Simple calc based motion in a plane Question

[zoxee]

Question:

http://gyazo.com/25303d1c24332bb51afb926867f04c41

My attempt thus far (i'm very new to physics so bare with me:

Take h to be the distance from where the ball is thrown up till the maximum point before it starts to decrease

when the object is at terminal velocity the acceleration is zero so:

using f = ma, we get $0 = kv_t^2 -mg$ $v_t^2 = mg/k$

initially using f=ma again:

$mv_0 \dfrac{dv_0}{dh} = -mg -kv_0^2$ rearranging and integrating I get $\dfrac{-m}{2k}ln|mg+kv_0^2| = h + c$ when h = 0 v_0 = 0 so: $c = \dfrac{-m}{2k}ln(mg)$ therefore I get $-\dfrac{m}{2k} ln(mg+kv_0^2) = h -\dfrac{m}{2k} ln(mg)$

when it returns using f = ma again:

$-mv_1 \dfrac{dv_1}{dh} = kv_1^2 -mg$ rearranging and integrating I get: $\dfrac{m}{2k} ln|kv_1^2 -mg| = h + D$ where D is a constant.

I am not sure how to find the constant D, or if what I am doing is correct at all.. any help would be good thank you

 

[Simon Bridge]

Just adjusting your notation there (i.e. $\small v_0$ is the initial veocity - so $\small dv_0/dt=0$)

... going up: $$v\frac{dv}{dy}=-g-\frac{k}{m}v^2\; : \; v(0)=v_0 \qquad ...(1)$$ ... going down $$v\frac{dv}{dy}=-g+\frac{k}{m}v^2\; :\; v(0)=v_1\qquad ...(2)$$

... that should clarify things.

Since you know $v_t^2=mg/k$, you can sub that in: $$\frac{k}{m}=\frac{g}{v_t^2}$$
Presumably you solved by changing variables etc.
If you define the maximum height reached to be y=h, then v(h)=0 :)

 

[zoxee]

Just adjusting your notation there (i.e. $\small v_0$ is the initial veocity - so $\small dv_0/dt=0$)

... going up: $$v\frac{dv}{dy}=-g-\frac{k}{m}v^2\; : \; v(0)=v_0 \qquad ...(1)$$ ... going down $$v\frac{dv}{dy}=-g+\frac{k}{m}v^2\; :\; v(0)=v_1\qquad ...(2)$$

... that should clarify things.

Since you know $v_t^2=mg/k$, you can sub that in: $$\frac{k}{m}=\frac{g}{v_t^2}$$
Presumably you solved by changing variables etc.
If you define the maximum height reached to be y=h, then v(h)=0 :)

im confused by your notation. For instance, when you put $v(0) = v_0$ does that mean I sub in 0 for all the v's? If so, where does the v_0 go? Also, how do you know $v(0) = v_1$

 

[Simon Bridge]

im confused by your notation. For instance, when you put $v(0) = v_0$ does that mean I sub in 0 for all the v's? If so, where does the v_0 go? Also, how do you know $v(0) = v_1$

There's a clue in that $dv/dy$ ... tells you that v is a function of y, where y is the height from the initial position.

There are two cases where y=0, once at the beginning of the trip, and once at the end.

 

[zoxee]

OK, here is my working, I got the correct answer:

considering the upward motion, and taking y to be positive when going up:

we have $mv \dfrac{dv}{dy} = -mg -kv^2$ re arranging we get $\dfrac{v}{-g-\frac{k}{m}v^2}dv = dy$ so $\frac{-m}{2k}log |-g -\frac{k}{m}v^2 | = y + c$ now taking this integral from $v_0$ to 0 (the highest point): $\left[\frac{-m}{2k}log |-g -\frac{k}{m}v^2 | \right ]_{v_0}^0 = \left[y \right ]_0^h$ thus $h = \frac{m}{2k} ln\left | \dfrac{-g -\frac{k}{m}v_0^2}{g} \right|$

now considering the downward motion:

