A motorcycle stunt rider plans to start from one end of a railroad flatcar, acce
ID: 2009480 • Letter: A
Question
A motorcycle stunt rider plans to start from one end of a railroad flatcar, accelerate toward the other end of the car, and jump from the flatcar to a platform. The motorcycle and rider have a mass of 381 kg and a length of 2.37 m. The flatcar has a mass of 2010 kg and a length of 21.7 m. Assume that there is negligible friction between the flatcar's wheels and the rails and that the motorcycle and rider can move through the air with negligible resistance. The flatcar is initially touching the platform. The promoters of the event have asked you how far the flatcar will be from the platform when the stunt rider reaches the end of the flatcar. What is your answer?
A motorcycle stunt rider plans to start from one end of a railroad flatcar, accelerate toward the other end of the car, and jump from the flatcar to a platform. The motorcycle and rider have a mass of 381 kg and a length of 2.37 m. The flatcar has a mass of 2010 kg and a length of 21.7 m. Assume that there is negligible friction between the flatcar's wheels and the rails and that the motorcycle and rider can move through the air with negligible resistance. The flatcar is initially touching the platform. The promoters of the event have asked you how far the flatcar will be from the platform when the stunt rider reaches the end of the flatcar. What is your answer?Explanation / Answer
The curious thing is that no speed or acceleration is given. This should get us thinking that it must not matter. If the rider accelerates quickly, there will be a big force on the flatcar, but for a small amount of time. If he accelerates slowly, there will be a small force on the flatcar, but it will act for a longer time. The center of mass of the flatcar motorcycle system stays where it is in the beginning of the problem. Where is the CoM to start? The center of mass of the flatcar is 21.7m/2 from the edge The CoM of the motorcyle is (21.7 -1/2*2.37)m from the edge [21.7m/2*2010 + (21.7 - 1/2*2.37)m*381]/(381+2010) = 12.39m from the platform. At the end of the problem the CoM is still 12.39m from the platform. Now we need to find where the flatcar is when motorcycle is at end of the flatcar, but the flatcar has moved, so the motorcycle is also this far from the platform. Again, there is algebra involved, which adds some spice to the problem. Let's call the distance from the platform 'x.' 12.39m = [2010kg*(21.7m/2 + x) + 381kg*(1/2*2.37 + x)]/(2010kg +381kg) [12.39*(2010+381) - 2010*21.7/2 - 381*2.37/2]/(2010+381) = x = 3.08m Check our work: is [2010*(3.08+10.85) + 381*(3.08 + 2.37/2)]/(2010+381) ?=? [2010*10.85 + 381*(21.7 -2.37/2)]/(2010+381) !Yes! both equal 12.39m Side note: My favorite physics professor once said that when the space shuttle (except he probably said Saturn V) takes off, the center of mass stays on the launch pad. Something to think about.
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