Momentum conservation: block-wedge problem

Click For Summary

Homework Help Overview

The problem involves a block sliding down a wedge, with the goal of finding the speed of the wedge when the block's speed relative to the wedge is given. The context is centered around momentum conservation principles in a system with two moving bodies.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the conservation of momentum in the horizontal direction and the complications arising from the wedge's motion. There is an exploration of breaking down the block's velocity into components and considering the instantaneous frame of the wedge.

Discussion Status

Some participants have identified mistakes in their initial assumptions regarding the relationship between the velocities of the block and the wedge. There is an ongoing exploration of different approaches to analyze the problem, with some guidance offered on avoiding confusion in notation.

Contextual Notes

Participants are navigating the complexities of relative motion and the implications of the wedge's movement on the block's velocity components. The discussion reflects a mix of attempts to clarify definitions and assumptions related to the problem setup.

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
 
Physics news 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
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
27
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
10
Views
3K
  • · Replies 92 ·
4
Replies
92
Views
10K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
28
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K