles is highlighted for solving the problem.
grewas8
Messages
16
Reaction score
0

Homework Statement


Three identical train cars, coupled together, are rolling east at 3.97 m/s. A fourth car traveling east at 4.59 m/s catches up with the three and couples to make a four-car train. A moment later, the train cars hit a fifth car that was at rest on the tracks, and it couples to make a five-car train. What is the speed of the five-car train?

Homework Equations



p=mv
Momentum is conserved


The Attempt at a Solution



Given- V(1,2,3)= 3.97 m/s
v(initial of 4)= 4.59

Find- v(1,2,3,4,5)
p at the instant 4 hits 1,2,3
p at the instant 5 hits 1,2,3,4

Cart 4- P=mv
P= 4.59m (initially)

Cart 1,2,3
P=mv
P=3m(3.97)=11.91m

When 4 collides with 1,2,3
p(of 4)=p(of 1,2,3)
4.59m=11.91m
Change in momentum=11.91m-4.59m=7.32m

When 1,2,3,4 collides with 5
P(1,2,3,4)=7.32m
P(1,2,3,4,5)=7.32m-mv

Really stuck after this.:confused: ..is anything i did above incorrect?
any help will be appreciated greatly :!)
 
Physics news on Phys.org
grewas8 said:
When 4 collides with 1,2,3
p(of 4)=p(of 1,2,3)
4.59m=11.91m
Change in momentum=11.91m-4.59m=7.32m

Huh? How can 4.59m=11.91m? You are making this much too complicated. What is the total momentum of the five cars before any collisions? How should this relate to the total momentum of the 5 car train after all of the collisions? So write down an expression for the final momentum in terms of the unknown final velocity and solve for it.
 
hi i was trying to do this question and i was wondering since all the masses are the same for the carts do we just ignore the masses?? then wouldn't the final velocity of the 5cars be the same as the 4cars?? can you please explain how one would go about doing this question??
 
No, it won't be the same. Use conservation of momentum. The momentum of the 4 cars is the same as the momentum of the 5 cars, but the velocity isn't because the mass has changed. In the future, try to post a new question instead of adding onto an old thread. You'll probably get a much quicker response.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
Replies
3
Views
7K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 3 ·
Replies
3
Views
11K
Replies
13
Views
2K
  • · Replies 5 ·
Replies
5
Views
9K
  • · Replies 12 ·
Replies
12
Views
2K