# Force and change in momentum in an inelastic/elastic collision

• AN630078
Thank you for your reply. Ok, so for question 2 if the collision is elastic then the kinetic energy before it rebounds is equal to the kinetic energy as it rebounds.1/2mv^2=1/2mv^2

#### AN630078

Homework Statement
Hello, I have a question below about elastic and inelastic collisions and momentum. I have answered the first question but I am struggling a little with the second question and would be very appreciative of any help. Although I have answered the first question I am uncertain whether my method would be correct anyhow.

A stream of water is pushed out through an opening of area 40 cm2 at a speed of 30 m/s and hits a wall (incident normally).
The density of water is 1000 kg/m^3

Question 1: What force is exerted on the wall if the water does not rebound?

Question 2: What would be the force if the water rebounds elastically?
Relevant Equations
Force= change in momentum
Question 1:

Since Force=Change in momentum = ∆P / ∆t
= mv / t

Momentum of the water coming out=mv
mv=ρVv=ρAv
Force d/dt (ρAv2t)=ρAv^2
Force = 1000*40*30ms^2
Force = 3.6*10^7 N

Question 2:
This is where I am confused because I understand in an elastic collision total kinetic energy and momentum are both conserved.
Therefore, Ek before =Ek after

However, the force previously calculated (equal to the change in momentum) was found when the water does not rebound.
Sorry, I think I am overthinking this and cannot formulate a sound conclusion as a result.
I know how to find the kinetic energy using Ek=1/2mv^2 (if that is at all useful here) but how would I find the force when the water rebounds elastically, since this would be different to the force calculated earlier? Sorry I am rather confused here!

AN630078 said:
Relevant Equations:: Force= change in momentum

mv=ρVv=ρAv
No, force is rate of change in momentum.
The equation above is dimensionally inconsistent. ρVv is ML-3 L3LT-1=MLT-1, whereas ρAv is MT-1. You seem to have just changed a V into an A.
AN630078 said:
Momentum of the water coming out=mv
That's just the momentum of some particular mass of water m.
AN630078 said:
Force = 1000*40*30ms^2
The 40 is in cm2, and the 30 is m/s, not m/s2.

Think of it this way... what mass of water comes out in one second? How much momentum does it have?

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• PeroK and etotheipi
The wall in this scenario is attached rigidly to the Earth, whose mass is many orders of magnitude greater than masses of the wall or the water. The momentum of the Earth-wall-water system is conserved, in both the elastic and inelastic cases, so long as you don't forget to take into account the momentum of the Earth.

Because the Earth is so massive, we can approximate that the Earth remains at rest. Momentum is not conserved in this regime, since you just need to remember that we're ignoring the very tiny velocity of the Earth which, when coupled with it's very large mass, does result in momentum conservation.

Let's also consider energy, still working in the stationary Earth regime. In the perfectly elastic collision, the kinetic energy of the water is exactly the same as what it was before it hit the wall. In the perfectly inelastic collision, the kinetic energy of the water reduces to zero, since it is dissipated to heat, sound, etc.

That's the context, anyway. In practice, you just need to notice that the change of momentum of the water in the elastic collision is double what it is in the inelastic collision. That's because in the elastic collision, $$\Delta \vec{v} = -2\vec{v}_0 \implies \Delta \vec{p} = -2m\vec{v}_0 \implies \frac{\Delta \vec{p}}{\Delta t} = -2\dot{m}\vec{v}_0$$whilst in the inelastic collision,$$\Delta \vec{v} = -\vec{v}_0 \implies \Delta \vec{p} = -m\vec{v}_0 \implies \frac{\Delta \vec{p}}{\Delta t} = -\dot{m}\vec{v}_0$$where I'm assuming the limit ##\Delta t \rightarrow 0##. Of course, you still need to make sure you have the correct mass flow rate ##\dot{m}##; see @haruspex's post above!

haruspex said:
No, force is rate of change in momentum.
The equation above is dimensionally inconsistent. ρVv is ML-3 L3LT-1=MLT-1, whereas ρAv is MT-1. You seem to have just changed a V into an A.

That's just the momentum of some particular mass of water m.

The 40 is in cm2, and the 30 is m/s, not m/s2.

Think of it this way... what mass of water comes out in one second? How much momentum does it have?
Thank you for your reply. Ok, I found the equation derived in a textbook in a similar problem but thought it to be peculiar. Typically, I would use force = change in momentum.

Would I use the mass flow formula to find the mass of water emerging in one second;
m = ρ v A
m = 1000 kgm^3 * 30 m/s * 4*10^-3 m^2
m = 120 kg s

Force=Change in momentum = ∆P / ∆t = mv / t
Force = 120 *30/1
Force = 3600 N ?

I am still uncertain of question 2 though.

AN630078 said:
I am still uncertain of question 2 though.
What is the momentum change per unit time of the water if it bounces off elastically? More specifically, how is the velocity of the water after the collision related to the velocity of the water before the collision in this case?

kuruman said:
What is the momentum change per unit time of the water if it bounces off elastically? More specifically, how is the velocity of the water after the collision related to the velocity of the water before the collision in this case?
Thank you for your reply. Ok, so for question 2 if the collision is elastic then the kinetic energy before it rebounds is equal to the kinetic energy as it rebounds.
1/2mv^2=1/2mv^2
I am not sure how to consider the change in momentum if it rebounds elastically nor the relation of the intial velcoity to the final velocity. I am sorry I do not know why I am having such difficulty with this.

