<li><p>An ice skater is spinning about a vertical axis with arms fully extended. If the arms are pulled in closer to the body, in which of the following ways are the angular momentum and kinetic energy of the skater affected?
Angular Momentum Kinetic Energy
(A) Increases Increases
(B) Increases Remains Constant
(C) Remains Constant Increases
(D) Remains Constant Remains Constant
(E) Decreases Remains Constant</p></li>
<li><p>If a particle moves in such a way that its position x is described as a function of time t by x = t^3/2, then its kinetic energy is propor¬tional to
(A) t^2 (B) t^3/2 (C) t (D) t^1/2 (E) t° (i.e., kinetic energy is constant)</p></li>
</ol>
<p>6 - C.</p>
<p>No net torque is done on the object, so that means the angular momentum must remain constant. That leaves only C and D. If a person’s arms are pulled closer, speed will increase. Since KE is a function of velocity, since velocity increased, and there’s no change in mass, KE must increase as well.</p>
<ol>
<li>KE = (mv^2)/2</li>
</ol>
<p>I can’t tell if you are doing x = t^(3/2) or x = (t^3)/2. So I’ll solve it both ways right quick.</p>
<p>Using the x = t^(3/2): v = dx/dt, therefore v = (x)'. Thus, v = [3t^(1/2)]/2.
As such, KE is proportional to v^2, so it is proportional to [9t^(1)]/4. You can cut out the coefficient (9/4), so it is proportional to t^1. Therefore the answer is C.</p>
<p>Using the x = (t^3)/2: v = dx/dt, therefore v = (x)'. Thus, v = [3t^2]/2.
As such, KE is proportional to v^2, so it is proportional to [9t^4]/4. You can cut out the coefficient (9/4), so it is proportional to t^4. Since this is not an answer choice, I’m going to go with the answer from above - C.</p>
<p>The answer to 6 is C, but it is a little more subtle. The angular momentum is (moment of inertia) * (angular speed), or L=I*w, and L remains constant. When the arms are pulled in, I decreases by some factor, so w must increase by the same factor for L to stay the same. Now, KE = 1/2 * I * w^2 … I goes down, but w goes up by the same factor and since w is squared, KE goes up.</p>
<p>Can someone explain problem number two at the following url please:</p>
<p><a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;
<p>Thanks!</p>
<ol>
<li><pre><code>A ball is dropped from a height of 10 meters onto a hard surface so that the collision at the surface may be assumed elastic. Under such conditions the motion of the ball is
</code></pre>
<p>(A) simple harmonic with a period of about 1.4 s<br>
(B) simple harmonic with a period of about 2.8 s
(C) simple harmonic with an amplitude of 5 m<br>
(D) periodic with a period of about 2.8 s but not simple harmonic
(E) motion with constant momentum</p></li>
<li><p>A bowling ball of mass M and radius R. whose moment of inertia about its center is (2/5)MR^2, rolls without slipping along a level surface at speed v. The maximum vertical height to which it can roll if it ascends an incline is? </p></li>
<li><p>A particle moves in a circle in such a way that the x ¬and y coordinates of its motion are given in meters as functions of time t in seconds by:
x = 5cos(3t) y = 5 sin (3t)</p></li>
<li><pre><code>Which of the following is true of the speed of the particle?
</code></pre>
<p>(A) It is always equal to 5 m/s. (B) It is always equal to 15 m/s.
(C) It oscillates between 0 and 5 m/s. (D) It oscillates between 0 and 15 m/s.
(E) It oscillates between 5 and 15 m/s.</p></li>
<li><pre><code>Two objects of equal mass hang from independent springs of unequal spring constant and oscillate up and down. The spring of greater spring constant must have the
</code></pre>
<p>(A) smaller amplitude of oscillation (B) larger amplitude of oscillation
(C) shorter period of oscillation (D) longer period of oscillation
(E) lower frequency of oscillation</p></li>
</ol>