AP Physics C: Mech Exam Practice

Test Preparation
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<p>So I was being a dork this afternoon, and I started practicing for the AP exam when I came across this problem, pulled from a previous year free-response exam:</p>

<p>An open top railroad car (initially empty and of mass M0) rolls with negigible friction along a straight horizontal track and passes under the sprout of a sand conveyor. When the car is under the conveyor, sand is dispensed from the conveyer in a narrow stream at a steady rate deltaM/deltat=C and falls vertically from an average height h above the florr of the railroad car. The car has an initial speed v0, and sand is filling it from time t = 0 to t = T. Express your answers to the following in terms of the given quatities and g.</p>

<p>a) Determine the mass M of the car plus the sand that it catches as a function of time t for 0 < t < T.</p>

<p>b) Determine the speed v of the car as a function of time t for for 0 < t < T.</p>

<p>c) i.) Determine the initial kinetic energy (Ki) of the empty car.
ii.) Determine the final kinetic energy (Kf) of the car and its load
iii.) Is kinetic energy conserved? Explain.</p>

<p>d.) Determine the expressions for the normal force exerted on the car by the tracks at the following times
i.) before t=0
ii.) for for 0 < t < T.
iii.) after t=T</p>

<p>Looks like something I've never seen before...anyone know how to go about solving this? Maybe I'm overcomplicating it...or maybe it's just that hard :-P.</p>

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<p>Funny you mention this question; it's on my problem set.</p>

<p>I'd like to know how to do this too, if anyone else can help!</p>

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<p>a)
M = Mo + Ct</p>

<p>b)
MoVo = MV
MoVo = (Mo + Ct)V
V = (MoVo)/(Mo + Ct)</p>

<p>c)
i) KE = 1/2 MoVo^2</p>

<p>ii) KE = 1/2 MV^2...plug in from part a) and b)
KE = 1/2(Mo + Ct)(MoVo/Mo+Ct)^2
KE = (Mo^2Vo^2)/2(Mo+CT)... make sure you plug in T (final time) for t</p>

<p>iii) No, because it's an inelastic collision</p>

<p>d) i) Fn = Mog
ii) Fn = Mog (weight of car) + (Ct)g (weight of sand) + Fs (extra force from sand)</p>

<p>Fs = change in momentum/ change in time (delta p/delta t)
(delta p/delta t) = ((delta M)(Vy))/delta t = CVy</p>

<p>use conservation of energy to find Vy
mgh = 1/2mVy^2...masses divide out
Vy = sq rt(2gh)</p>

<p>Fn = (Mo + Ct)g + C(sq rt(2gh))</p>

<p>iii) Fn = (Mo + CT)g</p>

<p>...yeah we did problem in class</p>

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<p>haha thanks man; yea, it seems to be quite the popular problem!</p>