Click here to go to the

By Eurostar (Eurostar) on Monday, May 12, 2003 - 04:50 pm: Edit |

I didn't even finish the free-response. The multiple choice was hard too. AHHHHH!!!!

I was happy about FRQ # 1 though. Did you convert Work=Fdx by differentiating x(t) and then plugging it in? Because that's what i did....

By Adamg (Adamg) on Monday, May 12, 2003 - 05:15 pm: Edit |

Was it just me or when you solved the theoretical model on Mech 3. you were off by around 80% at 300kg???? I figured it out to be 67m.

By Adamg (Adamg) on Monday, May 12, 2003 - 05:21 pm: Edit |

I left about 6 blank on the MC. I'm pretty sure I got all the check box parts on the FR wrong. At least those are only worth at most 3 points.

By Adamg (Adamg) on Monday, May 12, 2003 - 05:26 pm: Edit |

For Mech 1. I intergrated the power equation to find work.

By Eurostar (Eurostar) on Monday, May 12, 2003 - 05:27 pm: Edit |

How much do you think the regression line and approximation and the t= (sqrt3 ?) are worth for the 3rd problem?

5/15?

By Eurostar (Eurostar) on Monday, May 12, 2003 - 05:30 pm: Edit |

Oh...but do you think finding x=blah then changing it to dx=blah dt and then plugging that in to Fdx would work?

By Eurostar (Eurostar) on Monday, May 12, 2003 - 05:31 pm: Edit |

What did you say for the checked boxes?

I said Work done by student is MORE than net work because the friction causes a negative work...

By Miscanon (Miscanon) on Monday, May 12, 2003 - 05:57 pm: Edit |

For the power...I used the formula Power = F * v and we had already solved (in earlier parts) for F & v as functions of t...so just multiply the two quantities.

What else did you guys think of the exam...any specific MC or FR?

By Callmecollege (Callmecollege) on Monday, May 12, 2003 - 06:41 pm: Edit |

how did u find the toal work done from time 0 to 2 secs on free response question 1? what did you guys get for an answer?

By Alias (Alias) on Monday, May 12, 2003 - 07:10 pm: Edit |

i don't remember the answer, but to get work, i integrated the force. since the force was in terms of time, i put in 2 seconds for that.

By Callmecollege (Callmecollege) on Monday, May 12, 2003 - 07:18 pm: Edit |

could u use the work-kinetic energy thrm to find work, AHHHH, why did i use that!!!!!!!!

By Miscanon (Miscanon) on Monday, May 12, 2003 - 07:21 pm: Edit |

yes the work-kinetic energy theorem is correct. W = delta (K)

By Callmecollege (Callmecollege) on Monday, May 12, 2003 - 07:22 pm: Edit |

so then i got 200, but everyone i talked to got larger numbers...

By Bdk (Bdk) on Monday, May 12, 2003 - 07:50 pm: Edit |

I got 3000 J for net work using both integral of power and change in KE.

By Miscanon (Miscanon) on Monday, May 12, 2003 - 09:24 pm: Edit |

I got 3000 J also. For the theoretical vs. experimental...did you guys get around 67 or 68 m /s predicted by theoretical? (Air resistance I said)

By Miscanon (Miscanon) on Monday, May 12, 2003 - 09:26 pm: Edit |

Do any of you remember the two current carrying loops, one on top of the other? The current was flowing in the loops in the same direction...and the question was that if the top loop's current was cut off...what would be the induced emf in the bottom loop? I said that since less flux would go through the loop...the current speed would increase to counteract the decreasing magnetic flux...have any thoughts on this?

By Ndhawk (Ndhawk) on Monday, May 12, 2003 - 09:28 pm: Edit |

I got 3000 for net work, I got all of free response 1, and only parts of the next two, and then im a little shaky on mc

By Eurostar (Eurostar) on Monday, May 12, 2003 - 09:30 pm: Edit |

Wait..am I going crazy or could I have used N*m as a label for work? I got 3000 as the net work also.

By Miscanon (Miscanon) on Monday, May 12, 2003 - 09:33 pm: Edit |

N * m is okay to use but J is acceptable also. (1 J = 1 N * m)

By Adamg (Adamg) on Monday, May 12, 2003 - 09:34 pm: Edit |

I got 3000J using the integral of power. On that other one I got 67 m and said that not all the potential energy is converted to kinetic. The poll has potential energy still, and I also said the bucket has potential energy (in drawing 2 it was still above the ground.) I don't know if saying friction will cut it because I remember a past FR problem that asked why there was a difference and a friction response received zero points.

By Diconoclastx (Diconoclastx) on Monday, May 12, 2003 - 10:12 pm: Edit |

I got everything right on the free respones. If you need help ask me.

The answer for the calculated x value was 37 point something. Which is close to the real thing which was also 37.

And for the work one just take delta KE. Much easier, screw integrating.

By Miscanon (Miscanon) on Monday, May 12, 2003 - 10:47 pm: Edit |

Dinoclastx...my earlier post: Do any of you remember the two current carrying loops, one on top of the other? The current was flowing in the loops in the same direction...and the question was that if the top loop's current was cut off...what would be the induced emf in the bottom loop? I said that since less flux would go through the loop...the current speed would increase to counteract the decreasing magnetic flux...have any thoughts on this?

Do you have any thoughts on this?

By Webhappy2 (Webhappy2) on Monday, May 12, 2003 - 11:23 pm: Edit |

I said increase also on that one, for your reason.

For the difference one, I got around 60 or so...

Also, to find rho-zero, I had a very messed-up expression; on the final part of E/M Question 1, to find Qenclosed, I had to integrated w/ that messed-up expression!!!

I jacked the MC for the Newtonian for sure.. ran out of time.

For E/M, when you have a metal w/ +Q charge, and you bring a second metal in contact, and pull the second one out, does each one end w/ Q/2? In particular, is the answer to the question involving a sphere touching 2 other ones end in 3F/8?

By Webhappy2 (Webhappy2) on Monday, May 12, 2003 - 11:32 pm: Edit |

Another question:

What was the PE for the catapult one? The first part, for Diagram I.

I did something screwy like torque but not torque...

I said Mg*3m - 10kg * g * 3m

By Eurostar (Eurostar) on Tuesday, May 13, 2003 - 01:44 am: Edit |

yeah...it asked for an "equation" on that PE one...I just kinda said PE=3*(M+10)G. I probably screwed it up, but I thought it was a bad question in the first place. MGH is PE, and so M=M+10, G, H=3...dunno

By Miscanon (Miscanon) on Tuesday, May 13, 2003 - 07:12 am: Edit |

Eurostar...I got the same answer as you ( 3 * (M+10) * g). When it asked to "derive" that...did it mean simply show the integration of (mg) to produce U = mgh....how'd you guys interpret that?

Posting is currently disabled in this topic. Contact your discussion moderator for more information. |

Administrator's Control Panel -- Board Moderators Only Administer Page |