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Collision problem

  1. Oct 7, 2016 #1
    1. The problem statement, all variables and given/known data

    Two titanium balls approaches each other with same speed and then collides elastically. After the colision, one of the ball which mass is 300 g becomes motionless.
    Determine the mass of another ball

    2. Relevant equations
    Conservation of momentum and energy

    3. The attempt at a solution

    (I use (+) sign for the direction of 0.3 kg ball movement)

    p = p'
    0.3 v - m v = 0 + m v'

    E=E'
    (1/2)(0.3)v^2 + (1/2) m v^2 = (1/2) m (v')^2
    0.3 v^2 + 1/2(m)(v^2) = (1/2) m (v')^2

    There are three unknowns : m,v, and v' yet I can only find two equations.
    Please help
     
  2. jcsd
  3. Oct 7, 2016 #2

    Simon Bridge

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    You don't need to find all three values though ... you can eliminate v' and v at the same time by putting each expression in terms of v/v' (or set v'=1).
     
  4. Oct 7, 2016 #3
    0.3 (v/v') - m (v/v') = m

    0.3 (v/v')^2 + 1/2(m)(v/v')^2 = (1/2) m

    I get (v/v') = 1/3 and m=0.075 kg

    Is it correct? The solution manual says it is 100 gram, but I doubt it though since it's not a book, just a paper
     
  5. Oct 7, 2016 #4
    Please help
     
  6. Oct 7, 2016 #5

    Ray Vickson

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    The answer 100 gram is correct.

    Analyze the problem in the CM (center-of-momentum) frame, in which the initial total momentum = 0. So, if the particles have masses m1 and m2 and move with respective velocities v and -v in the lab frame, the velocities of particles 1 and 2 in the CM frame are v-u and -v-u, where u = velocity of CM frame as measured in the lab frame.

    Now, the nice thing about the CM frame is that for a perfectly elastic 2-particle collision the initial and final speeds of each particle (separately) remain unchanged; that is, each particle's kinetic energy remains unchanged by the collision. That means that the CM-frame velocity of particle after the collision is -(v-u) = u-v. Now transform back to the lab frame.
     
  7. Oct 7, 2016 #6
    Why should I use CM frame? Could you point out my mistakes on my approach?

    Alright, so
    v1' = -(v-u) = u-v
    v2' = -(-v-u) = v+u

    In the lab frame,
    v1' = u-v
    0 = u-v
    u = v

    v2' = v+u = 2v

    p' = 0 + 2mv

    p = p'
    0.3v - mv = 2mv
    m = 0.1 kg = 100 gram

    But, Why should I use CM frame? Could you point out my mistakes in my approach using just lab frame? When to use CM frame, when not to?
     
  8. Oct 7, 2016 #7

    SammyS

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    You ask for help.

    You are given help.

    You don't like the help you're given, even though it gives you the correct answer.

    Added in Edit:
    Well, OK ...
    How did you get (v/v') = 1/3 ?

    I get something else.
     
    Last edited: Oct 7, 2016
  9. Oct 7, 2016 #8

    ehild

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    You miss a "1/2" in the first term.
     
  10. Oct 7, 2016 #9

    Simon Bridge

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    ... I am in a different time zone to you sorry.
    ... you initially wrote:
    You probably just made an algebraic or arithmetic error - go through and check each step.
    [edit: confirmed above]

    To answer your last question: you do not have to use CM frame to do this problem.
    However, this frame is one where the maths is usually easier so you are less likely to make mistakes.

    Note: One of the advantages to providing assistance free of charge is that you get to give someone the assistance they need as opposed to the assistance they ask for. People will not always like the answers they get. Someone who could be relaxing in front of the TV with a hot SO and a shot of Irish, took the trouble to reply a certain way because they thought that person, and anyone else who googles to the thread in some future years from now, would need to hear it.
     
    Last edited: Oct 7, 2016
  11. Oct 8, 2016 #10
    Sorry, I was just curious about CM frame method since I hadn't learned it yet. But, thanks to Ray Vickson for making me know there's such method.

    Thanks for pointing out my mistake!

    Thank you so much! I've just learned CM frame because of the reply, and it really helps me in solving momentum problem.
     
  12. Oct 8, 2016 #11

    SammyS

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    Oh my !

    I thought that was merely a typo in @terryds's post, not in his working of the problem.
     
  13. Oct 3, 2017 #12
    Very simple using transfer of momentum for an elastic collision:

    Δp = 2μΔv where μ is the reduced mass (m1xm2/[m1+m2]) and Δv the relative velocity.

    Then 300v - 2 x 300m/(300+m) x 2v = 0 whence 1 - 4m/(300+m) = 0.
     
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