AN630078 said:
Typically, I would use force = change in momentum.
Force=Change in momentum = ∆P / ∆t = mv / t
The change in momentum is ∆P. (Average) Force=Rate of Change in momentum = ∆P / ∆t.
AN630078 said:
Would I use the mass flow formula to find the mass of water emerging in one second;
m = ρ v A
m = 1000 kgm^3 * 30 m/s * 4*10^-3 m^2
m = 120 kg s

Force=Change in momentum = ∆P / ∆t = mv / t
Force = 120 *30/1
Force = 3600 N ?
Yes, but let me tidy that up a bit:
In time Δt, mass ρAvΔt emerges at speed v, carrying momentum ∆P=ρAv2Δt. Thus ∆P / ∆t=ρAv2.

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• PeroK and AN630078
The equation you have
AN630078 said:
1/2mv^2=1/2mv^2
says nothing new. It is always true that A = A, 5 apples = 5 apples and so on. What are you really trying to say with this equation? Hint: A good habit to have is using different symbols for different entities.

• AN630078
haruspex said:
The change in momentum is ∆P. (Average) Force=Rate of Change in momentum = ∆P / ∆t.

Yes, but let me tidy that up a bit:
In time Δt, mass ρAvΔt emerges at speed v, carrying momentum ∆P=ρAv2Δt. Thus ∆P / ∆t=ρAv2.
Question 1;
So the change of momentum would be ∆P / ∆t=ρAv^2
∆P / ∆t=1000 kgm^3*4*10^-3 m^2*30^2 ms^-1
∆P / ∆t= 3600 N

Question 2; I am sorry but I am still confused.
So, I know I have stated it already but we know a collision is elastic if kinetic energy is conserved. Attempting to follow @kuruman advice of using different symbols for different entities this could be shown by;
1/2 m1 v1^2 i + 1/2 m2 v2^2 i = 1/2 m1 v1^2 f + 1/2 m2 v2^2 f

and momentum will also be conserved;
m1v1 i + m2v2i = m1v1 f + m2v2 f

The difference in the velocities of the two objects is called the relative velocity, which indicates how fast object 1 is moving relative to object 2. The magnitude of the relative velocity is the same before and after the collision, therefore, |v1i-v2i| = |vi f - v2f|, however, the relative velocity will have opposite signs before and after the collision;
(v1i-v2i)=-(vi f - v2f)

Using the first equation to show KE is conserved, would the value for the final velocity be negative, v=-30m^-1?

Moreover, since m=ρAvΔt
And from the law of the conservation of momentum FΔt=mΔv
Therefore, the force after the water rebounds elastically would be; F=mΔv/Δt
How would I find the change in velocity here, since the magnitude is the same before and after the collision v1 i = v2 f ? Would this be 30ms^1=-30ms^-1 so the difference would be 60ms^-1?

Sorry, clearly I am still struggling but I am trying to understand this problem. Would this approach be correct to find the force if the water rebounds elastically?

Yes, the change in velocity is twice what it was in the inelastic case.

PeroK said:
Yes, the change in velocity is twice what it was in the inelastic case.
Thank you for your reply. So would the force if the water rebounds elastically be:
F=mΔv/Δt
How would I find the mass, since I earlier found the mass flow rate to be 120kg/s, so would the mass be 120 kg?
Then F=120kg * 60ms^-1/1
F=7200 N

Also would Δt = 1 s?

AN630078 said:
Thank you for your reply. So would the force if the water rebounds elastically be:
F=mΔv/Δt
How would I find the mass, since I earlier found the mass flow rate to be 120kg/s, so would the mass be 120 kg?
Then F=120kg * 60ms^-1/1
F=7200 N

Also would Δt = 1 s?
The force must be twice what it was before. You can take ##\Delta t## to be any value.

PeroK said:
The force must be twice what it was before. You can take ##\Delta t## to be any value.
Why would the force be twice it was before?
Would it not be;
F= ∆P / ∆t=ρAv^2
∆P / ∆t=1000 kgm^3*4*10^-3 m^2*60^2 ms^-1 / 1
F= 14400 N?

AN630078 said:
Why would the force be twice it was before?
Would it not be;
F= ∆P / ∆t=ρAv^2
∆P / ∆t=1000 kgm^3*4*10^-3 m^2*60^2 ms^-1 / 1
F= 14400 N?
I thought you got it to be twice what it was in the inelastic case. ##F = 3600N## and ##F = 7200N## respectively.

I don't know what you're doing now. You got the right answer, I agreed, then you produced a different answer?

Sorry, I have just reread my posts, I had confused myself. Yes, I did find twice the original kinetic energy. Why would this be negative, is it because KE i = KE f
Therefore, KEi=3600 J and KEf = 7200 J
So 3600J=-7200J ?

AN630078 said:
Sorry, I have just reread my posts, I had confused myself. Yes, I did find twice the original kinetic energy. Why would this be negative, is it because KE i = KE f
Therefore, KEi=3600 N and KEf = 7200 N
So 3600N=-7200N ?
You don't measure KE in Newtons.

PeroK said:
You don't measure KE in Newtons.
Sorry joules! My mistake PeroK said:
You don't measure KE in Newtons.
Are you sure this is correct, I feel like there should be more to the problem somehow?

AN630078 said:
Are you sure this is correct, I feel like there should be more to the problem somehow?
This is chaos. In post #7 @haruspex gave you the formula:
$$F = \frac {\Delta P}{\Delta t} = \rho A v^2$$
From this you calculated:
$$F= 3600N$$
This was for the inelastic case. In the elastic case we have twice the change in momentum, hence twice the force:
$$F = 7200N$$
That's it. You don't need to do anything more.

• haruspex
AN630078 said:
So the change of momentum would be ∆P / ∆t=ρAv^2
As I tried to explain in post #7, that terminology is wrong.
As I wrote, the change in momentum is ∆P. It is not ∆P / ∆t.
∆P / ∆t is the rate of change of momentum, i.e. the force.