Solving Ball's Wall Impact Time w/Force & Mass

[An1MuS]

Homework Statement

A ball is being pulled by a force given by $F=kv^2$. Where "k" already includes the mass of the ball. After 10m it will run into a wall. The initial speed is 1m/s and there are no other forces involved. How much time does it take for the ball to hit the wall?

Homework Equations

$kv^2=\frac{dv}{dt}$

The Attempt at a Solution

So i know that this is a differential equation of separable variables.
$\frac{dv}{v^2}=kdt$
I probably should integrate both sides, but i don't know what limits i should use if any, and if I'm on the right track

$\int \frac{dv}{v^2}=\int kdt$

 

[Chestermiller]

If the velocity at time t = 0 is equal to zero, then the ball never moves, and an infinite time is required to reach the wall. If the initial velocity is v₀, then you can use that as the initial condition.

 

[Andrew Mason]

Homework Statement

A ball is being pulled by a force given by $F=kv^2$. Where "k" already includes the mass of the ball. After 10m it will run into a wall. How much time does it take for the ball to hit the wall?

Homework Equations

$F=ma$

$kv^2=\frac{dv}{dt}$

? I thought F = kv². dv/dt = a = F/m

The Attempt at a Solution

So i know that is a differential equation of separable variables.
$\frac{dv}{v^2}=kdt$
If i probably should integrate both sides, but i don't know what limits i should use if any, and if I'm on the right track

$\int \frac{dv}{v^2}=\int kdt$

You want to express position as a function of time. You have to know its initial speed. Is the initial speed (at 10 m from the wall) = 0? Is there any other force on the ball (eg. friction)? Can you state the whole problem as given?

AM

 

[gneill]

Don't forget to include the mass when you replace F with ma. Yes, k contains the mass, but it's a separate equation from F = ma. Thus $m a = k v^2$., and so $a = (k/m)v^2$.

 

[An1MuS]

? I thought F = kv². dv/dt = a = F/m
You want to express position as a function of time. You have to know its initial speed. Is the initial speed (at 10 m from the wall) = 0? Is there any other force on the ball (eg. friction)? Can you state the whole problem as given?

AM

The whole problem as given is actually a fluid mechanics problem where a ball is dropped and the drag force increases with speed squared. I just stated this one, because if i know how to solve this simpler case i should know how to solve the fluid mechanics one.

You guys are right, I updated the OP with the info you asked.

 

[Andrew Mason]

The whole problem as given is actually a fluid mechanics problem where a ball is dropped and the drag force increases with speed. I just stated this one, because if i know how to solve this simpler case i should know how to solve the fluid mechanics one.

You guys are right, I updated the OP with the info you asked.

If it is a drag force, the acceleration is negative. Are you sure that F = kv² and not F = -kv²?

AM

 

[An1MuS]

If it is a drag force, the acceleration is negative. Are you sure that F = kv² and not F = -kv²?

AM

In the fluid problem it's negative, but for the sake of learning, i put it positive to be simpler. Then it's just a matter of me putting a minus sign in the fluid exercise where it's needed. I made up this exercise avoiding fluid mechanics jargon so that more people could understand (and reply to) the exercise, and also because i just need to understand the underlying principle of solving this (simpler) exercise for me to solve the fluid mechanics one. Hope this makes sense...

 

[Arkavo]

If the velocity at time t = 0 is equal to zero, then the ball never moves, and an infinite time is required to reach the wall. If the initial velocity is v₀, then you can use that as the initial condition.

absolutely correct if initial velocity is 0 the force never gets a chance to act

 

[Andrew Mason]

In the fluid problem it's negative, but for the sake of learning, i put it positive to be simpler. Then it's just a matter of me putting a minus sign in the fluid exercise where it's needed. I made up this exercise avoiding fluid mechanics jargon so that more people could understand (and reply to) the exercise, and also because i just need to understand the underlying principle of solving this (simpler) exercise for me to solve the fluid mechanics one. Hope this makes sense...

You have to find the solution to this differential equation:

dv/dt = kv²/m

dv/v² = kdt/m

That should be easily solved using integration, using the value for the initial speed at t=0.AM

 

[An1MuS]

You have to find the solution to this differential equation:

dv/dt = kv²/m

dv/v² = kdt/m

That should be easily solved using integration, using the value for the initial speed at t=0.AM

"m" should already be inside k.

$kt=\frac {-1}{v_f}- (-\frac {1}{v_i})$ where $v_i=1$ But then where do the 10meters come in? And what about $v_f$?

 

[Chestermiller]

You should really be integrating with respect to x, rather than with respect to t, by substituting v(dv/dx) for dv/dt. That way you can find the velocity directly at x = 10.

 

[Andrew Mason]

"m" should already be inside k.

What does that mean? You have written F = kv². F = ma so a = kv²/m unless you want to dispute Newton's second law.

$kt=\frac {-1}{v_f}- (-\frac {1}{v_i})$ where $v_i=1$ But then where do the 10meters come in? And what about $v_f$?

Take Chestermiller's suggestion and integrate with respect to distance.

AM

 

[An1MuS]

What does that mean? You have written F = kv². F = ma so a = kv²/m unless you want to dispute Newton's second law. Take Chestermiller's suggestion and integrate with respect to distance.

AM

Y, you're right. So it should be

$F=mkv^2$ so that it gives $kv^2 = \frac {dv}{dt}$

Integrating in respect to x, is this it?

$\int kv^2 dx = \int v\frac{dv}{dx}dx$

And if it's this i don't know how to continue

Also i don't understand this substitution $\frac{dv}{dt}=v\frac{dv}{dx}$

 

[Chestermiller]

Y, you're right. So it should be

$F=mkv^2$ so that it gives $kv^2 = \frac {dv}{dt}$

Integrating in respect to x, is this it?

$\int kv^2 dx = \int v\frac{dv}{dx}dx$

And if it's this i don't know how to continue

Also i don't understand this substitution $\frac{dv}{dt}=v\frac{dv}{dx}$

No. That's not it. If

$kv^2 = v\frac{dv}{dx}$

then you can cancel a v out from both sides of the equation, and obtain:

$kv = \frac{dv}{dx}$

The solution to this differential equation is

$v=v_0e^{kx}$

The substitution $\frac{dv}{dt}=v\frac{dv}{dx}$ comes from $v=\frac{dx}{dt}$ so that $dt=\frac{dx}{v}$

$\frac{dx}{dt}=v_0e^{kx}$

So, dt=\frac{e^{-kx}}{v_0}dx
So, t=\frac{1-e^{-kx}}{kv_0}

 

[Andrew Mason]

Y, you're right. So it should be

$F=mkv^2$ so that it gives $kv^2 = \frac {dv}{dt}$

Integrating in respect to x, is this it?

$\int kv^2 dx = \int v\frac{dv}{dx}dx$

And if it's this i don't know how to continue

Also i don't understand this substitution $\frac{dv}{dt}=v\frac{dv}{dx}$

You have:

F = md²x/dt² = mdv/dt = mkv²

That can be rearranged to:

dv/v² = kdtSince v = dx/dt, dt = dx/v, so this becomes:

dv/v = kdx

Find the solution by integration and solve for x_f-x_i = 10m.

AM