Impulse or Momentum Challenging Question

[i_hate_math]

Homework Statement

A railroad car moves under a grain elevator at a constant speed of 4.50 m/s. Grain drops into the car at the rate of 420 kg/min. What is the magnitude of the force needed to keep the car moving at constant speed if friction is negligible?

Homework Equations

U=V+dV-Vrel , U is the velocity of dM, which is 0 in this case
dM/dt • Vrel = M • dV/dt where V is the velocity not volume

The Attempt at a Solution

So from the question, dM/dt is 420kg/min=70kg/s,
I played with the 1st rocket equation, was able to get (4.5+dV)*70=M*a=Force needed.
But how do I find dV with limited information on the system given?

 

[haruspex]

420kg/min=70kg/s

No it isn't.

U=V+dV-Vrel , U is the velocity of dM, which is 0 in this case
dM/dt • Vrel = M • dV/dt

Since you do not define all your variables, I cannot tell whether these equations are correct.
Consider the mass added to the car in one second. What is its gain in momentum? What rate of change of momentum does that imply? What is the relationship between force and rate of change of momentum?

 

[i_hate_math]

No it isn't.

Since you do not define all your variables, I cannot tell whether these equations are correct.
Consider the mass added to the car in one second. What is its gain in momentum? What rate of change of momentum does that imply? What is the relationship between force and rate of change of momentum?

Okay, i see. So the force required is the new applied force applied to the car so it moves at the same constant speed. i did it again with using the chain rule:
d(MV)/dt=V • dM/dt + M • dV/dt, where M • dV/dt is the thrust of the car before mass started to vary. then the additional force required is just V • dM/dt = 4.5 • 7 = 32.5 N

 

[i_hate_math]

Okay, i see. So the force required is the new applied force applied to the car so it moves at the same constant speed. i did it again with using the chain rule:
d(MV)/dt=V • dM/dt + M • dV/dt, where M • dV/dt is the thrust of the car before mass started to vary. then the additional force required is just V • dM/dt = 4.5 • 7 = 32.5 N

Does that look alright?

 

[haruspex]

Okay, i see. So the force required is the new applied force applied to the car so it moves at the same constant speed. i did it again with using the chain rule:
d(MV)/dt=V • dM/dt + M • dV/dt, where M • dV/dt is the thrust of the car before mass started to vary. then the additional force required is just V • dM/dt = 4.5 • 7 = 32.5 N

Small arithmetic error, but otherwise fine.

 

[i_hate_math]

Small arithmetic error, but otherwise fine.

aw bugger me its 31.5. thanks heaps

 

[i_hate_math]

Small arithmetic error, but otherwise fine.

Hi mate, could you please help me out with another question?? Here's the link to it
https://www.physicsforums.com/threads/challenging-question-on-impulse-or-momentum.864888/