Homework Help: Wedge-Air Collision Problem

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1. Jan 25, 2016

UchihaClan13

1. The problem statement, all variables and given/known data
Air of density ρ, moving with velocity v strikes normally on an inclined surface (having area A) of a wedge of mass m kept on a horizontal surface. Collisions are perfectly elastic (No loss of kinetic energy). Minimum coefficient of static friction between wedge and the horizontal surface, for the wedge to be stationary, is

2. Relevant equations
Force imparted to the wedge=ρ*A*v^2
Forces exerted are broken down into their respective components

3. The attempt at a solution
Breaking down the force (exerted by the air collisions)on the wedge into components we get,
ρ*A*v^2costheta + mg= N(the normal force exerted by the ground surface on the wedge)
Here i assumed that the wedge will be in equilibrium in the y-direction
and in the x-direction(after drawing the f.b.d of the wedge),one can see that
ρ*A*v^2sintheta - ffriction = m*a(here a is zero since in the question it says so)
now comes the hardest part
we know that
so for the minimum coefficient of static friction,i first assumed that the friction force wouldn't be equal to the limiting friction and got stuck with an inequality
Ffriction≤ [PLAIN]https://upload.wikimedia.org/math/9/3/9/939974a71dda1b83cce5ab82a2d2cec1.png(mg [Broken] + ρ*A*v^2costheta) hence i left this case
and now took the frictional force as limiting friction yielding [PLAIN]https://upload.wikimedia.org/math/9/3/9/939974a71dda1b83cce5ab82a2d2cec1.png=[B][B][B][B][B]ρ*A*v^2sintheta/([B][B][B]ρ*A*v^2costheta [Broken] + mg)
I think i am going wrong somewhere or i might have overlooked the "collisions are elastic" or i might have made a wrong assumption
[/B][/B][/B][/B][/B][/B][/B]
Anyways,as usual
Help is much appreciated!!:)[/B]

Last edited by a moderator: May 7, 2017
2. Jan 25, 2016

ehild

Is it right?
imagine a particle of mass m and velocity v collides into the wedge elastically. What is the velocity of the particle after the the collision? How much is the change of its momentum?

3. Jan 25, 2016

UchihaClan13

I don't know,for sure
And as for your second question,the particle which strikes the wedge elastically with a velocity v would rebound back with the same velocity v.
hence the change in momentum would be Pfinal - Pinitial = -mv-mv=-2mv or 2mv in the opposite direction

4. Jan 25, 2016

UchihaClan13

Wait
are you trying to imply that the change in momentum w.r.t time equals the impulsive force imparted by the wedge to the "supposed" block.
Therefore,by newton's third law ,the force exerted by the block on the wedge equals the impulsive force but in the opposite direction
But if you take a certain time of collision "t" into consideration then the mg force of the wedge gets a factor of t,doesn't it?
How do you proceed further?

5. Jan 25, 2016

ehild

Yes, you have to take the total change of momentum of the air flow which is twice the original momentum. During the collision, which is very short time for an individual air molecules, gravity plays no role.
The momentum imparted to the wedge in unit tine is equal the force exerted on it by the flow of air molecules. You need to take the vertical component of this force in addition to the weight of the wedge to calculate the normal force, as you did it in the OP.

Last edited: Jan 25, 2016
6. Jan 25, 2016

UchihaClan13

Thanks a lot ehild for that particular insight
I completely overlooked it
Now i am getting the correct answer
Let me write it down again(This time correctly)
Mass of air falling per second on the wedge=ρ*A*v (in 1 second,this much mass falls)
Hence momentum of the mass of air =ρ*A*v^2
final mometum of the mass of air after an elastic collision =- ρ*A*v^2 (as it's in the opposite direction)
therefore change in momentum in 1 second=-2ρ*A*v^2
therefore force imparted by the air on the wedgeρ*A*v^2
elementary vector resolutions and algebraic manipulations yield μ2ρ*A*v^2sinθ/(mg + ρ*A*v^2cosθ)
thanks again

Last edited by a moderator: Jan 26, 2016
7. Jan 25, 2016

UchihaClan13

sorry for that odd message Something went wrong

8. Jan 25, 2016

UchihaClan13

Here i considered the mass flow rate as mass per second
and calculated the change of momentum in 1 second
hence my time factor "t" is 1 second
I neglected the time of an individual collision since it's negligible
Is my reasoning correct?

Last edited by a moderator: Jan 26, 2016
9. Jan 26, 2016

ehild

It was correct, only the force exerted by the air flow was wrong.

10. Jan 26, 2016

UchihaClan13

Thanks once again :)