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# Calculating a ball of unit mass' velocity. watch

1. A ball of unit mass is constrained to roll along a straight track in such a way that overall
it experiences a resistive force which is inversely proportional to its velocity. Given that its
initial velocity c is positive, show that the velocity at some later time t has the form:

for some constant . Find an expression for the position of the particle at time t, and hence prove that the ball travels a distance
before coming to rest, when the model breaks down.

The answer for the first part of the question kind of makes it look like I have to use the constant acceleration formula , however I'm not sure how I would get the acceleration from this . Please help!
2. If the resistive force is inversely proportional to its velocity, it has the form -c/v for some positive constant c.

In which case, what is dv/dt? You get an easy differential equation. Solve it, and you'll get the first part out.
3. (Original post by Happy2Guys1Hammer)
A ball of unit mass is constrained to roll along a straight track in such a way that overall
it experiences a resistive force which is inversely proportional to its velocity. Given that its
initial velocity c is positive, show that the velocity at some later time t has the form:

for some constant . Find an expression for the position of the particle at time t, and hence prove that the ball travels a distance
before coming to rest, when the model breaks down.

The answer for the first part of the question kind of makes it look like I have to use the constant acceleration formula , however I'm not sure how I would get the acceleration from this . Please help!
Tne ac(dec)celeration is function of v with factor.

so

Solve for v
t=0 v=c

for the position

solve s with t=0 s=0
4. (Original post by ztibor)
Tne ac(dec)celeration is function of v with factor.

so

Solve for v
t=0 v=c

for the position

solve s with t=0 s=0
Okay so I've done:
. Is this right? I can't see where c comes into this.

Also, could you please explain why "" as I can't see the transition there.
5. (Original post by Happy2Guys1Hammer)
Okay so I've done:
. Is this right? I can't see where c comes into this.

Also, could you please explain why "" as I can't see the transition there.
You haven't included a constant of integration!

You're told that resistive force is inversely proportional to speed i.e. Fres = -B/v for some constant B. Also use the general Newton's 2nd law F = ma where a is acceleration and F is the total force acting.
6. (Original post by davros)
You haven't included a constant of integration!

You're told that resistive force is inversely proportional to speed i.e. Fres = -B/v for some constant B. Also use the general Newton's 2nd law F = ma where a is acceleration and F is the total force acting.
Okay but if I use a constant of integration, I will just get +c or whatever, but that's not the same c that v is equal to surely. Also I need it to be c^2. but it'll just be c. There's also a 1/2 in there which isn't in the final answer and there's no constant in the final answer. Are you sure this is the correct method because It doesn't seem to take shape to me.
7. (Original post by Happy2Guys1Hammer)
Okay but if I use a constant of integration, I will just get +c or whatever, but that's not the same c that v is equal to surely. Also I need it to be c^2. but it'll just be c. There's also a 1/2 in there which isn't in the final answer and there's no constant in the final answer. Are you sure this is the correct method because It doesn't seem to take shape to me.
You can always rescale your constants so k -> 2k (or k -> k/2). The c^2 is inside a square root sign, so when you put in t = 0 you get v = c whjch is the required initial condition.
8. (Original post by davros)
You can always rescale your constants so k -> 2k (or k -> k/2). The c^2 is inside a square root sign, so when you put in t = 0 you get v = c whjch is the required initial condition.
Oh okay I understand that now. Thank you very much .
9. (Original post by davros)
You can always rescale your constants so k -> 2k (or k -> k/2). The c^2 is inside a square root sign, so when you put in t = 0 you get v = c whjch is the required initial condition.
Okay so I've looked back through my answer and I really don't think I did the method correctly.

Here's what I've got:
. As .

Letting .

This doesn't quite seem right to me, is it correct?
10. (Original post by Happy2Guys1Hammer)
Okay so I've looked back through my answer and I really don't think I did the method correctly.

Here's what I've got:
. As .

Letting .

This doesn't quite seem right to me, is it correct?
It looks about right, but to make it more "convincing" I'd use something like A and B for your constant of proportionality and your constant of integration respectively, and then use the boundary conditions and the factor of 2 to get c and , rather than starting with the constants stated in the question and then making it look like you're fudging things by changing their values

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Updated: May 18, 2013
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