Calculating Velocity Post-Collision: White Snooker Ball

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In the collision between a moving white snooker ball and a stationary blue ball, the initial velocity of the white ball is 3.2 m/s, and the blue ball reaches a velocity of 2.77 m/s after the collision. The conservation of momentum principle must be applied in two dimensions, requiring the separation of momentum into x and y components. The initial calculation incorrectly assumed a direct momentum transfer without considering these components, leading to an erroneous result. The correct post-collision velocity of the white ball is 1.6 m/s, determined by properly applying momentum conservation in both directions. Understanding the two-dimensional nature of the problem is crucial for accurate calculations.
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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.

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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.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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