<p>If you answer this I will give you all the old AP Physics C tests, from 1974-1993, and 1997-2006.</p>
<p>damn it i can't solve it lol. Do you think you can still send me the questions please?
Here's what I would do though:
part B is just conservation of momentum (M1V1 + M2V2)= (M1+M2)Vf
The Vf would be the final velocity of the ball.
As soon as you know that, you can break velocity into its components to find the angle.</p>
<p>a. Mechanical Energy is Conserved</p>
<p>K.E At bottom = P.E At top
so
1/2<em>I</em>Omega^2=m<em>g</em>h ---> Sub m<em>l^2/3 for I (which is same as 3</em>1.2^2/3) and l/2 for h (cause center of gravity only moves up that much) and everthing else thats known
so
1/2<em>1.2^2</em>Omega^2=3<em>10</em>1.2/2
Solve for omega you get :
(i dont raelly think i need to show u how to solve that..)
Omega= 5 Radians/Second</p>
<p>b. KE conserved(stated in problem)</p>
<p>.5<em>m</em>vi^2=.5<em>m</em>vf^2+.5<em>I</em>Omega^2 ----> Sub All Numbers in like last time (and Omeaga as well)</p>
<p>.5<em>1</em>10^2=.5<em>1</em>vf^2+.5<em>1.2^2</em>5^2</p>
<p>once again solve for vf to get </p>
<p>vf= 8 m/s</p>
<p>c. this one is REALLY easy</p>
<p>Angular Momentum = m<em>v</em>r<em>sinTheta= m</em>v<em>r (in this case) = 1</em>10*1.2=12 kg m^2/s</p>
<p>d. Angular Momentum is Consereved:</p>
<p>m<em>v</em>r= I<em>omega+ m</em>v<em>r</em>cosTheta ----> sub numbers</p>
<p>12= 1.2^2*5+1.2CosTheta
solve to get
CosTheta =.5
Theta = 60 degrees</p>
<p>and er..t.hats all..phew that took long enough to type.....</p>
<p>oh and let me know if something isnt clear or w.e..</p>