# Physics problem 2D collisions

1. Apr 5, 2013

### x86

1. The problem statement, all variables and given/known data
In a game of billiards the 0.165 kg cue ball is hit toward the 0.155 kg eight ball which is stationary. The cue ball travels at a speed of 6.2 m/s and after impact rolls away at an angle of 40.0 degrees counterclockwise from its initial direction with a velocity of 3.7 m/s. Find the velocity of the eight ball after the collision.

2. Relevant equations
m1v1 + m2v2 = m1vf1 + m2vf2

3. The attempt at a solution
I cut the coordinate system up into the x y and z axis. In the y axis, there is no movement, as the ball stays on the table. So, we have x and z to work with. I drew a Cartesian graph with the axis 'X' and 'Y' where 'Y' represents 'Z'. I will also interpret the cue ball being shot up the Y axis into the eight ball, instead of being shot horizontally.

So I cut the motion up like this, in the y axis vectors:
0.165kg * 6.2 m/s + 0 = 0.155kgVf + 0.165kg*3.7m/s*sin40
Vfy = 4.4 m/s [up]

x vectors:
0 = 0.155kgVf + 0.165kg*3.7m/s * cos40
Vfyx = -3.0m/s

Now, I add the vectors using Pythagorean theorem and my result is 5.3 m/s [W 55 degrees N] or 5.3 m/s [35 degrees clockwise].

However, the book gives me the answer of 4.4 m/s [35.2 degrees clockwise].

This is the part that confused me, why is the book not taking the X forces into account? Its just using the answer I got in the Y coordinate system. This is confusing, because my teacher did the same thing in school and answered with "there is no movement in the Y axis".

The problem is that there are three dimensions: X,Y,Z and Y is set to 0, but we still have movement in the Z axis (which I represented as Y in this answer).

So I have no idea why its done like this, and am very confused. Did the book make a mistake, or am I wrong?

2. Apr 5, 2013

### tia89

Pay attention... when you compute what you call $V_f$, you have to consider that also the second ball goes away with an angle... therefore the correct equations should be

Along y:
$$m_c V_{0c}=m_c V_{fc}\cos(40)+m_8 V_f\cos\theta$$
Along x:
$$0=m_c V_{fc}\sin(40)-m_8 V_f\sin\theta$$

Now solve for $V_f$ and if you want also for $\theta$. In this way indeed $V_f$ comes 4.4 =)

Last edited: Apr 5, 2013
3. Apr 5, 2013

### tia89

mmmh... sorry I noticed something just now... your way is also correct because, even if you would have better to call in different ways $V_f$ in the equations in the two directions, it is the same thing (you are leaving implicit the $\sin\theta$ and $\cos\theta$ I am writing explicitely).

The problem is in the definitions... along your Y you have to put the $\cos$ (measuring the angles leaving from the Y axis) while along X you put $\sin$. In this way it works...

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