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Impulse exerted on objects

  1. Jan 11, 2009 #1
    1. The problem statement, all variables and given/known data
    Two small steel balls A and B have mass 0.6 kg and 0.2 kg respectively. They are moving
    towards each other in opposite directions on a smooth horizontal table when they collide
    directly. Immediately before the collision, the speed of A is 8 m s–1 and the speed of B is 2 m s–1. Immediately after the collision, the direction of motion of A is unchanged and the speed of B is twice the speed of A. Find
    (a) the speed of A immediately after the collision,
    (b) the magnitude of the impulse exerted on B in the collision.

    2. Relevant equations
    total momentum before collision = total momentum after collision
    p=mv
    Impulse = Ft = m(v-u)

    3. The attempt at a solution
    I've got the correct answer for (a), 4.4 m s-1 so I'll skip that bit. It's (b) that I don't get. It asks for the impulse exerted on B so that should be the impulse exerted by A

    Ft = 0.6(4.4 - 8) = 0.6(-3.6) = -2.16 N s

    since they ask for magnitude only

    [tex]/therefore[/tex] Impulse = 2.16 N s

    But the answer given calculates the impulse exerted by B and still gets the same answer

    Impulse = 0.2(2 + 8.8) = 2.16 N s

    Why?

    EDIT: Sorry don't know how to get the therefore symbol in LaTeX...my first time using it actually :D
     
    Last edited: Jan 11, 2009
  2. jcsd
  3. Jan 11, 2009 #2

    alphysicist

    User Avatar
    Homework Helper

    Hi Joe Jacobs,

    This is the impulse exerted on A. The impulse on an object is equal to its change in momentum. Notice that this impulse is negative, and it is the ball B that puts a negative force on ball A. (Ball A felt a negative force, because it had a positive velocity and then slowed down.)

    If you are asking why the magnitudes are the same, think about what Newton's third law says about this situation.
     
  4. Jan 11, 2009 #3
    That was the answer I was looking for. Thx
     
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