A crate of mass \(m\) sits atop a frictionless ramp that has a height and length \(L\). The ramp has mass \(2 m\), which is concentrated at the lower left corner, and it sits on a frictionless ice pond. The crate is released and slides down the ramp. By the instant that the crate reaches the bottom, how far has the ramp moved and in what direction? (A) It hasn't moved. (B) It has moved to the left a distance \(L / 3\). (C) It has moved to the right a distance \(L / 3\). (D) It has moved to the left a distance \(L / 2\). (E) It has moved to the right a distance \(L / 2\).

Initially, both the crate and ramp are stationary, so the total horizontal momentum of the system is zero.

When the crate reaches the bottom of the ramp, its velocity vector will be going down the ramp at an angle. The angle is given by \(\theta = tan^{-1}\frac{h}{L}\), where \(h\) is the height of the ramp and \(L\) is the length of the ramp. We can use the conservation of mechanical energy principle to calculate the velocity \(v_c\) of the crate when it reaches the bottom: \(\frac{1}{2}m v_c^2=mgh\) Solving for \(v_c\): \(v_c = \sqrt{2gh}\) Now, let's find the horizontal component of the velocity vector of the crate, \(v_{cx}\). We can do this using trigonometry: \(v_{cx} = v_c \cos(\theta) = \sqrt{2gh} \cos(\theta)\) The momentum of the crate in the horizontal direction is given by: \(p_{cx} = m v_{cx} = m\sqrt{2gh} \cos(\theta)\)

Since the total horizontal momentum of the system must remain constant and the initial momentum is zero, the horizontal momentum of the ramp \(p_{rx}\) should be equal and opposite to that of the crate: \(p_{rx} = -p_{cx} = - m\sqrt{2gh} \cos(\theta)\)

To find the horizontal displacement of the ramp, we should first calculate the horizontal velocity of the ramp, \(v_{rx}\) using the momentum formula: \(v_{rx} = \frac{p_{rx}}{m_{ramp}} = \frac{-m\sqrt{2gh} \cos(\theta)}{2 m}\) To determine the time it takes for the crate to reach the bottom, we will use the time formula applied to the crate: \(t = \frac{d}{v_c\sin(\theta)} = \frac{L}{\sqrt{2gh}\sin(\theta)}\) The horizontal displacement \(x\) experienced by the ramp can be calculated by multiplying the horizontal velocity of the ramp by the above time: \(x = v_{rx}\times t = \frac{-m\sqrt{2gh} \cos(\theta)}{2 m}\times \frac{L}{\sqrt{2gh}\sin(\theta)}\) \(x = - \frac{L}{2}\cot(\theta)\) since \(\cot(\theta)=\frac{adjacent}{opposite}=\frac{L}{h}\), \(x= -L\times\frac{L}{2h}\) In our problem, the ratio of height and the length of the ramp is \(L:2L\). Therefore, \(\frac{L}{2h} = \frac{L}{3}\). Plugging this value into the previous equation, we get: \(x = -\frac{L}{3}\) So, the final answer is: (B) The ramp has moved to the left a distance \(L/3\).