How Do You Calculate the Motion and Velocity of a Falling Object with Drag?

[keelejody]

i have a question but no mark scheme so i can't see where I am going wrong. a mass, m, is dropped with speed zero from point O at time t=0 after time t it has traveled x. the body is subject to acceleration due to gravity and drag -mkv.

(A) write the equation of motion:

ok so i know v=dx/dt

and I've said f=m(dv/dt)

so f=m(dv/dt)=-mg-mkv? because theyre opposite

I can't think what else to write, since this is 5 marks... unless i need sort this in terms of ODE's where g and k are constants?

(B) calculate velocity as a function of time, and the limiting velocity at very large time.

so i need v(t)? from v=dx/dt and this is where i use ODE's and work out V.

again i can't see where the marks come from this is worth 8.

 

[berkeman]

i have a question but no mark scheme so i can't see where I am going wrong. a mass, m, is dropped with speed zero from point O at time t=0 after time t it has traveled x. the body is subject to acceleration due to gravity and drag -mkv.

(A) write the equation of motion:

ok so i know v=dx/dt

and I've said f=m(dv/dt)

so f=m(dv/dt)=-mg-mkv? because theyre opposite

I can't think what else to write, since this is 5 marks... unless i need sort this in terms of ODE's where g and k are constants?

(B) calculate velocity as a function of time, and the limiting velocity at very large time.

so i need v(t)? from v=dx/dt and this is where i use ODE's and work out V.

again i can't see where the marks come from this is worth 8.

The accelerating force and the drag force are in opposite directions, so they should have opposite signs.

 

[gneill]

If you can write an expression for the total acceleration, you'll be away to the races.

 

[ashishsinghal]

so f=m(dv/dt)=-mg-mkv? because theyre opposite

drag is opposite to velocity. velocity is in direction of g. Right. hence you cannot write m(dv/dt)=-mg-mkv. also dv/dt is +ve, this means
m(dv/dt)= mg - mkv

 

[rwisz]

(A) The equation of motion for this scenario can be written as:

m(d^2x/dt^2) = -mg - mkv

This equation takes into account the acceleration due to gravity (-mg) and the drag force (-mkv), where k is a constant and v is the velocity of the mass at any given time.

(B) To calculate the velocity as a function of time, we can rearrange the equation to solve for v:

(d^2x/dt^2) = (-mg - mkv)/m

Using the chain rule, we can rewrite this as:

(dv/dt)(dx/dt) = (-mg - mkv)/m

Substituting v=dx/dt, we get:

(dv/dt)v = (-mg - mkv)/m

This is a separable differential equation that can be solved by integrating both sides with respect to t:

∫(dv/dt)v dt = ∫(-mg - mkv)/m dt

Integrating the left side with respect to v and the right side with respect to t, we get:

(v^2)/2 = -gt - (kv^2)/2 + C

Where C is an integration constant. Solving for v, we get:

v = √(2C/m - 2gt)/(1 + k)

To find the limiting velocity at very large time, we can take the limit as t approaches infinity:

lim v = √(2C/m)/1 = √(2C/m)

Since C is an integration constant, we can write it as C = v0^2, where v0 is the initial velocity (which is zero in this scenario). Therefore, the limiting velocity at very large time is:

v_lim = √(2v0^2/m) = √(2mg/k)

This means that as time goes on, the mass will approach a constant velocity of √(2mg/k).