1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    2.5*50+2.5*38=25*10+2.5*v
    v_x=78 m/s

    For y
    2.5*50+2.5*2.3=2.8*9.4+2.5*v
    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)
    (7730J+2410.96J)-(4930J+3131.61J)
    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

    gneill

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Finding velocity and work of a collision
Loading...