Collision Physics Homework: Calculating Speed of Center of Mass for Cart A and B

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SUMMARY

The discussion centers on calculating the speed of the center of mass for two carts involved in a collision. Cart A, with a mass of 0.20 kg, travels at 3.0 m/s, while Cart B, with a mass of 0.40 kg, moves at 2.0 m/s in the opposite direction. The correct calculation for the center of mass speed is 2.3 m/s, derived from the equation Vcom = [(0.2)(3) + (0.4)(-2)] / (0.2 + 0.4). The confusion arises from misinterpretation of the collision dynamics, particularly regarding the direction of motion.

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Homework Statement



Cart A, with a mass of 0.20 kg, travels on a horizontal air track at 3.0m/s and hits cart B,
which has a mass of 0.40 kg and is initially traveling away from A at 2.0m/s. After the collision
the center of mass of the two cart system has a speed of:
A. zero
B. 0.33m/s
C. 2.3m/s
D. 2.5m/s
E. 5.0m/s

Homework Equations





The Attempt at a Solution



Vcom = [(0.2)(3)+(0.4)(2)]/(0.2+0.4) = 2.3333

but ans says its 0.33 m/s? why?
 
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It seems to me that you might be mis-interpreting the question. If your description of the problem is accurate, then it's easy to see that the answer is not 0.33 m/s. If A bumps into B from behind, the speed of both must be faster than the original speed of B. However, if A and B collide head-on, their center of mass will move at 0.33 m/s.

If it is the case that B is traveling away from A as you say, then the answer should be 2.3 m/s.
 

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