# Collision Problem

## 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 seperated 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)

Related Introductory Physics Homework Help News on Phys.org
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.)

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 fucntions eliminate the need to do the initial and final subscripts on the velocities?