Determining v from momentum / impulse

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The discussion centers on determining the final momentum of a hockey stick after it strikes a ball, with the conclusion that the correct answer is (a). Initial assumptions about using conservation of momentum are challenged due to the influence of external forces like air resistance. It is clarified that for option (b) to hold true, the initial momentum of the hockey stick would also need to be known. The momentum change of the stick is linked to the impulse exerted on the ball. Ultimately, the final momentum of the hockey stick is equal to the initial momentum of the ball, emphasizing the importance of understanding these dynamics.
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Homework Statement
A ball, initially at rest, is struck by a hockey stick. It leaves the hockey stick at speed v. Which quantity, together with the mass of the ball, can be used to determine v?
a) the impulse of the force on the ball
b) the final momentum of the hockey stick
Relevant Equations
momentum, impulse
Answer is (a)

I thought it would be (b) due to conservation of momentum - so final momentum of the hockey stick is equal to the initial momentum of the ball. I assume this isn't correct because there are other external forces acting (air resistance?) Is that sound?
 
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For (b) to be correct, you will also need the initial momentum of the hockey stick. Then the momentum change of the stick will be equal (in magnitude) to the impulse of the force on the ball which brings us to (a).
 
g9WfI said:
final momentum of the hockey stick is equal to the initial momentum of the ball
Further to @kuruman's point, that wouldn't help even if it were true. You know the initial momentum of the ball; you need to find its final momentum.
 
Thread 'Chain falling out of a horizontal tube onto a table'
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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