Find velocities of both Pucks!?

Question

The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.040 kg and is moving along the x axis with a velocity of +5.5 m/s. It makes a collision with puck B, which has a mass of 0.060 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angles shown in the drawing.


http://www.webassign.net/CJ/07_32.gif

http://answerboard.cramster.com/physics-topic-5-114595-0.aspx

still cant figure it out

Answer 1

masses m1, m2 = .04 .06
pre-collision velocities v1i, v2i = 5.5 0
post-collision angles theta1, theta2 = 65 -37
post-collision velocities v1f, v2f:
P = m1*v1i + m2*v2i
v1f = P/((1/tan(theta1) - 1/tan(theta2))*m1*sin(theta1)) = 3.3839
v2f = P/((1/tan(theta2) - 1/tan(theta1))*m2*sin(theta2)) = 3.3974
The ref. is a collision calculator. To use, you must enter nonzero initial values (very small no. for 0) and final velocity & theta of one body.

Answer 2

This is all about the conservation of momentum. Since this is a 2-D problem, you have to look at the horizontal separately than the vertical.

Horizontally

ma*via + mb*vib = ma*vfa*cos(65) + mb*vfb*cos(37)
We know that vib = 0
ma*via = ma*vfa*cos(65) + mb*vfb*cos(37)

Vertically
We have no intial mometum vertically, so
0 = ma*vfa*sin(65) - mb*vfb*sin(37)

We have 2 equation and 2 unkowns, vfa and vfb.
Solve either equation for vfa or vfb and plug into the other equation.

From the vertical
vfa = mb*vfb*sin(37) / (ma*sin(65))

Plug that into the horizontal and solve for vfb
ma*via = ma*vfa*cos(65) + mb*vfb*cos(37)
ma*via = ma* mb*vfb*sin(37) / (ma*sin(65)) * cos(65) + mb*vfb*cos(37)
ma*via = mb*vfb*sin(37)/(tan(65) + mb*vfb*cos(37)
ma/mb * via = vfb*sin(37)/(tan(65) + vfb*cos(37)
ma/mb * via = vfb * (sin(37)/(tan(65) + cos(37)
vfb = (ma/mb * via) / ((sin(37)/(tan(65) + cos(37))

Now that you know vfb, simply plug and chug and find vfa.

Good luck.