Help Nash with Physics Final: 10 m/s^2 for Gravity

[Nash77]

Attached is my practice physics final. I need help! Below are my attempts at solutions... probably way off but maybe I just need a head start. Thanks for looking!

PS We are told to use 10 m/s^2 for gravity.

--Nash.


III.

1) at the top, all potential energy = mgh = 1*10*6 = 60
at the break point P: total energy = 60 = mgh + 1/2mv^2 = 1*10*1 + 1/2*1*v^2
60 = 10 + 1/2*1*v^2 50 = 1/2*1*v^2 10 = v
It is moving horizontally to the right when it breaks.
2) how far will the block travel?
Can I figure this out without time? I tried to use different formulas and got 5m, but I think that's wrong. I mixed up Vx with Ay so I think it's wrong.

IV.

KE of bullet = 1/2*0.01*100^2 = 50 J
I tried using conservation of energy, so KE for the bullet = 50 = KE for the Mass = 50 J
I got a height of 5.05 when this KE turns into PE for the Mass, but I'm sure it's not right to say KE is conserved... how do you know how much of it transfers to the Mass when the bullet gets lodged in the block of wood?

V.

1) Only two forces acting in the vertical direction, Weight acting down and Tension acting up. The weight = mg = 1*10 = 10 N. So the Tension should be the same, equal and opposite, since the rod is at rest. So Tension should be 10 N also, just acting up whereas weight is acting down. Is this correct, or do I need to take components of the tension?
2.) I said no force exerted by the wall, since there is no motion in the horizontal direction. Weight works down, so this shouldn't affect a rod attached perpendicular to the force of gravity.. but can there really be no force?
3.) Linear Velocity? moment of inertia = 1/3ML^2 = 1/3*1*1^2 = 1/3,
I know that Torque force = I*alpha (angular acceleration.) But how does angular acceleration relate to linear acceleration? How would I find either in this problem? Am I way off track here?

VI.

I'm going to try to work on this one now. Any suggestions would be much appreciated.

Thanks for having patience with all of this... PS Don't be too harsh.. Physics isn't my best subject if you couldn't tell ;) Still, I'm willing to learn!

 

[PhanthomJay]

Hi, Nash...you've got quite a few problems in this post, and you may not get help for all of them when posting them all at once. You'd be better off posting IV, V, AND VI in separate posts, showing your relevant equations in addition to the attempt you have made and questions you are asking.

For part III, the velocity and direction at the bottom of the swing looks good. To get the horizontal distance travelled, look in the y direction to find the time based on the vertical distance travelled.

 

[Nash77]

Thanks, I will do that right now.

While on this one thought, I've been working on IV. (bullet and block)

I found the momentum of the bullet, = mv = 0.01*100 = 1 N*s
This momentum should be conserved when the bullet is lodged in the block, so the system has the same momentum, 1 N*s, so 1 = (M+m)V = (0.99 + 0.01)V so the new Velocity is 1 m/s.

The system of block and bullet will reach the greatest angle when all energy is potential energy, so at PE = (M+m)gh = 1*10*h, solved for h and found it to be 0.05m.

I know I need to make a triangle somehow to find the angle, but I can't visualize it. How do I find the angle from these two heights?

 

[PhanthomJay]

Thanks, I will do that right now.

While on this one thought, I've been working on IV. (bullet and block)

I found the momentum of the bullet, = mv = 0.01*100 = 1 N*s
This momentum should be conserved when the bullet is lodged in the block, so the system has the same momentum, 1 N*s, so 1 = (M+m)V = (0.99 + 0.01)V so the new Velocity is 1 m/s.

Yes, very good

The system of block and bullet will reach the greatest angle when all energy is potential energy, so at PE = (M+m)gh = 1*10*h, solved for h and found it to be 0.05m.

you're on a roll, also correct.

I know I need to make a triangle somehow to find the angle, but I can't visualize it. How do I find the angle from these two heights?

Don't visualize it, draw a sketch. If h is 0.05 m, then the vertical distance to the pivot is .95 m, the length of the string is 1 m , and use trig to get the angle.

 

[Nash77]

Thanks!

Another equation that worked out was replacing h (0.05) for L(1-cos theta) from an energy equation PE= mgr(1-cos theta). Any idea where that equation comes from?