*******Official AP Physics B/C Discussion/Prep Thread*******

[vader1990]

^yatta, don't these questions usually tell you what the moment of inertial formula is is: like for a sphere isn't it I=0.5mr^2 or something?

[Yatta!]

For a solid ball the moment of inertia is 2/5 m r^2.

The total KE energy of the ball when you reach the bottom of the slope is 1/2 m v^2 + 1/2 I w^2, but because it is rolling without slipping, w*r = v so KE = 1/2 m v^2 + 1/2 (2/5 m r^2)*(v^2/r^2) = 7/10 m v^2.

Total KE = the change in PE so 7/10 m v^2 = gmh = 100m.

Solving for v, you get the final velocity to be 11.952m/s, so the correct answer is B.

To the Physics B'ers: don't worry about this kind of question, it will not appear on your test.

vader1990: Yes usually the question gives away the moment of inertia of any object but I find it useful to know the moment of inertia of some common objects off the top of my head.

[einstein7th]

Nice question Yatta - got some people.

Physics C'ers, don't forget about rotational KE - it's pretty tricky. =D

[drnoeyedea]

I was bout to say...you people forgot rotational kinetic.
Answer is indeed 11.95 as yatta said. i admit, i was a bit puzzled when i didnt see a radius. Then i looked at the equation sheet and r*w = v.

How about this question.

A thin polyvinyl chloride rod of length 5m has a velocity of 3000 m/s as it passes through a magnetic field of strength 400T that is directed into the page. What is the direction of the current flow in the rod?

[A-Card]

anyone have some old ap exams online that they can direct me too?

[einstein7th]

@drnoeyedea, is the rod going from left to right through the field?

If so, the current is flowing up.

Here's a good one, B/C both good for:

Two objects, one of mass 3kg and moving with a speed of 2 m/s and the other of mass 5gh and speed of 2 m/s, move toward each other and collide head-on. If the collision is perfectly inelastic, find the speed of the objects after the collision.

A .25 m/s
B .5 m/s
c. .75 m/s
d. 1 m/s
e. 2 m/s

Enjoy.

[vader1990]

(E)

m1v1 + m2v2 = Vf(m1 + m2)
Vf = m1v1 + m2v2 / (m1 + m2)
Vf = (6 + 10) / (3 + 5) = 16 / 8 = 2 m/s

NEW PROBLEM:

An Original Gangsta is initially on a circular platform that is free to rotate without friction about its center. The gangsta jumps off tangentially to shank another OG who called his mama fat; this sets the platform spinning. As the shanking gangsta jumps off tangentially what quantities are conserved.

I. Angular Momentum
II. Linear Momentum
III. Kinetic Energy
IV. Mechanical Energy

[snipez90]

ohshi, I'm too used to boxes on an inclined plane. Anyways isn't linear momentum always conserved?

[einstein7th]

@Vader - Incorrect bro - ntnt. Lol, I got it wrong at first too, take a look again.

[vader1990]

^lol....is it |-0.5|= 0.5 (since the objects are moving towards each other one has + velocity and the other has -)...lol...I need to be more careful

@snipez90, linear momentum, is not conserved in this case...hejumps off tangentially, this is a hint in the question try again (liner =/= angular)

EDIT: MOMENTUM IS CONSERVED...but which kind?

[snipez90]

m1v1 + m2v2 = (m1+m2)vf => (3kg)(2m/s)-(5kg)(2m/s) = (8kg)vf => vf = -.5 m/s. Speed = |-.5| = .5 m/s. Answer: B?

*EDIT* damn, dude I need to study up on rotational

[fignewton]

I had this on another thread, but in case you missed it:

A charge -1 C is stuck at x=0. A charge +4 C is stuck at x=2 m. You need to place a positive charge Q somewhere along the x-axis so that it doesn't move. No glue, staples, nails etc. are available.

It should be placed at:

a) x = 2/3 m
b) x = -1 m
c) x = -2 m
d) x = 5 m
e) x = -1/2 m

[snipez90]

C, first notice that it must be placed on the negative x-axis. Now let x be the distance between the origin and the position of the particle on the negative x-axis.

Now the force on Q due to the -1C charge is denoted by kQ/x^2 and directed to the right. The force on Q due to the 4C charge is denoted by 4kQ/(x+2)^2 and is directed to the left. Fnet = 0, so equate the opposite forces, solving the quadratic yields x = 2, and thus Q's position is -2.

[vhoidz]

does anyone know any good tricks to memorizing certain formulas, if so, please share

[snipez90]

vhoidz, I don't know any specific tricks but dimensional analysis may often point you in the right direction when it comes to recalling formulas.

Anyways new question:

Suppose that an electron (charge -e) could orbit a proton (charge +e) in a circular orbit of constant radius R. Assuming that the proton is stationary and only electrostatic forces act on the particles, which of the following represents the kinetic energy of the two-particle system?

let k = 1/(4pi(e0))
a) k(e/R)
b) (k/2)(e^2/R)
c) -(k/2)(e^2/R)
d) k(e^2 / R^2)
e) -k(e^2 / R^2)