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Edwardo_Elric
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Homework Statement
A hockey puck B rests on a smooth ice surface and is struck by a second puck A, which was originally traveling at 40.0m/s east and is deflected 30 degrees from its original direction. Puck B acquires a velocity of -45 degrees with the original velocity of A.
THe pucks have the same mass.
a.) Compute the speed of each puck after the collision.
b.) What fraction of the original kinetic energy of puck A is dissipated during the collision?
Homework Equations
[tex]m_{A}v_{A1x} + m_{B}v_{B1x} = m_{A}v_{A2x} + m_{B}v_{B2x}[/tex]
[tex]m_{A}v_{A1y} + m_{B}v_{B1y} = m_{A}v_{A2y} + m_{B}v_{B2y}[/tex]
The Attempt at a Solution
a.)
m_{A} = m_{B}
For the x- component
m_{A}(40.0m/s) = m_{A}(V_{A} cos(30) + m_{A} V_{B} cos(315)
40.0m/s = V_{A} cos(30) + V_{B} cos(315) (Eq 1);
For the y- component
0 = m_{A}(V_{A}sin(30) + V_{B}sin(315))
(V_{A}sin(30)) = (V_{B}sin(315))
V_{A} = (V_{B}sin(315)) / [sin(30)] (Eq 2);
substituting the x- component:
40.0m/s = (V_{B}sin(315)) / [sin(30)] + V_{B} cos(315)
V_{B} = 18.86 m/s
V_{A} = 26.667 m/s
Yep i got this problem right... same as back of book
b.) HEres my problem:
What fraction of the original kinetic energy of puck A is dissipated during the collision?
what does it mean when the kinetic energy is dissipated?
K1 = 1/2(mA)(V_{A})^2 = 1/2 ( 40m/s)^2
K2 = 1/2(mA)(V_{A})^2 + 1/2(mA)(V_{B})^2 = 1/2(mA)(18.86)^2 + 1/2(mB)(26.667)^2
the answer is 0.196
thank you very much