If a 1305 kg train car on a one dimensional track, moving at 4 m/s north, hits and bounces off a second train car with a mass of 3172 kg , moving at 10 m/s south, find the magnitude of the final speed of the first car (the second car will have a different speed, most probably in the opposite direction).
If a 1305 kg train car on a one dimensional track, moving at 4 m/s north, hits and bounces off a second train car with a mass of 3172 kg , moving at 10 m/s south, find the magnitude of the final speed of the first car (the second car will have a different speed, most probably in the opposite direction).
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![If a 1305 kg train car on a one-dimensional track, moving at 4 m/s north, hits and bounces off a second train car with a mass of 3172 kg, moving at 10 m/s south, find the magnitude of the final speed of the first car (the second car will have a different speed, most probably in the opposite direction).
**Hint:** One-dimensional problem, call north positive. Use the "hit and bounce" or elastic formula that goes with car #1
\[
v'_1 = \frac{m_1 - m_2}{m_1 + m_2} v_1 + \frac{2m_2}{m_1 + m_2} v_2
\]
\[
v'_2 = \frac{2m_1}{m_1 + m_2} v_1 - \frac{m_1 - m_2}{m_1 + m_2} v_2
\]
Also, a positive answer would indicate motion to the north and a negative answer would show southward motion - leave positive/negative in your answer.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbbba0909-7b4f-4e80-8e24-06dfda2d061e%2Ff9bc2507-6617-41ae-bd6a-bf18eb294f29%2Ff9t7rje_processed.png&w=3840&q=75)
Transcribed Image Text:If a 1305 kg train car on a one-dimensional track, moving at 4 m/s north, hits and bounces off a second train car with a mass of 3172 kg, moving at 10 m/s south, find the magnitude of the final speed of the first car (the second car will have a different speed, most probably in the opposite direction).
**Hint:** One-dimensional problem, call north positive. Use the "hit and bounce" or elastic formula that goes with car #1
\[
v'_1 = \frac{m_1 - m_2}{m_1 + m_2} v_1 + \frac{2m_2}{m_1 + m_2} v_2
\]
\[
v'_2 = \frac{2m_1}{m_1 + m_2} v_1 - \frac{m_1 - m_2}{m_1 + m_2} v_2
\]
Also, a positive answer would indicate motion to the north and a negative answer would show southward motion - leave positive/negative in your answer.
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