Momentum conservation: block-wedge problem

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The discussion focuses on solving the block-wedge problem using momentum conservation principles. The initial attempt incorrectly assumed the block's velocity in the x-direction as scos(theta), leading to an erroneous calculation of the wedge's speed. By switching to the instantaneous frame of the wedge and directly analyzing the block's speed v, the correct speed of the wedge was derived as mvcos(theta)/(M+m). Participants emphasized the importance of clarity in notation to avoid confusion in future calculations. The conversation highlights the significance of understanding relative motion in physics problems.
Krushnaraj Pandya
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Homework Statement


A block of mass m slides down a wedge of mass M and inclination theta from rest. All the surfaces are smooth. Find the speed of the wedge when the speed of the block w.r.t to wedge is v.

Homework Equations


V(c.m.)=m1v1+m2v2/(m1+m2)

The Attempt at a Solution


Conserving momentum in horizontal direction - I got the speed of the wedge as -mscos(theta)/M where s is the velocity of block w.r.t ground. I can't figure out how to incorporate relative velocity in this formula and get it in terms of v
 
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Krushnaraj Pandya said:
I got the speed of the wedge as -mscos(theta)/M where s is the velocity of block w.r.t ground.
This isn’t correct. Since the wedge is also moving, s is not directed along θ.

You could break s up into sx and sy then say Vwedge =sxm/M ... but sx is not scosθ

I’m thinking we should be able to avoid such subtleties by ignoring s and working directly with v.

If we take the instantaneous-frame of the wedge, the block moves at speed v and along the angle θ. Analyzing momentum in this frame seems fruitful.
 
Nathanael said:
This isn’t correct. Since the wedge is also moving, s is not directed along θ.

You could break s up into sx and sy then say Vwedge =sxm/M ... but sx is not scosθ

I’m thinking we should be able to avoid such subtleties by ignoring s and working directly with v.

If we take the instantaneous-frame of the wedge, the block moves at speed v and along the angle θ. Analyzing momentum in this frame seems fruitful.
I realize my mistake in assuming s in x direction as scos(theta) and the mistakes with my approach...I got the speed of the wedge as mvcos(theta)/M+m, beautiful solution- thank you
 
Krushnaraj Pandya said:
I realize my mistake in assuming s in x direction as scos(theta) and the mistakes with my approach...I got the speed of the wedge as mvcos(theta)/M+m, beautiful solution- thank you
Well done! You are quick.

Just a tip, you should type it as mvcos(theta)/(M+m) to avoid confusion.
 
Nathanael said:
Well done! You are quick.

Just a tip, you should type it as mvcos(theta)/(M+m) to avoid confusion.
Thanks :) I'll keep that in mind from now on
 
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