Challenging Mechanics Questions: Help Needed!

[Auron87]

We have been given a set of questions to do. Most of these I have managed fairly easily but there are some which I am really stuck on and am getting quite annoyed with now as I have been working at them for ages! Help on any of these would be really appreciated:

The Saturn V rocket which launched the Apollo space missions had the following specifications:
mass at lift-off = 3.0 x 10^6 kg
velocity of exhaust gases = 1.0 x 10^4 m/s
initial rate of fuel consumption at lift-off = 3.0 x 10^3 kg/s

1.(a) Calculate:
(i) the force (thrust) produced at lift-off
(ii) the resultant force acting on the rocket at lift-off

2.(a) An empty railway truck of mass 10000kg is traveling horizontally at a speed of 0.5m/s. (had to calculated the momentum and kinetic energy - not sure if this will help with the bit I am stuck with!)
Sand falls vertically into the truck at a constant rate of 40kg/s. Calculate the additional horizonal force which must be applied to the truck if it is to maintain a steady speed of 0.5m/s.

Thanks for anyones help - I can usually manage these kind of questions but I'm really stuck for some reason!

 

[ShawnD]

Originally posted by Auron87
The Saturn V rocket which launched the Apollo space missions had the following specifications:
mass at lift-off = 3.0 x 10^6 kg
velocity of exhaust gases = 1.0 x 10^4 m/s
initial rate of fuel consumption at lift-off = 3.0 x 10^3 kg/s

1.(a) Calculate:
(i) the force (thrust) produced at lift-off

Ft = mv

Derive both sides

F = (\frac{dm}{dt})(v)

(ii) the resultant force acting on the rocket at lift-off

I think this means to include gravity. It should be whatever you got from part (i) - gravity.

2.(a) An empty railway truck of mass 10000kg is traveling horizontally at a speed of 0.5m/s. (had to calculated the momentum and kinetic energy - not sure if this will help with the bit I am stuck with!)
Sand falls vertically into the truck at a constant rate of 40kg/s. Calculate the additional horizonal force which must be applied to the truck if it is to maintain a steady speed of 0.5m/s.

Ft = mv

Derive both sides

F = (\frac{dm}{dt})(v)

 

[Auron87]

I've not met that formula before but I do understand where its from but how can you use it because you don't know time or is it necessary to work out the time or something??

 

[ShawnD]

dm/dt is a variable, not a division of 2 or 4 variables

In the first question, dm/dt is 3.0 x 10^3 kg/s

In the second question, dm/dt is 40kg/s