momentum help

[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 }

[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(\fr act{\pi}{6})
and
0=m_1v_{1f}\sin(\theta_2)+m_2v_{2f}\sin(\frac{\pi} {6})

[himanshu121]

[o)]

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

[Doc Al]

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

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?

[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

[Doc Al]

Originally posted by 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
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.

[bard]

can someone help me in simplifying these eqautions? thnx

[Doc Al]

Originally posted by bard
can someone help me in simplifying these eqautions? thnx
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?

[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

[Doc Al]

Originally posted by 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
Check your numbers. I got:
.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.

[bard]

Hey Doc Al,

thnx for helping me through this process :).

I got \theta_{2}=-33

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

[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 :)