# billiard ball Momentum help

1. Dec 11, 2003

### bard

A billiard ball of mass $$m_{A}=0.400kg$$ moving with a speed $$v_{A} =1.8m/s$$ strikes a second ball, initially at rest, of mass $$M_{B}=0.500kg$$. As a result of the collision, the first ball is deflected off at an angle of $$30\deg$$ with a speed of $$v'_{A}=1.1 m/s$$.

a) taking the x-axis as the positive direction of motion of ball A, write down the equations expressing the conservation of momentum for the components in the x and y directions seperatley.

B)Solve the equations for the speed, $$v'_{B}$$, and the angle, $$\theta'_{2}$$ of ball b. Do not assume the collision is elastic.

my work

$$equations$$

$$m_{A}v_{A}=m_{A}v'_{A}\cos\theta'+ m_{B}v'_{b}\cos\theta'_{2}$$

$$0=m_{A}v'_{A}\sin\theta'+m_{B}v'_{b}\sin\theta'_{2}$$

Last edited: Dec 11, 2003
2. Dec 11, 2003

### Wooh

I tried my hand at it, but bear with me, I didn't have my calc handy so I couldn't simplify anything, so I had to carry a bunch of phrases. As far as answers, I got $$\theta_{2} = \sin^{-1} \frac{.5(1.1)\sin (\frac{\pi}{6})}{-.4(1.8)}$$ For Vfinal for the initially moving ball, I got...
$$V_f=\frac{.4(1.8) - .5(1.1)\cos(\frac{\pi}{6})}{.4\cos(\theta_{2})}$$
Like I said though, I didn't have a good way to test it out or anything, but that it was I gots, where $$\theta_{2}$$ is in radians, and in standard position. The whole $$2\pi-\theta_2$$ thing was unaccurate because it will output $$\theta_2$$ in standard position or as a negative, and doing subtracting it from $$2\pi$$ will merely warp the results. Try it out and tell me if I was close or not.
[/tex] because the signs are different between the ref angle, which is what $$\theta_2$$ is, and the actual angle, which is the aforementioned angle.
To get, I basically plugged into $$m_1v_o=m_1v_{1f}\cos(\theta_{2})+m_2v_{2f}\cos(\fract{\pi}{6})$$
and
$$0=m_1v_{1f}\sin(\theta_2)+m_2v_{2f}\sin(\frac{\pi}{6})$$

Last edited: Dec 12, 2003
3. Dec 12, 2003

### himanshu121

coefficient of restitution is 1 for elatic collisions which will also help u

4. Dec 12, 2003

### Staff: Mentor

Your equations look fine to me. What's the problem? If you plug in the numbers, you'll get two very simple equations. Did you try it?

5. Dec 12, 2003

### bard

well im not sure whether this problem has a definite answer(as in 1.2 or 3.4 etc)or whther im suposed to solve in terms of variables

6. Dec 12, 2003

### Staff: Mentor

If they meant you to solve it in terms of variables, why did they bother giving you all those values? Just plug in the numbers, then rewrite the two equations. You'll have two (simple) equations with two unknowns.

7. Dec 12, 2003

### bard

can someone help me in simplifying these eqautions? thnx

8. Dec 12, 2003

### Staff: Mentor

What are you looking for? An algebraic simplification without having to use the given information? Not going to happen. Did you try plugging in the numbers?

9. Dec 12, 2003

### bard

yes i plugged in numbers and i got

$$.068=v'_{B}\cos\theta'_{2$$

$$-.72=v'_{B}\sin\theta'_{2}$$

thats all i can do to simplify

Last edited: Dec 12, 2003
10. Dec 12, 2003

### Staff: Mentor

$$.68=v'_{B}\cos\theta'_{2}$$

$$-.44=v'_{B}\sin\theta'_{2}$$

These equations are easy to solve! To get the angle, divide them to get $$tan\theta'_{2}$$. Then plug back in to get $$v'_{B}$$. It won't get much easier than that.

11. Dec 12, 2003

### bard

Hey Doc Al,

thnx for helping me through this process :).

I got $$\theta_{2}=-33$$

$$v'_{B}=.81 m/s$$

12. Dec 12, 2003

### Wooh

I had made my classical error of using the reference angle for $$\theta$$ rather than the true angle, and I swapped a sign or two when I did it on paper. Sorry, at least when I reworked it my answers matched up :)