# Momentum conservation: block-wedge problem

• Krushnaraj Pandya
In summary, the conversation discusses finding the speed of a wedge when a block slides down it from rest at an inclination theta. The solution involves analyzing momentum in the instantaneous frame of the wedge and results in a speed of mvcos(theta)/(M+m). A tip is given to avoid confusion when typing the equation.
Krushnaraj Pandya
Gold Member

## 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

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

## 1. What is momentum conservation?

Momentum conservation is a fundamental principle in physics that states that the total momentum of a closed system remains constant, regardless of any external forces acting on the system.

## 2. How does momentum conservation apply to the block-wedge problem?

In the block-wedge problem, the total momentum of the system (consisting of the block and the wedge) remains constant as long as there are no external forces acting on the system, such as friction.

## 3. What is the equation for momentum conservation in the block-wedge problem?

The equation for momentum conservation in the block-wedge problem is: m1v1i + m2v2i = m1v1f + m2v2f, where m1 and m2 are the masses of the block and wedge, v1i and v2i are the initial velocities of the block and wedge, and v1f and v2f are the final velocities of the block and wedge.

## 4. How can momentum conservation be used to solve the block-wedge problem?

By using the equation for momentum conservation, the initial and final velocities of the block and wedge can be determined. This information can then be used to calculate other quantities, such as the acceleration of the block and wedge or the distance traveled by the block.

## 5. What are some real-life applications of momentum conservation in the block-wedge problem?

Momentum conservation in the block-wedge problem can be applied in situations such as collisions between objects on an inclined plane or in the motion of a car on a ramp. It is also used in engineering and design, such as in the construction of ramps and other structures that involve objects moving on an incline.

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