Solve Mass of unknown atom in Collision Problem

[bpw91284]

Homework Statement

An atom (m = 19.0 u) makes a perfectly elastic collision with another atom at rest. After the impact, the atom travels away a
a 52.9° angle from its original direction and the unknown atom travels away at a −50.0° angle. What is the mass (in u) of th
unknown atom?

Homework Equations

Conservation of Momemtum
Law of Sines

The Attempt at a Solution

Atom A is the m_a=19 u...

m_a*v_a,i=m_a*v_a,f + m_b*v_b,f

Draw triangle, use law of sines...

sin(45.8)/(m_b*v_b,f)=sin(84.2)/(m_a*v_a,i)=sin(50)/(m_a*v_a,f)

4 eqns (bottom one can be separated into 3 eqns correct?), 4 unknowns (m_b, v_a,i, v_a,f, and v_b,f)

Tried to solve in my TI-89 and it did not work (note I'm not in radian mode or anything stupid)

 

[Saketh]

Use conservation of momentum and conservation of kinetic energy to find the answer.

(Hint: Write the conservation of momentum expressions in the x- and y-directions. Note that the y-momentum before the collision is 0, so it has to be 0 after.)

 

[bpw91284]

Use conservation of momentum and conservation of kinetic energy to find the answer.

(Hint: Write the conservation of momentum expressions in the x- and y-directions. Note that the y-momentum before the collision is 0, so it has to be 0 after.)

So like...?

X-direction

m_a*v_a = m_a*v_a*cos(52.9) + m_b*v_b*cos(50)

0.5*m_a*(v_a)^2 = 0.5*m_a*[v_a*cos(52.9)]^2 + 0.5*m_b*[v_b*cos(50)]^2

Y-direction

0 = m_a*v_a*sin(52.9) + m_b*v_b*(-sin(50))

Don't need to do KE in y-direction because already have 3 equations for 3 unkowns...
Does doing the trig functions eliminate the need to do the initial and final subscripts on the velocities?