A block of mass m is placed on a triangular block of mass M, which in turn is placed on a horizontal

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A block of mass m is placed on a triangular block of
mass M, which in turn is placed on a horizontal surface
as shown in figure (9-E21). Assuming frictionless
surfaces find the velocity of the triangular block when
the smaller block reaches the bottom end.

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It can also be done by conserving COM acceleration.😊

neelamsingh
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I have solved using Kinematics same answer

Thanks 😊

mayankkr
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How you have taken velocity of m v2cos - v1, from which frame you are solving question

RyanO
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Bhaiya aapko kaise pata chala ki kis frame se velocity di hai please bataiye

ayushsinghkaintura
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Bro just take the velocity wrt to ground at the start then answer will be simpler and sin alpha will not be used

gamingzeraora
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Sir this question is from which book or exam? Please answer 🙏

purusood
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Sir can't we directly say that V2 (in block's frame) is root(2gh)

aqstv
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why is it important to consider relative velocity of block wrt wedge instead ground?

ranvijayrathore
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The answer is wrong in hcv the key is different from your key

devscorner
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y k along to external force h phir kinetic energy y k along kyu liye

ammarsaba
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Relative calculate karte samay v1 minus ke saath hona chahiye tha na

rohanraj