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Collisions in 2-dimensions

  • #1
can anyone offer up some help? i've been working on this forever with no success.

On a frictionless surface, a 0.35 kg puck moves horizontally to the right (at an angle of 0°) and a speed of 2.3 m/s. It collides with a 0.23 kg puck that is stationary. After the collision, the puck that was initially moving has a speed of 2.0 m/s and is moving at an angle of −32°. What is the velocity of the other puck after the collision?

(____m/s, _____°)


any help would be very appreciated. :-)
 

Answers and Replies

  • #2
Doc Al
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Hint: What quantity is conserved during the collision?
 
  • #3
Is the speed of the second puck after the collision:

v = 0,9191 i - 1,613 j

?
 
  • #4
Hootenanny
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shogunultra said:
Is the speed of the second puck after the collision:

v = 0,9191 i - 1,613 j

?
I believe the question requires the answer in the form of a speed with a direction (degrees above east)
 
  • #5
yaminohohenheim said:
can anyone offer up some help? i've been working on this forever with no success.

On a frictionless surface, a 0.35 kg puck moves horizontally to the right (at an angle of 0°) and a speed of 2.3 m/s. It collides with a 0.23 kg puck that is stationary. After the collision, the puck that was initially moving has a speed of 2.0 m/s and is moving at an angle of −32°. What is the velocity of the other puck after the collision?

(____m/s, _____°)


any help would be very appreciated. :-)
After my calculations, I got v = 1.4 ms-1 and in the direction of 49.2 deg in the positive x and positive y quadrant. It's better to use conservation of kinetic energy to find the resultant velocity of the stationary puck after collision. Then use conservation of total momentum in the vertical displacement to find the angle of deviation.
 
  • #6
Doc Al
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Folks, provide help, not complete solutions.

For this problem, the only thing you know is conserved is momentum. No need to assume that kinetic energy is conserved.
 

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