The script for an action movie calls for a small race car (of mass 1500 kg and l
ID: 2233569 • Letter: T
Question
The script for an action movie calls for a small race car (of mass 1500 kg and length 3.0 m) to accelerate along a flat-top boat (of mass 3500 kg and length 20 m), from one end to the other, where the car will then jump the gap between the boat and a somewhat lower dock. You are the technical advisor for the movie. The boat will initially touch the dock, as in the figure below. The boat can slide through the water without significant resistance and both the car and the boat can be approximated as uniform in their mass distribution. Determine what the width of the gap will be just as the car is about to make the jump,Explanation / Answer
The idea is that the center of mass of the car-boat systemwill stay in the same place, relative to the dock. Calculate theposition of the cm for the beginning, when the car is at the rightside, and the end, when the car is at the left side. Use the centerof the boat as x = 0.beginning: the formula for center of mass, x of cm = (1 / total mass) ( mboat xboat + mcar x car ) = = (1 / 6000 ) ( 4500 * 0 + 1500 * 7) = 1.75 meters to the right of the center of the boat . (note: the position of the center of the car is 7 meters from the center of the boat because half theboat is 8.5 meters and half the car is 1.5 meters,therefore xcar is 8.5 -1.5 = 7) using the exact same calculation, the center of mass at the end is 1.75 meters to the left ofthe center of the boat! So the center of mass moves 1.75 + 1.75 = 3.50 meters relative to the boat... but it stays in the same place relative to the dock. Therefore, the boat must move 3.50 meters away from the dock. So the gap as the car is about to jump will be 3.50 meters The idea is that the center of mass of the car-boat systemwill stay in the same place, relative to the dock. Calculate theposition of the cm for the beginning, when the car is at the rightside, and the end, when the car is at the left side. Use the centerof the boat as x = 0.
beginning: the formula for center of mass, x of cm = (1 / total mass) ( mboat xboat + mcar x car ) = = (1 / 6000 ) ( 4500 * 0 + 1500 * 7) = 1.75 meters to the right of the center of the boat . (note: the position of the center of the car is 7 meters from the center of the boat because half theboat is 8.5 meters and half the car is 1.5 meters,therefore xcar is 8.5 -1.5 = 7) using the exact same calculation, the center of mass at the end is 1.75 meters to the left ofthe center of the boat! So the center of mass moves 1.75 + 1.75 = 3.50 meters relative to the boat... but it stays in the same place relative to the dock. Therefore, the boat must move 3.50 meters away from the dock. So the gap as the car is about to jump will be 3.50 meters