Two blocks \(A\) and \(B\) of same mass \(M\) are connected with each other with an ideal string of length \(2l\) passing over an ideal pulley. The block \(A\) is connected to a light pan \(C\) with an ideal string as shown in figure. A particle of mass \(\frac{M}{2}\) is dropped on pan from height \(\frac{l}{2}\) as shown. If collision between particle and pan is plastic, acceleration of \(B\) just after the collision is?

Ans is \[\frac{g}{9}\].

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## Comments

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TopNewest@Raghav Vaidyanathan @Tanishq Varshney @Kushal Patankar Help!

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@satvik pandey i have made a note on this problem inviting views of other brilliant members(bc now even i am confused :/ )

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Thanks for making it! :)

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Why wasn't \(T_1\) impulsive.😓

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Can you please post the solution to Angular width of Refracted Beam!

Raghav Vaidyanathan

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@Raghav Vaidyanathan

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ok

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@Nishant Rai

Although your answer is correct, the FBDs that you have drawn are not. You have missed out the gravitational force on \(A\). Due to this reason, you have skipped using \(T_1 -Mg=Ma\)

In reality, at the moment the mass strikes the pan:

\(T_1 = Mg + M\frac {v^2}{l}\)

And hence: \(T_1 -Mg=Ma=M\frac {v^2}{l}\)

Otherwise, you can also use constraint relations to get to same result.

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@satvik pandey

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Raghav Vaidyanathan

i have done the same thing, (this solution is not mine, this question came in mock paper of some coaching institute, they have provided this solution)

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Okay, but they aren't fully wrong, we can get everything without FBD itself...

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I have a confusion. I think it is assumed that just after the collision block A performs a circular motion. But as soon as the block a A gets a tangential velocity it would have centripetal as well as a radial acceleration. How can we ignore that?

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I think you guys would like to see this discussion.

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My Answer is a little different..

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