A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

ques

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    For an elastic collision of a mass \[m_{1}\] moving with a velocity \[v\] with a mass \[m_{2}\] moving with speed \[v_{2}\] after collision I'm getting the expression for v2 as \[v_{2}=\frac{2m_{1}v}{m_{1}+m_{2}}=\frac{2v}{(1+\frac{m_{2}}{m_{1}})}\] Is this correct?

  2. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I do not completely understand your question, which seems to be missing some information. We need initial velocities of both objects to find the final velocities. Denoting initial velocities u1, u2, and final velocities v1 and v2, then we get from conservation of momentum: \(m_1 u_1+m_2 u_2 = m_1 v_1+m_2 v_2\) ........(1) Equation (1) applies whether collision is elastic or not. Also, sign is important, so velocities are positive in a given direction, assuming the collision is rectilinear. then, for an elastic collision, there is no loss of energy, so equating kinetic energies and multiplying the equation by a factor of two (to get rid of the 1/2 on both sides): \(m_1 u_1^2+m_2 u_2^2 = m_1 v_1^2+m_2 v_2^2\) ........(2) These two equations will enable you to solve for v1 and v2, given u1 and u2, or vice versa.

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

    Sorry I meant that \[m_{2}\] is at rest and the velocity \[v_{2}\] is it's velocity after collision \[m_{1}v^2=m_{1}v_{1}^2+m_{2}v_{2}^2\] eliminating v1 using equation of kinetic energy im getting \[v_{2}=\frac{2m_{1}v}{m_{1}+m_{2}}=\frac{2v}{(1+\frac{m_{2}}{m_{1}})}\]

  4. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Yes, your expression is correct. From momentum: \(v1=-(m2~v2-m1~v)/m1\) substitute into the energy equation, \(m1~v^2=m1~(-(m2~v2-m1~v)/m1)^2+m2~v2^2\) and solving for v2 gives: \(\Large v_2=\frac{2~m_1~v}{m_1+m_2}\) (sorry, was too lazy to put subscripts in the intermediate steps!)

  5. 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.