1998 E&M C3 Solution

1998 E&M C3 Solution - Distribution of points (a) 4...

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Unformatted text preview: Distribution of points (a) 4 points Using Newton’s 2“Cl law along the ramp, F g sin0 — Fm = ma , where Fg 2 force of gravity, and Fm = magnetic force At constant speed, a = 0, so Fg sin6 — Fm = O For the following expression for Fm 1 point F m = F g sin 6 For substituting F g = mg 1 point F m = mg sin 6 For substituting Fm = IZB 1 point MB = mg sin 6 For correct answer 1 point I = mg sin 0 [B (b) 4 points d l8! — —& dt For correct expression for ¢M 1 point ¢M = fo For the time deriviative of ¢M 1 point d 1M1. = dt For correct use of Ohm’s law to find current 1 point 8 = [R I = E = fl R R For equating this expression to the expression for I from part (a) and solving for 1) 1 point B! 1) _ mg sin 9 R B! U = ng Sln 9 3222 Distribution of points E & M 3 (continued) (0) 2 points For correct expression for power 1 point P = 1 2R For correction substitution of the expression for I from part (a) 1 point P _ (ng2 sin2 6)R _ 32% (d) 3 points For either of the following two expressions 1 point mfl : mg sin0 — MB dt dv . 82621) —— = gsmB — dt mR For rearranging terms in either expression to set up intergral 1 point For example, do 32621) mR gsinH— For integrating this expression, and applying 0(0) = 0 to get answer 1 point ngsin0 Bzflzt of =————— l—ex — () 3,6, [ p( M (e) 2 points For correct answer 1 point Yes, the final speed of the bar decreases For any reasonable justification 1 point For example; Since the two resistors are in parallel across the emf in the bar, the new effective R ng sin 6 resistance is 3-. And since 1) = Blgz from part (b), then ifR decreases, the speed decreases. ...
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1998 E&M C3 Solution - Distribution of points (a) 4...

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