chain question

[pringless]

A 1520 kg rocket has 4860 kg of fuel on board. The rocket is coasting through space at 94 m/s and needs to boost its speed to 348 m/s. It does this by firing its engines and ejecting fuel at a relative speed of 807 m/s until the desired speed is reached. How much fuel is left on board after this manuever?

[himanshu121]

Apply the Conservation of Momentum in inertial reference frame with which the rocket is moving

[ShawnD]

this work is all wrong

[Doc Al]

Originally posted by pringless
A 1520 kg rocket has 4860 kg of fuel on board. The rocket is coasting through space at 94 m/s and needs to boost its speed to 348 m/s. It does this by firing its engines and ejecting fuel at a relative speed of 807 m/s until the desired speed is reached. How much fuel is left on board after this manuever?
You have to treat the rocket as a system of variable mass. As the fuel is ejected, it exerts a thrust v_{rel}\frac{dM}{dt} on the rocket. Give it a shot (you'll need a little calculus).

ShawnD, note that 807 m/s is the relative speed: it is not the speed of the exhaust with respect to the ground.

[himanshu121]

Yes Shawn D u have to apply momentum conservation w.r.t ground

[ShawnD]

Ok then lets try this again. I'll change the frame of referance to the ship so that my initial is 0.

initial momentum:
p = (1520 + 4860) * 0
p = 0

relative momentum of the fuel:
p = -807x

new momentum of ship:
p = (1520 + (4860 - x )) * (348 - 94)
p = (1520 + 4860 - x) * 254
p = (6380 - x) * 254
p = 1620520 - 254x


final = initial:
1620520 - 254x - 807x = 0
1620520 - 1061x = 0
x = 1527kg of fuel spent



look better?

[Doc Al]

Originally posted by ShawnD
Ok then lets try this again. I'll change the frame of referance to the ship so that my initial is 0.

No. The ship is a non-inertial frame. Its speed (and mass) are changing. See my previous post. (I'll post a solution tomorrow.)

[ShawnD]

You souldn't have to factor that in. The velocity of the fuel is always the same speed relative to the ship.

[Doc Al]

Originally posted by ShawnD
The velocity of the fuel is always the same speed relative to the ship.
Well, no. The ship is speeding up. As each bit of fuel is ejected, its speed relative to the ship is the same. But the speed of the ship is different for each bit.

[ShawnD]

Al can you please explain what you mean? I said the fuel is always the same speed relative to the ship. You said no, then you said it's relative speed is the same (what I said).

[Doc Al]

Originally posted by ShawnD
Al can you please explain what you mean? I said the fuel is always the same speed relative to the ship. You said no, then you said it's relative speed is the same (what I said).
Sure. Each bit of fuel, as it is ejected, moves at the same speed relative to the ship. For example, at time t=1 a bit of fuel m1 is ejected. It moves at 807 m/s relative to the speed of the ship at time t=1. But the ship is accelerating. At time t=100 (say) the bit of fuel m1 is no longer moving at 807 m/s relative to the ship, since the ship is now moving faster. So... all those bits of fuel are moving at different speeds. Make sense?

[ShawnD]

You are looking at fuel it has already spent, I'm looking at fuel it is currently spending.

[Doc Al]

Originally posted by ShawnD
Oh i see what you're saying. Jesus that is one hard question.
Yep. It's meant for a calculus-based college course, not high school. For your amusement, here's how I would solve this problem:

Let M stand for the mass of the ship+remaining fuel, v for the speed of the ship. Each bit dM of fuel ejected exerts a force on the ship:
v_{rel}\frac{dM}{dt}=Ma=M\frac{dv}{dt}

Rearranging,
v_{rel}\frac{dM}{M}={dv}

Integrating from the initial speed&mass to the final,
v_{rel}\ln{\frac{M_f}{M_i}=v_f-v_i

Plugging in some numbers:
\ln{\frac{M_i}{M_f}=\frac{348-94}{807}

Finally,
\Delta M= 1723

So, of the original 4860 kg of fuel, 3137 kg is left. (Unless I made an arithmetic error.)

But the real question is: Why is this thread titled "chain question"? [:D]