Hi, I need help with a question on a practice exam. I got the first two questions, but I have no clue on where to start as to figuring out how far the cart has moved. Anyone have any suggestions? Thanks!

A cart of mass M = 9 kg rolls without friction on a horizontal surface. It is attached through a freely pivoting initially-horizontal massless rod of length L to a ball of mass m = 3 kg. The system is initially at rest when the ball is released. The pendulum swings down and to the left, and at the bottom of its swing the ball is observed to have a velocity of 3.5 m/sec to the left.

http://online.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys211/oldexams/exam2/fa10/fig3.gif

What is the speed of the cart when the ball is at the bottom? 1.17 m/s
What is the length L of the pendulum? 83 m
How far to the right has the cart moved, when the ball is at the bottom??

Here are a couple different ways to think about it:

1) You figured out that the cart will be moving to the right at 1/3 the speed of the pendulum moving to the left (because of the 3kg/9kg mass ratio). But there was nothing magic about the fact that the pendulum happened to be at the bottom of its arc. What if the pendulum had only dropped 45 degrees? The horizontal component of its speed would be smaller than at 90 degrees, but the cart would still be traveling to the right at 1/3 of that speed (regardless of what it was). Hence, even though the leftward speed of the pendulum changes throughout it's flight, the rightward speed of the cart is always 1/3 as much. Hence, no matter how the speed of the pendulum varied, the cart traveled to the right only 1/3 as far as the pendulum traveled left.

2) Keep in mind that the horizontal center of mass of the system will not have changed. Since the 3 kg mass has moved 83 cm to the left, how far has the 9 kg cart moved to the right to compensate? 1/3 that far, right?

The answer according to the practice test is L/4 or 1/4 of the length of the pendulum. I see how L/3 is possible but how does L/4 work?

Oops. Sorry about that. :o

I forgot that since that pendulum is tied to the cart, the pendulum's horizontal travel isn't L, it's L-x, where x is the distance that the cart moved.

Let's try that again:

\Delta x_{cart}=\frac{\Delta x_{pendulum}}{3}

\Delta x_{pendulum}=L-\Delta x_{cart}

Now substituting the first equation into the second:

\Delta x_{cart}=\frac{L-\Delta x_{cart}}{3}

\Delta x_{cart}=\frac{L}{4}