Which law is more appropriate to explain the uplift of a rocket or the recoil of a gun?

Question

i am doubted whether rocket principle and recoil is better explained by law of conservation of momentum or newton's third law. please help me in a detailed manner

Answer 1

Newton's third law - action/action.

In the case of a rocket, fuel is burned. This creates a force. The force of the fuel's burning pushes against the top of the storage container that the fuel is in. This force pushes the rocket up. The burning fuel is expelled.

Answer 2

Newton's Third Law
For every action, there is an equal and opposite reaction.

For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.

I think newton's third law is a better fit to explain rocket principle and recoil

Answer 3

As many of the longer answers correctly indicate...Newt's three laws are interrelated. They can all be invoked to describe the rocket and gun behavior. So, to answer one law or the other is misleading. It's something like the glass is half full or half empty issue.

Check this out p = mv, momentum p is just mass times its velocity. This applies to a rocket M and a bullet m; so that P = Mv and p = mV can be used to describe their respective momenta.

But dp/dt = dm/dt v + m dv/dt is the force on that mass. dp/dt is just the differential of momentum over time. In you haven't had calculus yet, this simply means the change in momentum as time passes.

Now note: dp/dt = m dv/dt if mass is a constant, like for a bullet. That results because dm/dt = 0, mass is not changing, that bullet stays at m = constant. And as we all know acceleration a = dv/dt the change in velocity over time. Thus, we have dp/dt = m dv/dt = ma = f. Hey, Newt's second law

If we consider the rocket, dp/dt = dm/dt v + m dv/dt and dm/dt <> 0 so the first term stays in. Why would mass not be a constant...because the rocket is losing/burning fuel over time and fuel is a major part of its mass. Even so, this is the force equation when mass is not a constant. It's still Newt's second law.

Finally, built on f = ma; where f is a net force, we have f = ma = 0 = sum(all forces). Thus, for a non-accelerating body, the net forces (f) must equal zero. Which is simply one way of saying, for every force there must be a counter force of equal, but opposite reaction. Newt's third law.

So there you have it...Newt's three laws, the momentum law was used to derive the force law, and the force law was used to derived the equal, but opposite law. Which one you choose to apply to a specific rocket or bullet/gun case depends on what kind of answer you are looking for. But, in the end, you will be applying all three of Newton's laws because they are interrelated.

Answer 4

It's momentum conservation.......or transfer of momentum as I'm sure some of the fuel momentum is not all burned and transferred/imparted to the rocket.

Clearly this is an action of momentum. As the fuel burns and is expelled the expelled mass causes the rocket to propel forward (momentum). The fuel is not pushing against anything, as a person pushes, say a wagon or a swing.

Answer 5

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Answer 6

Newton's third law of motion. It states that to every action there is an equal and opposite reaction. To calculate the velocity and recoil velocity, or the masses law of coservation of momentum is dealt with.

Answer 7

a systems momentum is conserved.
in a recoil, the gun and bullet are initially at rest.
so when the bullet accelerates to a certain muzzle velocity the gun itself also accelerates to a certain velocity. since it is much heavier than a bullet, its velocity is less.
same with a rocket. the rocket's total momentum is zero. so when it shoots out gas and whatnot, to maintain momentum, it moves in the opposite direction.

Answer 8

Conservation of momentum is a more general version of Newton's 3rd law, as it can be applied to both classical and non-classical physics.

As such however you choose to do it, you will be using the conservation of momentum.

Answer 9

If you look at the derivation for the rocket equation, you'll find it is entirely based on conservation of momentum. Specifically, you equate the change in rocket velocity ot the amount of fuel expended and the speed at which it is ejected from the engine

du = - uf*dM/M u = speed of rocket, uf = speed of fuel, dM = change in rocket mass (fuel expended), M=rocket mass

OR Mdu = -ufdM ---> conservation of momentum.

If you integrate teh first equation, you get the rocket equation with no external forces acting:

u = uf - ln(Mb/M0) where Mb = mass of rocket without fuel, amd M0 = total mass at lif-off