<p>We can work through an exam together on CC posting answers and explanations on how to do certain problems. My Physics B teacher has been absent for much of the year and admits he is already not that great of a teacher so I have been self studying for the exam. I have the 1998 College Board Released Exam and would really like it if everyone who wanted to participate would work through the exam together and post their answers. I also have the answers to the test but no explanation so I thought it would be a good idea to get the reasoning behind each one I miss. As of now I can only answer probably 50% of the questions correctly, so I'll start the thread off by answering and explaining all of the problems I can do as I go.</p>

<p>This pdf doesn't have the answers but I have a PDF on my computer with the answers on it. I can't find an online version with the answers right now but if someone does, please post it! If no one has success finding it, I can email it to you.</p>

<p>1) B : The two objects presumably have different masses, but speed due to gravitation is not affected by mass. Momentum is (mass)<em>(velocity); kinetic energy is (1/2)</em>(mass)<em>(velocity)^2; change in potential energy deals with (mass)</em>g*height, so the other answers are wrong.</p>

<p>2) C : power = work/time; work = Force<em>distance…so 700</em>8/10 = 560 watts</p>

<p>3) E : inelastic collision-- mv+Mv = mv’+Mv’ => mv + 0(because M is at rest) = (m+M)v’
solving for v’ = mv/(M+m)</p>

<p>4) C : I only know how to explain this with plugging in values for the mass and velocity. mv must equal mv’ because of conservation of momentum. If we pretend the original mass was 2 kg and velocity was 2 m/s and the final mass was 3 kg, then 2<em>2 = 3</em> v’. Simplifying this makes v’ = 1.5 m/s which is less than the original.</p>

<p>5) C : gotta hate roman numeral questions…but this one is pretty easy. Power is measured in watts and is defined by the equation Power = work/time. Work is measured in joules so joules per second and watts works.</p>

<p>6) E : angular momentum = r * mv. 4* (2*3) = 24 kg m/s^2</p>

<p>9) well rate = change/time.
rate=power
w/t =power
w= Force x Distance.
Therefore:
Power = Force x Distance /Time
Velocity = Distance/ Time
So, RAte= Force x Velocity
rate= umgv choice A</p>

<ol>
<li><p>A.The rate of work done is power. Power = W/t= F(fric)x/t = F(fric)v = umgv.</p></li>
<li><p>A.The lines are caused by electron transitions from higher energy levels to lower energy levels, causing energy to be emitted. These levels have different principal quantum numbers. Thus, the answer is A.</p></li>
<li><p>B. The oil drop experiment proved that charges are quantized. The charge-mass ratio is simply a calculation and has nothing to do with waves. Thus II is the only valid experiment.</p></li>
<li><p>A. Charge must be conserved. Number of nuclei and protons do not have to be conserved (think of beta emissions).</p></li>
<li><p>A. charge is 0 so by F=qE F=0.</p></li>
<li><p>E. E=V/d 2000=E1=V/d. E2=2V/(d/5) = 10V/d = 10(2000) = 20000 E.</p></li>
<li><p>A. 1 ohm and 3 ohm resistor are in series, so Req = 1 +3 = 4 ohms. 4 ohms and 2 ohms resistor are in parallel, so Req= 1/((1/4)+(1/2))= 1/(3/4) = 4/3 = 1 + (1/3) = A</p></li>
<li><p>E. Voltage is the same in parallel resistors, but the bottom has a smaller resistance than the Req of the top, so E.</p></li>
<li><p>C You have one vector directed to the left and one directed straight down, so C is the only choice that completes the right triangle in the appropriate direction.</p></li>
<li><p>D. U=qV. V=kQ/R = kQ/d. V(total)= kQ/d + kQ/d = 2kQ/d. therefore, U= 2kqQ/d.</p></li>
<li><p>C. flux = BA = 2T (5 *10^-2 * 8 *10^-2) = 8 *10^-3</p></li>
<li><p>D. P=IV= 4(120)= 480 Watts = .48 kW. .48 kW (10cents/kW * hour) (2hours) = 9.6 cents.</p></li>
</ol>

<ol>
<li>(A) because work is area under the curve.</li>
<li>(A) because pressure and temperature are directly related. PV = nRT. Or common knowledge.</li>
</ol>

<ol>
<li><p>C. Work done by system is negative. Heat added to system is positive. Therefore, 400 J- 100J = 300 J.</p></li>
<li><p>Skip, not part of the curriculum anymore.</p></li>
<li><p>B. Image is at the focal length, so f=R/2 = 1/2 or .5</p></li>
<li><p>E n1v1=n2v2 or n1/n2=v2/v1. Index of refraction for water is > than index of refraction for air so n2>n1 and that means v2<v1 so the speed in water is less than the speed in air. v=f(lambda) so if v decreases than lambda must decrease. E.</p></li>
<li><p>D. 1/f = 1/p+1/q. 1/9=1/6 + 1/q -> q= -18 so M=- (-q/p) or 3.</p></li>
<li><p>B. Speed is constant and amplitude actually decreases.</p></li>
<li><p>D. Image is real because 1.5f > f (virtual if distance is smaller than f). 1/p + 1/q =1/f -> 1/f=1/q + 1/1.5f -> q=3f so hi = 3/1.5 ho = 2ho -> larger</p></li>
<li><p>A. Rays 1 and 3 are parallel so the index of refraction must be equal by snells law.</p></li>
<li><p>Not part of curriculum.</p></li>
</ol>

<p>Whoa I had no idea people actually replied! I stopped posting my answers on here and just stopped working on it! I shall start again and thanks for clarifying those for me & anyone else who happens to see this!!!</p>