Question

Consider a rocket in space that ejects burned fuel at a speed of v_ex= 2.0 km/s with respect to the rocket. The rocket burns 10 % of its mass in 290 s (assume the burn rate is constant).

(a) What is the speed of the rocket after a burn time of 145.0 s? (suppose that the rocket starts at rest; and enter your answer in m/s)
b)What is the instantaneous acceleration a of the rocket at time 145.0 s after the start of the engines

Answer 1

time is not related for first part? in velocity formula there is no time, so we should not worry about 290s and 145s for first part? Thanks

Answer 2

The center of gravity of the fuel/rocket system does not move. There is no net force on the rocket-fuel system. The rocket goes right, the fuel goes left. The momentum of the fuel to the left is equal and opposite to the momentum of the rocket to the right. The rocket pushes fuel left as hard as the fuel pushes the rocket right
Look here:
http://www.ux1.eiu.edu/~cfadd/1350/09Mom/Rock.html

Answer 3

You should use the equation dv = -v_ex*ln(m_f/m_i). You need to determine how much fuel is burned throughout the first 145 seconds, than substract that number from the initial mass to get m_f.

After that, to find a, you need to find the derivative of the above equation.
(You could you perhaps Wolfram Alpha for that)

Answer 4

Hi billy, It is given that 10% of mass in 290 sec and asking in 145s. I used m_0/m= 10/9 which is in 290sec I guess. and I got wrong ans. I have only one submission left. I don't know how to find m_0/m in 145sec. Please can you do it pls?

Answer 5

For a?

Derivative with respect to what?

We do not know mass, just ratio, so assume x= ration, then derivative of ln(x) =1/x = 1/ratio?

Answer 6

it's simple
10%/2 because 145=290/2 and it said "assume the burn rate is constant" so V=Vex*ln(1/(1-%/2))
for your case
V=2000*ln(1/(1-0.05))
a=V/145

Answer 7

That is average a, I ended up using the formula to calculate a speed 0.001 s before the given time then dV/dT to get instantaneous a, which then got the right answer

Answer 8

@Joe,
I've tried something like you've said... But I don't know if I'm right...
I found velocity at t=125 and t=124.999, and found 113.139 and 113.140. So instant acceleration would be somthing abount 0.9312 m/s^2 (roughly, 1 m/s^2)?
I could understand your saying.

Answer 9

Hi Anon or Joe. Rocket Acceleration The following are the variables: Vex=1500m;rocket burns 5% of its mass in 280 seconds. speed v after 140 seconds?
V= Vex*ln(1/(1-0.025))
V= 1500*ln(1/(1-0.025))
V= 1500*0.0253178
V=37.97612 m/s
a=V/140=37.9767/140=0.271267 m/s^2
Please check if you could be so kind. I have run out of options. Thanks

Answer 10

Hil,
I haven't tried this one yet... but I don't know if a is an "average" cause a changes as function of mass... you see? As the rocket goes foward, it loses mass, but it keeps throwing mass out, so it has an "average" thrust... so, thrust won't change. Fth = ma... as m goes down... a goes up. So, as far as I can tell, it isn't an average. \:
Even though I know somehow what has to be done, I just can't come up with some answer. \:

Answer 11

@Hil if you had calculated correctly it seem to be right