Momentum Conservation: Bullet enters a block

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The discussion centers on the application of conservation of momentum in a scenario involving a bullet entering a block. The initial confusion arises from the need to determine the variable V, which is dependent on height h. A participant clarifies that the equation involving l, h, and a is derived from Pythagoras' theorem. By expanding the equation and simplifying, the relationship becomes clearer. The explanation ultimately resolves the confusion regarding the variables involved.
Shreya
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Homework Statement
I am not able to solve the second part of the question.
Please refer the image below.
V refers to the final velocity after collision, v is the initial velocity of bullet. L is the length of string and h is the height that the block rises to.
Relevant Equations
Conservation of Momentum
I can understand that using conservation of momentum, we can find v. But we need V for that. The equation for V involves h and so we need h. But I am not able to comprehend the equation involving l,h and a. The question doesn't specify what a is.
Please be kind to help
 

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That’s just Pythagoras’ theorem.
$$
l^2 = (l-h)^2 + a^2
$$
Expand the (l-h) square and cancel ##l^2## on both sides.
 
Last edited:
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Orodruin said:
That’s just Pythagoras’ theorem.
Thanks @Orodruin! I get it now.
 
Last edited:
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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