Rocket Problem -- Solve using the Rocket Equation

[mannerplay]

Homework Statement

Halliday Principles of Physics 10th edition
The mass of the rocket is 50 kg, the mass of the fuel is 450 kg
the rocket's maximum v rel=2 km/s
if R=10kg/s, what velocity does the rocket moves when it consumes all its fuel?
solve when the acceleration of the rocket is 20 m/s^2

Relevant Equations

Ma=-Rv (R is the ratio which a rocket's loss of its mass and v is the relative speed of the fuel to the rocket)

I tried the second rocket equation

vf = vi + v rel * ln(Mi/Mf)

but it gives out approximately 4900 m/s for the answer
but the answer is 4160 m/s

 

[jbriggs444]

What are $v_i$, $v_\text{rel}$, $M_i$ and $M_f$?

You claim to have put them into the equation: $v_f=v_i + v_\text{rel} \times ln(\frac{M_i}{M_f})$ but I cannot tell what inputs you used to that equation.

If I reverse-engineer the supposedly correct answer, I come up with a different problem where the mass of the fuel is only 350 kg.

I have not been able to reverse-engineer your answer to see what you did. That is why it is important to show your work.

 

[erobz]

if R=10kg/s, what velocity does the rocket moves when it consumes all its fuel?
solve when the acceleration of the rocket is 20 m/s^2

For starters if we are using the relevant equation as written $R = -10 ~ \rm{\frac{kg}{s}}$
OR we leave $R$ positive and note that $v_{e/R} = -2000 \rm{ \frac{m}{s} }$. Either way we note that the acceleration of the rocket is in the direction of motion for the rocket - not against.Next, I believe either the question is flawed OR they are two separate questions.

1) What velocity does the rocket moves when it consumes all its fuel?

2) What is its velocity when the acceleration of the rocket is 20 m/s^2? ( Flawed )

EDIT: Question 2 is flawed because the acceleration of the rocket is not less than $40~ \rm{ \frac{m}{s^2} }$ over the duration of powered flight.

$$ \left. \frac{dv}{dt} \right|_o = \frac{10~\rm{\frac{kg}{s}} 2000~\rm{ \frac{m}{s}} }{ 450~\rm{ kg } + 50~\rm{kg}} = 40~ \rm{ \frac{m}{s^2} } $$

$$ \left. \frac{dv}{dt} \right|_f = \frac{10~\rm{\frac{kg}{s}} 2000~\rm{ \frac{m}{s}} }{ 50~\rm{kg}} = 400~ \rm{ \frac{m}{s^2} } $$

Obviously that last one (the final acceleration at burnout) is a touch unrealistic!

Didn't realized this thread was dated...

 

[berkeman]

Thread closed temporarily for Moderation...

 

[berkeman]

Update -- this thread was probably doomed from the start because of the flawed OP question. A number of unhelpful replies have been deleted and the thread will remain closed now. Thanks.