# Thread: A 1D Force Problem

1. ## A 1D Force Problem

Hi, this came up as a practice problem for a midterm. Assistance would be awesome.

Thanks!

A particle of mass m starts from rest at a distance d from the origin and subsequently moves towards the origin due to the presence of an attractive inverse square law force F = -(k/x^2)i. When does the particle reach the origin?

I'm thinking of starting with F=ma, but I don't know where to go from there.

2. ## Acceleration

Hello calculusfrk
Originally Posted by calculusfrk
Hi, this came up as a practice problem for a midterm. Assistance would be awesome.

Thanks!

A particle of mass m starts from rest at a distance d from the origin and subsequently moves towards the origin due to the presence of an attractive inverse square law force F = -(k/x^2)i. When does the particle reach the origin?

I'm thinking of starting with F=ma, but I don't know where to go from there.
I haven't time for a complete answer here, but to start off, I think you'll need to use acceleration = $v\frac{dv}{dx}$, so that you can separate the variables. So $F = ma$ gives

$-\frac{k}{x^2}=mv\frac{dv}{dx}$

$\Rightarrow -k\int\frac{dx}{x^2}=m\int v\,dv$

This integrates OK. Then use $x=d, v=0$ to find the constant of integration. Then use $v = \frac{dx}{dt}$; separate the variables again, and (provided the integration is then OK) you should be there.

3. ## there you go...

F = ma = -k/x^2
==> a = -k/(m*x^2) ---------------eq1

from kinematics, distance x travelled in time t
x = u*t + (1/2)*a*t^2 where u = 0
==> x = (1/2)*a*t^2 = (1/2)* (-k/(m*x^2)) *t^2 (using eq1)
==> x = (-k * t^2 / (2*m) )^(1/3) ----eq2

again from kinematics, definition of velocity
dx/dt = v and v = u + a*t where u = 0
==> integral (dx) = integeral(t*v dt) ( limits from t = 0 to t = t1, x from -d to 0)
==> ingegral dx = integeral( - (1/2)^(2/3) * (k/m)^(1/3) t^(-1/3) dt )
unpon integration and applying limits that at t= 0 x = -d and t = t1 x = 0)
you get t1 = (2d/3)^(3/2) * (m/k)^2........

Please check the mathematics again, my answer might be wrong but the procedure is right.... the proceduer is what you need to understand.....Please get back to me if you dont understand any thing.....

4. Originally Posted by vinay
F = ma = -k/x^2
==> a = -k/(m*x^2) ---------------eq1

from kinematics, distance x travelled in time t
x = u*t + (1/2)*a*t^2 where u = 0
==> x = (1/2)*a*t^2 = (1/2)* (-k/(m*x^2)) *t^2 (using eq1)
==> x = (-k * t^2 / (2*m) )^(1/3) ----eq2

again from kinematics, definition of velocity
dx/dt = v and v = u + a*t where u = 0
==> integral (dx) = integeral(t*v dt) ( limits from t = 0 to t = t1, x from -d to 0)
==> ingegral dx = integeral( - (1/2)^(2/3) * (k/m)^(1/3) t^(-1/3) dt )
unpon integration and applying limits that at t= 0 x = -d and t = t1 x = 0)
you get t1 = (2d/3)^(3/2) * (m/k)^2........

Please check the mathematics again, my answer might be wrong but the procedure is right.... the proceduer is what you need to understand.....Please get back to me if you dont understand any thing.....
Vinay, the suvat equations you used here are for constant acceleration problems so the procedure is NOT correct.

5. ## oops my apologies to all

oops my apologies to all

Thank you "the doc" for pointing out

Let me check and come back

6. ## Further to Grandad's post

Grandad was correct in his method.

An alternative approach that would not require integrating the acceleration term would be to use the physical grounds that as the particle starts at rest it's final KE must be equal to the PE it gains in the 'fall'. Hence you would arrive at the same equation as grandad with

$\frac{1}{2} m v^2 = - \int_{d}^{x} \frac{k}{{x'}^2} \, \mathrm{d} x'$

i.e.

$\frac{1}{2} m v^2 =k \left(\frac{1}{x} - \frac{1}{d} \right)$

where $v$ is the velocity at $x$ and $x'$ is simply a dummy variable in place of $x$ so we can use the limits to take care of the constant of integration.

So rearranging the above equation we have that

$\sqrt{ \frac{x}{d-x} } \frac{dx}{dt} = \sqrt{ \frac{2k}{md} }$

thus integrating wrt the dummy variable $t'$ in place of $t$ using the limits $x'=d$ at $t'=0$ and $x'=0$ at $t' = t$ we have that

$\int_{d}^{0} \sqrt{ \frac{x'}{d-x'} } \, \mathrm{d}x' = \int_{0}^{t} \sqrt{ \frac{2k}{md} } \, \mathrm{d} t' = \sqrt{ \frac{2k}{md} } t$.

Now to deal with the integral wrt $x'$ let $x' = d \sin^2 \theta$ so

$\mathrm{d} x' = 2 d \sin \theta \cos \theta \, \mathrm{d} \theta$

hence

$\sqrt{ \frac{2k}{md} } t = d \int_{-\frac{\pi}{2}}^{0} 2 \sin^2 \theta \, \mathrm{d} \theta$

$\iff \sqrt{ \frac{2k}{md} } t = d \int_{-\frac{\pi}{2}}^{0} \left(1- \cos 2 \theta \right) \, \mathrm{d} \theta = \frac{\pi}{2} d$.

Finally this gives us that

$\color[rgb]{0,0,1} \boxed{t = \frac{\pi d}{2} \sqrt{\frac{md}{2k}}}$ .

I'll leave it to you to check in case I made any algebraic errors!

7. ## algebric correction

There is a small error in the above algebric solution

Here is the right one . t = 0.5 * pi * d * sqrt ( m*d / (2*k) )

Which can also be checked by dimentional analysis.
d = [L]
m =[M]
k = [ML3T-2]

RHS = [T]
LHS = [L] * [M L / (ML3T-2)]^.5 = [T]

So the right answer would be ans given by the don * sqrt(d)

I really appreciate the physical understanding of "the doc"

8. Yes, I was being lazy as I don't transcribe from paper and couldn't be bothered checking the algebra hence the deliberate caveat at the end asking for someone else to check the solution for algebraic errors.

So thanks Vinay!