Trouble with a Rocket Propulsion question (Variable Mass & Momentum)

AI Thread Summary
The discussion revolves around solving a rocket propulsion problem involving variable mass and momentum. The user set the upward direction as positive and derived a differential equation for velocity, but encountered discrepancies in the final velocity equation. The correct solution should reflect the initial condition of zero velocity at time zero, which was overlooked during integration. A key error identified was not properly evaluating the lower limit of the integral. The user appreciates the guidance in correcting their approach to the problem.
vparam
Messages
17
Reaction score
3
Homework Statement
A fully fueled rocket has a mass of 21,000 kg, of which 15,000 kg is fuel. The burned fuel is spewed out the rear at a rate of 190 kg/s with a speed of 2800 m/s relative to the rocket. If the rocket is fired vertically upward calculate its final velocity at burnout (all fuel used up). Ignore air resistance and assume g is a constant 9.80 m/s^2.
Relevant Equations
M * dv/dt = ∑F_ext + v_rel * dM/dt
I chose to set the upwards direction to be positive and dM/dt = R = 190 kg/s, so I can solve the problem in variable form and plug in. With the only external force being gravity, this gives

M(t) * dv/dt = -M(t) * g + v_rel * R

where M(t) is the remaining mass of the rocket. Rearranging this equation gives:

dv/dt = -((v_rel * R)/M(t)) - g

Since R is constant, M(t) = M_0 - R * t, where M_0 is the initial mass of the rocket. Plugging in gives:

dv/dt = -((v_rel * R)/(M_0 - R * t)) - g.

Solving as a separable differential equation, I arrived at the answer (assuming v = 0 at t = 0):

v(t) = -g * t + v_rel * ln(M_0 - R * t).

However, after plugging in values, I'm not able to get the correct answer. The solution instead has a different equation for v(t):

v(t) = -g * t + v_rel * ln(M/M_0).

Any help about where I could be going wrong with the physical setup or the math of this problem would be much appreciated. Thanks in advance!
 
Physics news on Phys.org
vparam said:
dv/dt = -((v_rel * R)/(M_0 - R * t)) - g.

Solving as a separable differential equation, I arrived at the answer (assuming v = 0 at t = 0):

v(t) = -g * t + v_rel * ln(M_0 - R * t).
Note that your result for v(t) does not satisfy the initial condition v = 0 at t = 0.

When you integrated ## \dfrac {dt}{M_0 - Rt}## from ##t = 0## to ##t = t##, did you forget to evaluate at the lower limit ##t = 0##?
 
Last edited:
Yes, it looks like that's where I went wrong. Thank you for your help!
 
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top