# Atwood machine problem [closed]

Sorry for the bad drawing, but I hope that this will help you get a hold of the problem.

Consider an Atwood Machine with a total of two blocks, a mass less pulley, ideal string. One block rests on the floor, while the other one is at a height (H). Now, the string near the block that rests on the surface is slack. So, the other block falls freely, and later induces a jerk in the other block. How do I calculate the initial velocity of the two block just after the string is taut.

My approach was: 1) Calculate the velocity of the block in motion (initially) at the point when the is ALMOST taut. 2) Now conserve mechanical energy b/w the point where string was almost taut but impulse wasnt generated, and the point where impulse was generated, and the second block had JUST started motion.

But, my book says, I should conserve Linear Momentum between the same two points. I think this is wrong, because, The string that holds the pulley in place, will have an impulsive tension as soon as the impulse is generated in the string that joins the two masses.

According to you, what is right, and why?

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## closed as off-topic by Brandon Enright, John Rennie, JamalS, Kyle Kanos, jinaweeMay 22 '14 at 13:11

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representing your problem with figure might help us to understand it better – Sigma Nov 21 '13 at 15:02

The answer is that both mechanical energy conservation and linear momentum conservation are not valid.

Linear momentum can't be conserved because there will be an impulse exerted on the system by the pulley. This impulse will be in the upward direction because notice how the tensions in the string on the pulley are downward, so the pulley will apply a force upwards to counteract this impulse.

Mechanical energy can't be conserved because we don't know whether all the impulses are conservative or not.

What you can do is consider an impulse $J$ on both the blocks upwards, and use change in momentum equals the impulse applied.

Thus you have $$J=m_1 v_{\mathrm{final}}$$ for the first block, and $$J=m_2 v_{\mathrm{final}}- m_2 v_{\mathrm{initial}}$$ for the second free falling block.

$v_{\mathrm{final}}$ will be the same for both blocks because they are constrained by the string. With these two equations, you can find $v_{\mathrm{final}}$

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Ok, just 1 doubt, why can you not conserve energy when the point of application of the impulse (pulley) has zero displacement in the time $dt$? I mean, therefore work done will be zero on the system... – Saurabh Raje Nov 22 '13 at 3:55
The impulse might have deformed the pulley a bit(compressed it), like an inelastic collision. That deformation will produce heat, which leads to loss of mechanical energy. – udiboy1209 Nov 22 '13 at 7:33