The collision between two pucks

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SUMMARY

The discussion focuses on solving a two-dimensional collision problem involving two pucks, A and B, with masses of 0.294 kg and 0.588 kg, respectively. Puck A moves at an initial velocity of 5.55 m/s along the x-axis, while puck B is initially at rest. The final speeds calculated for puck A and puck B after the collision are 3.41 m/s and 2.57 m/s, respectively. The solution involves applying conservation of momentum in both the x and y components, using the equations m1v0 = m1v1' cos θ1 + m2v2' cos θ2 and 0 = m1v1' sin θ1 - m2v2' sin θ2.

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Homework Statement


Puck A has a mass of 0.294 kg and is moving along the x-axis with a velocity of 5.55 m/s. It makes a collision with puck B, which has a mass of 0.588 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. Find the final speed of
Puck A
Puck B

Homework Equations


I have spent an hour on this problem. I have two different eqns one for x and one for y, respectively:
m1v0 = m1v1 cos 65 + m2v2cos 37
and
0 = m1v1 sin 65 - m2v2 sin 37



The Attempt at a Solution


I solved for v2 from the eqn for y and tried to plug that value into the eqn for x. This did not work and I am running out of ideas and patience. Please help! There is also a pic for this problem..
 

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I think you accidentally messed up you arithmetic when solving the question. Your X and Y equations are correct so you've set up your question correctly. i think you have just made a mistake when solving because the substitution of V2' of you Y equation into your X equation is the correct thing to do.

I got V1' = 3.41 m/s; V2' = 2.57 m/s; And these numbers make sense in my mind
 
Your numbers worked out perfectly. I don't understand where i am going wrong, because i am not getting your numbers. can you do a step by step please? id really appreciate it.
 
X-component: m1v1 = m1v1' cos 65 + m2v2'cos 37
Y-Component: 0 = m1v1' sin 65 - m2v2' sin 37

Y: 0 = m1v1' sin 65 - m2v2' sin 37
m1v1' sin 65 = m2v2' sin 37
m1v1' sin65/(m2sin37) = v2' -> Plug in numbers
(0.294 kg)*(v1')*(sin65)/[(0.588 kg)(sin37)] = v2'
0.75298*v1' = v2' -> substitute into the X

m1v1 = m1v1' cos 65 + m2v2'cos 37
(0.294 kg)(5.55 m/s) = (0.294 kg)v1'(cos65) + (0.588 kg)(0.75298v1')(cos37)
1.6317 kg*m/s = 0.12425*v1' kg + 0.3536*v1' kg
1.6317 kg*m/s = 0.4778*v1' kg
v1' = 3.41 m/s -> substiture back into Y

0.75298*v1' = v2'
0.75298*(3.41 m/s) = v2'
v2' = 2.57 m/s

So this was my solution... a lot messier on the computer but i hope you can find where you and i differed in our solutions and this can help you avoid the same mistake in the future.

Cheers
 
Thankyou so much! This was such a help to say the least
 

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