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Final Velocity (involving momentum)

  • Thread starter grouchy
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[SOLVED] Final Velocity (involving momentum)

1. Homework Statement

A 0.329-kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.222-kg puck that is intitially moving along the x-axis with a velocity of 2.24 m/s. After the collision, the 0.222-kg puck has a speed of 1.24 m/s at an angle of 24 degrees to the positive x-axis. (a) Determine the velocity of the 0.329-kg puck after the collision. (b) Find the fraction of the kinetic energy lost in the collision.


3. The Attempt at a Solution

i tried m1 = 0.329kg ----- m2 = 0.222kg

m1v1(i) + m2v2(i) = m1v1(f) + m2v2(f)

v1(f) = [m2v2(i) - m2v2(f) cos theta] / m1
v1(f) = 0.7471105 m/s (this was wrong)

I assume I need to find the x and y components and then use vector sumation but I have NO idea how to find the y...
 

rl.bhat

Homework Helper
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Let v2(f) makes an angle theta with positivde x-axis. Take the compondents of v1(f) and v2(f) along x and y- axis. Y camponents miust be equal and opposite because v1 is along x-axis. So you can wright two equations. 1) m1v1 + m2*0 = m1v1(f)cos(24) + m2v2(f)cor(theta)
2) m1v1(f)sin(24) = m2v2(f)sin(theta) . Rewrigth the equation(1) as
m1v1 -m1vi(f)cos(24) = m2v2(f)cos(theta)....(.3) Square eq. 2 and 3 and add. After solving you will get v2(f)
 

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