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Homework Help: Finding velocity and work of a collision

  1. Oct 27, 2016 #1
    1. The problem statement, all variables and given/known data
    Two 2.5 kg bodies, A and B, collide. The velocities before the collision are ##\vec{v}_A = \left( 50 \hat{i} + 50 \hat{j} \right)~m/s## and ##\vec{v}_B = \left( 38 \hat{i} + 2.3 \hat{j} \right)~m/s##. After the collision, ##\vec{v}'_A = \left( 10 \hat{i} + 9.4 \hat{j} \right)~m/s##. What are (a) the x-component and (b) the y-component of the final velocity of B? (c) What is the change in the total kinetic energy (including sign)?

    2. Relevant equations

    3. The attempt at a solution
    So i decoupled the x and y components and found the components of the velocity of B seperately
    For x
    v_x=78 m/s

    For y
    v_y=42.9 m/s

    I then stayed with the decoupled approach to find the work done in each direction
    for x:
    initial KE= 1.25*50^2+1.25*38^2=4930J
    final KE= 1.25*10^2+1.25*78^2=7730J

    for y:
    initial KE= 1.25*50^2+1.25*2.3^2= 3131.61J
    final KE= 1.25*9.4^2+1.25*42.9^2=2410.96J

    Subtracted the initial KE from final KE to find work (ignore the unnecessary parentheses as they give me a sense of organization)
    2080.35 J
    This cannot be right can it? How can the system gain KE?
    Last edited by a moderator: Oct 27, 2016
  2. jcsd
  3. Oct 27, 2016 #2
    It seems like the velocities haven't shown up. Here is a direct screenshot of the problem:
    Phy 1.PNG Phy 1.PNG
  4. Oct 27, 2016 #3
    This is a strange problem. How can you have object A moving in the +x direction at 50 m/s, and object B moving in the +x direction at 38 m/s, then, after the collision, object A is moving in the +x direction at only 10 m/s - unless energy was imparted to object A by some other means than the collision?

    One other possibility is that sometimes I just get confused when I look at some of these problems.
  5. Oct 27, 2016 #4
    The only way I can see that happening is if B has a larger mass than A, but it is stated that their mass are equal. Anyway I used the solutions and they were correct. Unless I am missing something, this is a very peculiar problem indeed.
  6. Oct 27, 2016 #5
    Even if B was infinite mass, the velocity of A could not change that much. If B was an infinite mass, and A and B had a perfectly elastic collision, A would bounce off of B with, at most, equal and opposite x component of the velocity. So since the difference is initially +12 m/s, the final difference would be -12 m/s - or an absolute velocity of (38-12) = 26 m/s.

    Maybe B had some type of explosive reactive armor. That must be it. :)
  7. Oct 27, 2016 #6


    User Avatar

    Staff: Mentor

    More prosaically, perhaps the problem's values were auto-generated by software (so everyone gets a unique question) and the author didn't put in sufficient reality checking.
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