Momentum conservation: block-wedge problem

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 3K views
Krushnaraj Pandya
Gold Member
Messages
697
Reaction score
73

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
 
on Phys.org
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