# 2D Conservation of Momentum

## 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?

## The Attempt at a Solution

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kuruman
Homework Helper
Gold Member
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?

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)

v1x = $$\frac{5}{m}$$ - v2x