<p>ohmahgaw… d’oh!! thanks tariq. I can’t get over how my teacher has never ever applied calculus in physics…</p>
<p>TariqM, since you have all the exams, can you help me out with the problems I listed above? 1988 MC Exam: Problems 8,9,27,29,30,35</p>
<p>
</p>
<p>I appreciate the sarcasm, haha… but I’m self-studying for this exam.</p>
<p>i’m with you mulberrypie, all year i wondered why we were required to be in BC calc to take physics and then i saw the practice tests. turns out despite my teacher never mentioning it, calculus is a big part of physics. ugghhh</p>
<p>I don’t know, there might not be much sarcasm in that post - my teacher rarely used calculus in our class!</p>
<p>I’m sick and tired of studying for this test, but I’m about to force myself to do another practice test.</p>
<p>
That genuinely sucks. But I wasn’t being sarcastic ahah, Tariq really did answer one of my questions and my teacher really didn’t apply calculus… it was just a coincidence that I also said d’oh :P</p>
<p>0987654321… I KNOW RIGHT</p>
<p>Oh, my bad then 
sorry if I seemed bitter or something; didn’t mean to :P</p>
<p>i’m taking the mechanics part tomorrow… to be honest going into the test i’m going to fail. my teacher hasn’t done squat, hell he even fixes all our grades to make it look like he does a good job. i will be very surprised if a pass</p>
<ol>
<li> Two people are initially standing still on frictionless ice. They push on each other so that one person, of mass 120 kg, moves to the left at 2 m/s, while the other person, of mass 80 kg, moves to the right at 3 m/s. What is the velocity of their center of mass?
<a href=“A”>B</a> Zero <a href=“B”>/B</a> 0.5 m/s to the left<br>
(C) 1 m/s to the right<br>
(D) 2.4 m/s to the left<br>
(E) 2.5 m/s to the right</li>
</ol>
<p>Help?? >_< (I suck at Center of Mass questions … )</p>
<p>just think of it after one second. P1 will be 2 meters left, P2 will be 3 meters right.
(80x3)+(120X-2)=0 the center of mass is in the same place aka its not moving so v=0.</p>
<p>Xcom=M1R1+M2R2…</p>
<p>A moon of mass m orbits a planet of mass 49m in an elliptical orbit. When the moon is at point A its distance from the center of the planet is rsubA and its speed is vsub0. When the moon is at point B, its speed is 5vsub0.</p>
<p>When the moon is at a point A, the distance from the moon to the center of mass of the planet-moon system is most nearly
A) 1/50 rsubA<br>
B) 1/7 rsubA<br>
C) 1/2 rsubA<br>
D) 6/7 rsubA<br>
*E) 49/50 rsubA</p>
<p>When the moon is at point B, the distance from the moon to the center of the planet is most nearly
A) 1/25 rsubA<br>
*B) 1/5 rsubA<br>
C) 1/sqrt5 rsubA<br>
D) rsubA<br>
E) sqrt5 rsubA</p>
<p>@ansar: when there’s no external force acting on the system, there’s no acceleration for the system as a whole so the center of mass of the system can’t move anywhere (if the two person are staying still initially).</p>
<p>Two identical stars, a fixed distance D apart, revolve in a circle about their mutual center of mass. Each star has a mass M and speed V. G is the universal gravitational constant. Which of the following is a correct relationship among these quantities?</p>
<p>A) v^2=GM/D
B) v^2=GM/2D
C) v^2=GM/2(D^2)
D) v^2=MGD
E) v^2=2G(M^2)/D</p>
<ol>
<li> A particle moves in the xy plane with coordinates given by
x = A cos wt and y = A sin wt,
where A = 1.5 meters and w = 2.0 radians per second. What is the magnitude of the particle’s acceleration?
(A) Zero (B) 1.3 m/(s^2) (C) 3.0 m/(s^2) (D) 4.5 m/(s^2) <a href=“E”>B</a> 6.0 m/(s^2)**</li>
</ol>
<p>ansar, when you take the second derivative of x and y, you notice that x has cosine in it and y has sine in it.
Set t to a whole number, so sine(wt) = 0, so the magnitude would just be the x component of acceleration.</p>
<p>^^^0987654321, what is the answer to that question? (A?)</p>
<p>I have a bad feeling for tomorrow :/.</p>
<p>it’s b…i have no idea why tho</p>
<p>Thanks agpyn3w</p>
<p>@0987654321
It’s b because you have to remember that the force between the two is Gm^2/D, but the F=mv^2/r and the r in this case is 1/2D. So Gm^2/(D)^2=mv^2/(D/2). Thus Gm/2D=v^2</p>
<p>Or think of it this way:</p>
<p>X<em>com = SUM(M</em>i * x<em>i) / M</em>tot</p>
<p>Taking the derivative with respect to time we get…</p>
<p>V<em>com = SUM(M</em>i * v<em>i) / M</em>tot</p>
<p>So</p>
<p>V_com = [120<em>(-2) + 80</em>3] / 200 = 0</p>
<p>EDIT: Whoops… hit refresh and forgot to hit next page… this is reply to the last question on page 10!</p>