2003 Physics C Mechanics FRQ

Test Preparation AP Test Preparation
1

<p>I’m stumped on the third FRQ of the 2003 exam. Part B, subpoint III of the third question asks to derive the equation for the velocity of the projectile as it leaves the cup. In the answers posted on CB, there is one step where the velocity of the counterweight bucket is equal to v(sub x)/6. Can someone explain how they came up with this relationship? Is it because 2 / 12 = 1/6, which is the ratio of the distance from the center of mass of the counterweight bucket to the pivot to the distance from the center of mass of the projectile bucket to the pivot?</p>

<p>Many thanks for your help! :)</p>

2

<p>Because they are attached, the angular velocity of both ends of the catapult is equal. Tangental velocity = radius * angular velocity. For the counterweight bucket, that’s v(cw) = (2 m)<em>w (pretend that w is an omega), and for the other side, its v = (12 m)</em>w. Therefore v(cw) = v/6.</p>

<p>Hope that helps.</p>

3

<p>ah, that makes sense. Thanks!</p>