<p>Hey guys, please take a look at:</p>
<p>Those are a few questions which I came across that I did not know how to answer/disagreed with the answer. The correct answer is highlighted in yellow. In some questions, the Bold text is work that I did which arrived at a different answer.</p>
<p>Thanks for the help.</p>
<p>*) first one is easy: force is -dU(x)/dt so you get -12x+4@x=3; so -36+4 = -32 so B</p>
<p>*)conservation of mechanical energy----> Ki+Ui=Kf+Uf</p>
<p>*)don't try to overthink things; just stick to the formula and you'll be fine....</p>
<p><em>)angular momenmtum (L) = m</em>w<em>r^2; where w=angular speed which is v/r, and r is the distance; so in this case r = a; so m</em>a^2*v/a; so simplify; mva</p>
<p>*) you have to split up the interval where the a=0; and whhen a-not aero, and v is not constant; you did all at the same time, and hence got the wrong answer; split up the interval from 0 to 1; and then from 1 to 11</p>
<p>*) conservation of momentum all the way</p>
<ul>
<li><p>Thanks, I had forgotten that definition of Force/P.E!</p></li>
<li><p>The potential energy in the system is in the spring gun, but how does one account for that?</p></li>
</ul>
<p>*I'm still not convinced--the wording asks for the time it takes to go back to the initial position, x=0 which is crossed once during a single cycle, right? I guess it is a matter of semantics.</p>
<p>*I am not familiar with L for objects not undergoing rotation. Doesn't r change as the particle approaches (0,a) ? I get answer choice D.</p>
<ul>
<li>I don't quite understand. I split the interval between 0,1 and 1,11. So at time 1, the particle is at position 24 meters, right? Then from t=1 to t=11 (10 seconds) the particle is decelerated at a rate of 6m/s/s. so</li>
</ul>
<p>pos.final = pos.init + .5a(t-1)^2</p>
<ul>
<li>D'Oh! I read the problem to mean the center of mass of the system as one mass approaches the other. I didn't perceive the problem as a collision. I was doing m1x1+m2x2 / (m1+m2) = Center of Mass and then doing implicit differentiation to get v terms out.</li>
</ul>
<p>Looking back, I left it out of my previous post: Thanks-a-million, vader!</p>
<p>^no prob. :)
I'll try to clear up the doubts you still have on the above questions once I get back from school today...I don't have much time to type right now; my bus is here....<em>runs for bus</em> :)</p>
<p>For question 28, when you calculate the new position at t=11 while a=-6, don't forget to include the initial velocity at t=1. When you throw a ball up in the air, the position at some later time depends on how fast you threw it initially.</p>
<p>The formula you want is pos. final=pos. init + 24<em>(t-1) + .5</em>a*(t-1)^2 ...</p>
<p>The cycle that takes time T for Q24 is: highest - middle - lowest - middle - highest. In general T is the time for the object to come back to the same place with the same velocity (in this case, at the top with v=0).</p>
<p>fignewton, thanks for the clarification, I included the initial velocity and got the correct answer. I was being dumb and was thinking that velocity was only "applied" between time t=0 and t=1, but dur the velocity isn't a force, so it continues after t=1.</p>
<p>I understand the concept of the cycle in SHM, I just think that maybe the question should be worded better because it asks for the position.</p>
<p>All that's left is 8 and 26</p>
<p>Number 8 is a bit of algebra. Use conservation of momentum to find the gun velocity in terms of the projectile velocity. Sub that into 1/2<em>(m2-m1)</em>v(gun)^2, look for a 1/2<em>m1</em>v^2 and replace with K ...</p>
<p>Hint for 26: constant speed means ... Fnet=0! For a point particle, that means angular momentum is constant, so it is the same at x=x0 as it is at x=0.</p>