Conservation of angular momentum help

[sunumen]

http://img215.imageshack.us/my.php?image=q1db9.jpg

I haven't any suggested answer on this problem

The most important thing is that I really don't know how to start to deal with this problem.

< I am lack of knowledge on angular momentum<<I don't know is this needed in solving this problem>> >

I guess that using momentum in the first part finding angular velocity
and use conservation of energy in solving angular displacement??

Can anyone give me some hints?
Thankyou.

 

[chaoseverlasting]

Yes. You need to know conservation of angular momentum. $$\vec{L}=\vec{L_{cm}}+\I\vec{w}$$.

More simply, as the net torque on the system is negligible, $$m_1v_1r_1=m_2v_2r_2$$. From this you find $$v_2$$ and the kinetic energy of the system, from which you can find the height and the angle obtained etc.

 

[sunumen]

chaoseverlasting, thanks for your reply.
I have tried to solve the problem in this way.
m$$_{1}$$v$$_{1}$$r$$_{1}$$=m$$_{2}$$v$$_{2}$$r$$_{2}$$
0.02x300x0.4 = 1.02x v$$_{2}$$x(0.4+0.5)
v$$_{2}$$=2.614m/s

angular velocity= v/r = 2.614/0.9
= 2.9 rad/s

===========================
K.E. loss = P.E. gain
1/2 mv$$^{2}$$ = mgh

1/2 x 2.614$$^{2}$$ = 9.8 x h
h=0.349m

cos (angle) = (0.9-h)/0.9
angle = 52.24 degree

===========================
There are some questions I want to clear
I hope that someone can answer me~~Thx

1~the 1.02(kg) which I highlighted is the sum of the bullet and mass.
Am I incorrect to consider both of them?
I should use 1kg only or 1.02 kg?
If I use 1.02 kg , I think I should consider the center of mass after the bullet embedded...but it is quite troublesome.

2~is there any other solution to find out the maximum angular displacement?
I feel that using conservation of energy may not get the exact solution

moreover, Is there a solution that the angular I found in part a can be used again to solve the angular displacement??

<Speak again, I haven't learn angular momentum before >.< >

Thx any help.