Conservation of Momentum and Kinetic Energy in 2D Collisions

[Foghorn]

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 traveling 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

P_x = (m₁ * v_f1 * cosX) + (m₂ * v_f2 * cosY)

P_y = (m₁ * v_f1 * sinX) + (m₂ * v_f2 * sinY)

KE before collision = KE after collision

KE_f = (.5)(m₁)_vf1² + (.5)(m₂)_vf2²

The Attempt at a Solution

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

P_x = 2 kg * 2.5 m/s = 5.0 kg*m/s
P_y = 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 = v_f1² + v_f2²

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?

 

[kuruman]

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?

 

[Foghorn]

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*v_1x² + v_1y²) + (.5*m₂*v_2x² + v_2y²)

Taking your advice, I basically got:

v_1y = v_2y

v_1x = \frac{5}{m} - v_2x

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