A community for students.
Here's the question you clicked on:
 0 viewing
chris215
 one year ago
Ball A moving along a straight line at 2 m/s collides elastically with ball B of same mass moving at 3 m/s. What is the velocity of the balls A and B after collision?
chris215
 one year ago
Ball A moving along a straight line at 2 m/s collides elastically with ball B of same mass moving at 3 m/s. What is the velocity of the balls A and B after collision?

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0mass= force times acceleration

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you need conservation of momentum and conservation of kinetic energy equations. Let \(v_{A1}\) be initial velocity for ball A, \(v_{A2}\) be final velocity for ball 2, \(v_{B1}\) be initial velocity for ball B, and \(v_{B2}\) be final velocity for Ball B Momentum Conservation (P = mv, momentum = mass * velocity) \[P_1=P_2\] \[mv_{A1}+mv_{B1}=mv_{A2}+mv_{B2}\] All the masses are the same so they cancel \[v_{A1}+v_{B1}=v_{A2}+v_{B2}\] Plug in numbers for \(v_{A1}\) and \(v_{B1}\) \[23=v_{A2}+v_{B2}\] \[1=v_{A2}+v_{B2}\] KE = ½mv² Do more or less the same thing to create a conservation of kinetic energy equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how about this way since it's multiple choice dw:1435874747318:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops dw:1435875089386:dw

chris215
 one year ago
Best ResponseYou've already chosen the best response.0wait but none of the answers represent that
Ask your own question
Sign UpFind more explanations on OpenStudy
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
 Engagement 19 Mad Hatter
 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.