Getting wrong answer in an (angular) impulse momentum problem

AI Thread Summary
The discussion centers on resolving an impulse momentum problem in an angular context, where the initial answer of 12mu/(3+cos2α) was consistently obtained. The participant struggled with their approach, which involved treating the rod arrangement as rigid, but was advised to consider it with a free joint instead. This adjustment led to the correct solution. The participant acknowledged their oversight regarding forum rules and expressed gratitude for the assistance received. The conversation highlights the importance of correctly interpreting physical arrangements in problem-solving.
Divya
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Homework Statement
Three masses m, 2m and 3m are connected by two massless and rigid rods
of length l which are currently perpendicular to each other, as shown in the
figure. If the masses initially travel at velocity u towards a vertical wall and
mass m undergoes a collision with the wall, determine the impulse delivered
by the wall to mass m if the final horizontal velocity of mass m is zero. There
is no friction between the wall and mass m. Assume that the tensions in the
rods are strictly longitudinal (because they are massless).
Relevant Equations
impulse = change in momentum
angular impulse = change in angular momentum
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WhatsApp Image 2022-08-20 at 11.18.12 AM.jpeg

I have tried this same approach three times and I got the same answer. I can't figure out what's wrong. Btw answer is 12mu/(3+cos2α)
And yes, sorry for my shitty handwriting. If you can't understand the reasoning behind any step then please let me know.
 
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Divya said:
sorry for my shitty handwriting
Forum rules state that images are for diagrams and textbook extracts. Please type in your working, preferably in LaTeX.
Also, it helps if you provide some explanation of your approach and define all variables.

It looks to me as though you are taking the rod arrangement as rigid. It only says initially a right angle. I would take it as having a free joint.

In my experience, it rarely gains anything to bother finding the mass centre of such an assemblage anyway.
 
I am really sorry for not seeing the rules of forum before posting ( I am new here). Also thanks for your help. The 'free joint' interpretation gave the right answer.
 
Divya said:
I am really sorry for not seeing the rules of forum before posting ( I am new here). Also thanks for your help. The 'free joint' interpretation gave the right answer.
well done
 
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