1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: 2-d momentum question

  1. Nov 10, 2007 #1
    1. The problem statement, all variables and given/known data

    A steel ball of mass 0.50 kg, moving with a velocity of 2.0 m/s [E], strikes a second ball of mass 0.30 kg, initially at rest.The collision is a glancing one, causing the moving ball to have a velocity of 1.5 m/s [E 30 N] after the collision. Determine the velocity of the second ball after the collision.

    2. Relevant equations

    Momentum = final momentum
    m1v1 +m2v2 = m1v1 +m2v2
    (c)^2 = (a)^2 + (b)^2 - 2(a)(b)cos C

    3. The attempt at a solution

    m1 = 0.50 kg
    v1 = 2.0 m/s[E]
    m2 = 0.30 kg
    v2 = 0 m/s
    v1 (final) = 1.5 m/s [E 30 N]
    v2 (final) = ?

    Ptotal = Ptotal (final)
    m1v1 + m2v2 = m1v1 + m2v2
    (0.50)(2.0[E]) = (0.50)(1.5[E 30 N]) + (0.30) (v2(final))
    1.0 [E] = 0.75[E 30 N] + 0.30(v2)

    Diagram i drew


    (c)^2 = (a)^2 + (b)^2 - 2(a)(b)cos C
    ((0.30)(V2))^2 = 0.75^2 + 1^2 -2(0.75)(1) cos 30
    0.09V2^2 = 2.5625 - 0.866025403
    V2 = square root(7.738606633)
    i used sign law to find out the equation
    V2(final) = 2.78[E 10.4 S]

    After doing all this work i went to the back of the book and the answer was 1.7 m/s[E 47 S] i hate when this happens:(
    if somebody can help me out i will really appreciate it

    thanks alot
  2. jcsd
  3. Nov 10, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    Separate it into N-S (y) and E-W (x) components. The net momentum in the y direction = 0, since initially there is no momentum in the y or transverse direction.

    Also consider conservation of energy.
  4. Nov 12, 2007 #3
    can somebody please show me how extacly i can solve this problem using components, i dont know where to start:frown:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook