Calculating Velocity Post-Collision: White Snooker Ball

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SUMMARY

The discussion focuses on calculating the post-collision velocity of a white snooker ball after colliding with a stationary blue ball of identical mass (0.173 kg). Initially, the white ball moves at 3.2 m/s, while the blue ball achieves a velocity of 2.77 m/s post-collision. The correct calculation for the white ball's velocity after the collision is 1.6 m/s, achieved by applying the conservation of momentum in two dimensions and breaking down the momentum into x and y components. The initial attempt incorrectly calculated the white ball's velocity as 0.43 m/s due to neglecting the two-dimensional nature of the problem.

PREREQUISITES
  • Understanding of momentum conservation principles
  • Familiarity with two-dimensional vector decomposition
  • Basic knowledge of snooker ball physics
  • Ability to apply equations of motion in collision scenarios
NEXT STEPS
  • Study the principles of conservation of momentum in two dimensions
  • Learn how to decompose vectors into x and y components
  • Explore collision types and their impact on momentum transfer
  • Practice similar problems involving elastic and inelastic collisions
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding momentum conservation in collision scenarios, particularly in sports physics contexts like snooker.

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Homework Statement


This diagram shows a white snooker ball moving with a velocity of 3.2 \ ms^{-1}. The white ball is on a collision course with a stationary blue ball of identical mass of 0.173 \ kg. The player is aiming to propel the blue ball into the pocket along the path shown.

COM.jpg

The player succeeds in propelling the blue ball along the path shown. The blue ball has a velocity of 2.77 \ ms^{-1} after the collision with the white ball. Calculate the velocity of the white ball after the collision.

Homework Equations


P=mv (I think)

The Attempt at a Solution


As momentum is conserved, I would have momentum before equals momentum after hence...

(0.173)(3.2) = (0.173)(v) + (0.173)(2.77) \implies v = 0.43 \ ms^{-1}

But this is incorrect. Can someone explain how I would tackle this question and point out my mistake. Thanks in advance. :smile: (The correct answer is v = 1.6 \ ms^{-1}.)
 
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Note that this is a two dimensional problem. In this case you have to break up the momentum of each ball into its x and y components. Apply conservation of momentum in each direction.
 
This question must be solved in two dimensions. Therefore separate the momentum before and after into x, y components. As a hint since both balls have the same mass, you can cross the mass out of your conservation equations and deal only with velocities. Remember to use the angle given to find the x and y velocity of the blue ball after the collision.
 

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