Question

You stand inside of a small cart attached to a heavy-duty spring, the spring is compressed and released, and you shake back and forth, attempting to maintain your balance. Note that there is also a sandbag in the cart with you. What effect does dropping the sandbag out of the cart at the equilibrium position have on the maximum speed of the cart?

Can someone please explain why it has no effect on the max speed of the cart?

Answer 1

If you drop the sandbag off the side, at the very moment you release it, it has the same velocity as the cart.

Hence when you drop it, it takes kinetic energy with it and the amount it takes is exactly proportional to its mass. Hence the kinetic energy of you and the cart is lowered by exactly the right amount that will not change your speed.

In mathematical language, let m be the mass of the sandbag and M the mass the cart and everything else in it; from now on, I'll just call the cart and everything else the cart.

Suppose at the equilibrium point of the spring, the cart and sandbag have kinetic energy E. Then by definition of kinetic energy,
\[ E = \frac{1}{2}(M+m)v_{original}^2 \]
where v is the speed.

Now, if you drop the sandbag over the side, at the moment it also has speed v, so its kinetic energy is
\[ KE_{sandbag} = \frac{1}{2}mv_{original}^2 \]
Therefore the remaining kinetic energy for the cart is
\[ E - KE_{sandbag} = \frac{1}{2}Mv_{original}^2 \]
Carefully note that we derived this expression by subtracting one energy from the other; we did not calculate it from the definition of Kinetic energy.

But let us now do so. Then the kinetic energy of the cart is
\[ KE = \frac{1}{2}Mv_{cart}^2 \]

These two expressions must be equal
\[ \frac{1}{2}Mv_{cart}^2 = \frac{1}{2}Mv_{original}^2 \]
therefore the speed of the cart must be exactly the same as the speed it was before the sandbag was dropped.