How Is Impulse Calculated for a Rocket with Changing Mass?

[Alameen Damer]

Homework Statement

A 1990 kg rocket is loaded with 102 kg of propellant. It exhausts the propellant in a burn of 25s. The rocket starts at rest, and moves with a speed of 240 m/s after the burn. Determine the impulse.

Homework Equations

Impulse (p) = m1vf-m1vi

The Attempt at a Solution

How would one calculate the impulse given that the mass changes because of the loaded propellant which is used up by the time the mass hits the final velocity.

 

[haruspex]

Homework Statement

A 1990 kg rocket is loaded with 102 kg of propellant. It exhausts the propellant in a burn of 25s. The rocket starts at rest, and moves with a speed of 240 m/s after the burn. Determine the impulse.

Homework Equations

Impulse (p) = m1vf-m1vi

The Attempt at a Solution

How would one calculate the impulse given that the mass changes because of the loaded propellant which is used up by the time the mass hits the final velocity.

The question seems ambiguous. Is it the total impulse generated by the fuel that's wanted, or the effective impulse on the rocket? The second is quite easy.

 

[Alameen Damer]

I would assume the effective impulse on the rocket, given that they talk about it speeding up, etc. If this is the case, using (p) = m1vf-m1vi, do I include the 102 kg propellant as part of the mass?

 

[haruspex]

I would assume the effective impulse on the rocket, given that they talk about it speeding up, etc. If this is the case, using (p) = m1vf-m1vi, do I include the 102 kg propellant as part of the mass?

The description speaks of rocket and fuel as distinct, so no, it would not include the 102kg. That's what makes the question easy. If yiu had to include the 102kg you would not know what to put for final velocity. Relative to the ground, fuel expelled early would have a different velocity from that expelled later.

 

[HallsofIvy]

The basic rule is that force equals the derivative of momentum. If the mass is constant, that is "mass times the acceleration" but if mass is not constant, it is (mv)'= m'v+ mv'.

 

[haruspex]

The basic rule is that force equals the derivative of momentum. If the mass is constant, that is "mass times the acceleration" but if mass is not constant, it is (mv)'= m'v+ mv'.

Halls, I'm not alone in regarding that formulation as misleading. It treats mass as an arbitrary variable, as though a moving object can acquire mass out of nowhere. In reality, the mass must come from or go to somewhere else. For the equation to work, that mass must not bring in or take out any momentum, i.e. when not part of the object it must be at rest in the reference frame. I wrote a homily on this in section 6 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-momentum-impacts/.

Anyway, if I'm reading the question correctly, there is no need to get tangled up with that. The information regarding the mass of fuel is redundant. (But maybe there are more parts to the question.)