A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 4 years ago

A 1,600 kg train car rolling freely on level track at 16 m/s bumps into a 1.0 × 103 kg train car moving at 10.0 m/s in the same direction, and the two latch onto each other and continue together. What is their final speed? 16 m/s 6.0 m/s 10 m/s 14 m/s

  • This Question is Closed
  1. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Again, use Conversation of Momentum. The momentum of the two trains before bumping into each other is the same as it is afterwards. Can you write down the momentum before they bump?

  2. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    whats the formula again?

  3. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is the ans 16m/s

  4. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thnx....jamesj left me hangin

  5. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    jamesj wants you to try to understand it yourself

  6. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i know but I asked for the formula n he left

  7. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    If i knew the formula Id do it on my own...he was helping me...n I appreciate it....jus need more help still

  8. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    total momentum before collision=total momentum after collision

  9. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thnx...again

  10. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Correction=the ans is 14m/s not 16 Didn't notice the power '1x103' write it like 1x10^3

  11. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yea...but I didnt wanna say you were wrong...still helped though :)

  12. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @physicsme, don't give the answer. Give the formula. Momentum = (mass) x (velocity)

  13. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thats all i really wated was the formula...

  14. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wanted*

  15. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay, actually I was practicing different questions so just wanted to solve it and check the answer.

  16. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Hence, for example the momentum of the first train before the collision is p1 = m1.v1 = (1600 kg)(16 m/s) = 25,600 kg.m/s The momentum of the second train is p2 = m2.v2 = (103 kg)(10 m/s) = 1,030 kg.m/s [btw, is that mass of the train car right? Looks wrong. If so, fix this calculation] Now after the collision, they have the same velocity v' and their combined mass is m = m1 + m2 Now set the sum of the momentum before = momentum afterwards

  17. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.