Find the speed and direction of the second puck after the collision?

Question

A hockey puck moving along the +x axis at 0.6 m/s collides into another puck that is at rest. The pucks have equal mass. The first puck is deflected 34 below the +x axis and moves off at 0.46 m/s. Find the speed and direction of the second puck after the collision. [please use positive for counterclockwise (above the +x axis) and negative for clockwise (below the +x axis)]:

Answer 1

I think Tung went astray when he typoed 34 as 39. Also, v2cos(theta) isn't 0.24 but 0.22.
I get v2 = 0.3376 m/s, theta2 = 49.64 deg.
Setting m1=m2=1 kg,
Px = m1v1 = 0.6 kg-m/s
v1x' = v1cos(-34 deg)
P1x' = mv1x' = 0.3814
v1y' = v1sin(-34 deg)
P1y' = mv1y' = -0.2572
P2x' = Px - P1x' = 0.2186
P2y' = -P1y' = 0.2572
theta2 = arctan(P2y'/P2x') = 49.64 deg
v2 = P2y'/(sin(theta2)*m2) = 0.3376 m/s

Answer 2

you already know that momentum is conserved and that it fairly is a vector sum M1V1 = M1V2 + M2V3 now the skill is likewise conserved a million/2M1V1^2 = a million/2M1V2^2 + a million/2 M2V3^2 yet the two hundreds are equivalent so cancel them alongside with the a million/2's V1^2 = V2^2 + V3^2 feels like the Pytagorean theorem, would not it? so as quickly as you draw a diagram and cancel out the equivalent M's you have the vector sum of V2 and V3 similar to V1 and we actually pronounced that those fulfill the Pythagorean theorem for this reason the climate, V2 and V3 make a precise attitude with one yet yet another so V2 is 39 deg under x and V3 is 51 deg above x this project in simple terms exists because of the fact the masses are equivalent this additionally occurs on a pool table

Answer 3

the conservation of momentum
P1 + P2 = P1' + P2'
P: vector of momentum
': mean that before collision
P2 = 0
P1x = mv1x = mv1'x + m v2'x
0.6 = 0.46 cos(34) + v2'x cos (alpha)
v2' cos(alpha) = 0.24
0 = v1'y + v2'y
0 = 0.46 sin(39) + v2' sin(alpha)
v2' sin(alpha) = - 0.29
v2' = 0.38 m/s
alpha = -50 degree