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2D Conservation of Momentum

  • Thread starter Foghorn
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  • #1
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Homework Statement



Object A has a mass of 2.0 kg and an initial velocity of 2.5 m/s. It strikes Object B, which is at rest and has a mass of 2.0 kg as well.

After the collision, the objects travel in different directions, with Object B travelling at an angle of 44 degrees from its original position.

What is the velocity of Object B after the collision and what is the displacement angle of Object A from the point of collision?

Homework Equations



Momentum before collision = momentum after collision

Px = (m1 * vf1 * cosX) + (m2 * vf2 * cosY)

Py = (m1 * vf1 * sinX) + (m2 * vf2 * sinY)

KE before collision = KE after collision

KEf = (.5)(m1)vf12 + (.5)(m2)vf22

The Attempt at a Solution



I first calculated the components of momentum and kinetic energy before impact.

Px = 2 kg * 2.5 m/s = 5.0 kg*m/s
Py = 0 kg*m/s
KE = .5 * 2 kg * 2.5 m/s = 6.25 J

Then, I setup equations relating the objects post-impact to the momentum and energy they should have.

5 = 2*vf1*cosX + 2*vf2*cos44
0 = 2*vf1*sinX + 2*vf2*sin44

6.25 = vf12 + vf22

I've tried using substitution to solve for one of the variables, but each time I end up getting arcsines within cosines equaling sines. And I really don't know how to solve from there.

Am I at least on the right track? Should I solve for the angle first? Does it matter?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
kuruman
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This is good so far. Now suppose you took the first two (momentum conservation) equations and you solved for the components of the momentum of mass 1 in terms of the other quantities. In other words, put everything that has subscript 1 on one side and everything that has subscript 2 or is constant on the other side. What do you get?
 
  • #3
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Awesome. Thanks!

I super simplified the problem following your post:

M = m1v1 + m2v2

Mx = m1v1x + m2v2x = 5
My = m1v1y - m2v2y = 0

KE = .5m1v12 + .5m1v22

KE = (.5*m1*v1x2 + v1y2) + (.5*m2*v2x2 + v2y2)

Taking your advice, I basically got:

v1y = v2y

v1x = [tex]\frac{5}{m}[/tex] - v2x

Plugged it all in and got my answers! Thanks again.
 

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