Which equation should be used to calculate impulse: I=F(t) or I=m1v1-m0v0?

[LT72884]

Just going over some basic equations to help someone out and i have noticed that i get 2 different answers when using 2 different but very similar equations for impulse.

first eq. I=F*t

lets say that i have a small rocket motor that has a force of 5.95 newtons and a impulse of 2.5 n-s. therefore t would be 0.42 second burn time.
burnout weight of rocket is 0.0357kg

now. f = ma so a = f/m. f = (thrust - burnout weight of rocket) so a = (5.95N-(0.0357*9.81))/0.0357 = 156.8627

now multiply by time to get velocity and its 65.88 m/s or 216 FPS

ok, now when i use I=m1v1 and let m=w/g, --> I=(w/g)v1 --> v1=Ig/w

i get 206 fps. not a big difference but im wondering which is the better approach to take?

thanks

 

[jack action]

Your second equation reads like this:
$$v=\frac{I}{m}$$
Here, I'm not sure where 206 fps (62.79 m/s) comes from, as this implies $m = \frac{2.5}{62.79}=0.0398\text{ kg}$ or $I = (0.0357)(62.79) = 2.24\text{ N.s}$.

Your first equation reads like this:
$$v= a\Delta t = \frac{F_y}{m}\frac{I}{F}$$
If $F_y$ would be equal to $F$, then both equations would be identical. Somehow, in your first equation, you decided otherwise, such that:
$$v = \frac{F_y}{F}\frac{I}{m} = \left(1-\frac{mg}{F}\right)\frac{I}{m}$$

 

[LT72884]

for second equation it would be v = (2.5*9.81)/(0.0357*9.81) = 70.02m/s = 229 fps, not 206. good catch. however, if i do it in freedom units, i do get 206

v= (0.5056*32.2)/0.0787Lbs = 206.86 f/s

for eq 1 its a =(thrust-rocket weight)/mass

then v = u +at u =0

a= 156.86m/s^2
t= 0.42 seconds

v = 65.88m/s or 216f/s

doing a quick search, it looks like this person is doing the same approach as me for the acceleration.

https://www.grc.nasa.gov/www/k-12/V...text=The acceleration of the rocket,a = F / m).

 

[jack action]

In your first equation, you should use the net force everywhere, like so:
$$\Delta t =\frac{I}{T-mg} = \frac{2.5}{5.95-(0.0357)(9.81)} = 0.4464\text{ s}$$
$$a = \frac{T-mg}{m} = \frac{5.95 - (0.0357)(9.81)}{0.0357} = 156.8567\text{ m/s²}$$
$$v = a \Delta t = (0.4464)(156.8567) = 70.02\text{ m/s} = 229\text{ fps}$$

 

[nasu]

lets say that i have a small rocket motor that has a force of 5.95 newtons and a impulse of 2.5 n-s. therefore t would be 0.42 second burn time.
burnout weight of rocket is 0.0357kg

Do you mean that the force is constant dring the burning period? What is the meaning of this number?
If this is the case, the acceleration is not constant, as the mass of the rocket changes.

 

[LT72884]

Do you mean that the force is constant dring the burning period? What is the meaning of this number?
If this is the case, the acceleration is not constant, as the mass of the rocket changes.

without getting into ODEs and numerical methods, the 5.95 newtons is the amount of thrust the small rocket motor produces. Most likely constant for rough estimates of zero drag flight

i found this article which uses the smae method

https://www.grc.nasa.gov/www/k-12/V...text=The acceleration of the rocket,a = F / m).

 

[LT72884]

In your first equation, you should use the net force everywhere, like so:
$$\Delta t =\frac{I}{T-mg} = \frac{2.5}{5.95-(0.0357)(9.81)} = 0.4464\text{ s}$$
$$a = \frac{T-mg}{m} = \frac{5.95 - (0.0357)(9.81)}{0.0357} = 156.8567\text{ m/s²}$$
$$v = a \Delta t = (0.4464)(156.8567) = 70.02\text{ m/s} = 229\text{ fps}$$

good point. I do agree with that. I need to take account of the net force everywhere, not just acceleration. This works if i know the impulse, and force, which i do, but if i am only given impulse and time, then i have to find the thrust.

EDIT: for the first equation, i can not use 5.95N like you did because that 5.95 is based on the 0.42 seconds. IE 2.50 = F(0.42 sec) therefore F is 5.95. however, thats the thrust of just the motor, now i subtract the rocket weight and then re solve for time and it works out. I think i follow you now. Thanks

 

[nasu]

without getting into ODEs and numerical methods, the 5.95 newtons is the amount of thrust the small rocket motor produces. Most likely constant for rough estimates of zero drag flight

i found this article which uses the smae method

https://www.grc.nasa.gov/www/k-12/VirtualAero/BottleRocket/airplane/rktalo.html#:~:text=The acceleration of the rocket,a = F / m).

Then you cannot assume constant acceleration when you calculate the final velocity. You need to integrate over the burning time. And you need to know how the mass varies with time, in order to do this.

 

[LT72884]

Then you cannot assume constant acceleration when you calculate the final velocity. You need to integrate over the burning time. And you need to know how the mass varies with time, in order to do this.

this is just for simple estimates. I do know i would have to integrate etc, but for simplicity, you can use the big 3 to solve for velocity.

Vf = Vi = at
Vi = 0 so Vf is acceleration * time

i know the acceleration of the rocket motor i9n basic simple terms a= (Thrust - weight)/mass according to the nasa article that i found after i had been messing around for a bit.

this is all for zero drag bare bones simple rocket flight.

I do have the Cd based on the Reynolds number, altitude of launch, surface roughness, induced drag, fin drag, tail drag, etc but that is way beyond the scope of this exercise for this high schooler im helping haha.

 

[nasu]

This has nothing to do with drag, Reynolds number, etc. It's just that the "big 3" works only for constant acceleration, which you don't have because the mass is not constant.

But even considering a constant mass situation, your first method of evaluation is wrong. The change in momentum is not equal with the impulse of the motor alone. The force in Newton's second law is the net force.
So, the correct equation is $$(F-mg)\Delta t=\Delta (mv)= mv_{final}-mv_{initial}$$
which is the same with what you get by going through acceleration, just arranged a little differently.
It's actually the same equation in both cases, if you write it correrctly.

 

[LT72884]

This has nothing to do with drag, Reynolds number, etc. It's just that the "big 3" works only for constant acceleration, which you don't have because the mass is not constant.

But even considering a constant mass situation, your first method of evaluation is wrong. The change in momentum is not equal with the impulse of the motor alone. The force in Newton's second law is the net force.
So, the correct equation is $$(F-mg)\Delta t=\Delta (mv)= mv_{final}-mv_{initial}$$
which is the same with what you get by going through acceleration, just arranged a little differently.
It's actually the same equation in both cases, if you write it correrctly.

correct, i understand what you are saying :) but rocket motors are so small that 3.5g of changing mass is almost negligible.

yes, this is for zero drag modeling only. I was just mentioning i can find Cd various ways etc.

ill have to look at your reply again. I did find a book online last night that did it exactly the way i did it. James Barrowman also does it this way, and a thousand other way haha

but from what i see with your equation, its the same as what Jack had posted, which is what i ended up using anyway.

t = I/(t-mg) where I is the change in momentum

 

[nasu]

This is Newton's second law. There is no controversy about it.
You may neglect the force of gravity (if it is much less than the thrust force) when you write the net force but then you do the same for both ways of solving if you want to get the same result.

 

[LT72884]

correct, there is no controversy :)

i thank you for your help aand guidance. it is much appreciated.