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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