Jumping from a moving wagon momentum problem -- find Buffy's weight

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Buffy's weight cannot be directly calculated without a clear understanding of the physics involved, specifically the relationship between mass and weight. The discussion highlights the confusion between mass (measured in grams or kilograms) and weight, which is the force due to gravity. The scenario describes Buffy jumping from a moving wagon, affecting both her and the wagon's velocities. To solve for Buffy's weight, the correct physics principles and formulas must be applied, but the original poster has not specified which calculations were attempted. Clarifying the approach and equations used is essential for resolving the problem.
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Homework Statement
Buffy is rolling along in her 11.8 kg wagon at 3.8 m/s (in the positive direction) when she jumps off the back. She continues to move forward at 1.8 m/s relative to the ground. This causes her wagon to go speeding forward at 8.87 m/s relative to the ground. How much does Buffy weigh?
Relevant Equations
n/a
I have attempted to plug in grams, kiliograms, and even pounds but it was wrong.
 
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Jujubee37 said:
Homework Statement:: Buffy is rolling along in her 11.8 kg wagon at 3.8 m/s (in the positive direction) when she jumps off the back. She continues to move forward at 1.8 m/s relative to the ground. This causes her wagon to go speeding forward at 8.87 m/s relative to the ground. How much does Buffy weigh?
Relevant Equations:: n/a

I have attempted to plug in grams, kiliograms, and even pounds but it was wrong.
Attempted to plug in into what? We do not know what formula or even concept you are trying to use. As such it is impossible to pinpoint where you are going wrong.
 
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My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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