$mv \dfrac{dv}{dy} = -mg + kv^2$ so $\dfrac{v dv}{-g + \frac{k}{m}v^2} = dy$ and integrating with limits, i.e. from the maximum height h, to 0, from v = 0 to v = v_1 : $\left [\frac{m}{2k} ln \left| -g + \frac{k}{m} v^2 \right| \right]_0^{v_1} = \left [y \right]_h^0$

so $-h = \frac{m}{2k} \left (ln \left | \dfrac{-g + \frac{k}{m}v_1^2}{g} \right | \right )$

so equating these h's we get $\frac{m}{2k} \left (ln \left | \dfrac{-g + \frac{k}{m}v_1^2}{g} \right | \right ) = -\frac{m}{2k} ln\left | \dfrac{-g -\frac{k}{m}v_0^2}{g} \right|$ taking the exponentials of both sides and using the fact that $\frac{k}{m} = \frac{g}{v_t^2}$ we get $-1 + \dfrac{v_1^2}{v_t^2} = \dfrac{1}{-1 - \frac{v_0^2}{v_t^2}}$ which goes on to give the required result

I still have a question however: When I considered the downward motion I was taking upwards to be positive, so it's obvious the acceleration is negative. If I use $mv \dfrac{dv}{dy} = -mg + kv^2$ it works and I get the required answer but if I use $-mv \dfrac{dv}{dy} = -mg + kv^2$ I don't get the correct answer. Why do I not put in the negative sign for acceleration?

 

[Dick]

OK, here is my working, I got the correct answer:

considering the upward motion, and taking y to be positive when going up:

we have $mv \dfrac{dv}{dy} = -mg -kv^2$ re arranging we get $\dfrac{v}{-g-\frac{k}{m}v^2}dv = dy$ so $\frac{-m}{2k}log |-g -\frac{k}{m}v^2 | = y + c$ now taking this integral from $v_0$ to 0 (the highest point): $\left[\frac{-m}{2k}log |-g -\frac{k}{m}v^2 | \right ]_{v_0}^0 = \left[y \right ]_0^h$ thus $h = \frac{m}{2k} ln\left | \dfrac{-g -\frac{k}{m}v_0^2}{g} \right|$

now considering the downward motion:

$mv \dfrac{dv}{dy} = -mg + kv^2$ so $\dfrac{v dv}{-g + \frac{k}{m}v^2} = dy$ and integrating with limits, i.e. from the maximum height h, to 0, from v = 0 to v = v_1 : $\left [\frac{m}{2k} ln \left| -g + \frac{k}{m} v^2 \right| \right]_0^{v_1} = \left [y \right]_h^0$

so $-h = \frac{m}{2k} \left (ln \left | \dfrac{-g + \frac{k}{m}v_1^2}{g} \right | \right )$

so equating these h's we get $\frac{m}{2k} \left (ln \left | \dfrac{-g + \frac{k}{m}v_1^2}{g} \right | \right ) = -\frac{m}{2k} ln\left | \dfrac{-g -\frac{k}{m}v_0^2}{g} \right|$ taking the exponentials of both sides and using the fact that $\frac{k}{m} = \frac{g}{v_t^2}$ we get $-1 + \dfrac{v_1^2}{v_t^2} = \dfrac{1}{-1 - \frac{v_0^2}{v_t^2}}$ which goes on to give the required result

I still have a question however: When I considered the downward motion I was taking upwards to be positive, so it's obvious the acceleration is negative. If I use $mv \dfrac{dv}{dy} = -mg + kv^2$ it works and I get the required answer but if I use $-mv \dfrac{dv}{dy} = -mg + kv^2$ I don't get the correct answer. Why do I not put in the negative sign for acceleration?

$mv \dfrac{dv}{dy}=ma=F$ regardless of whether you are going up or down, you don't need to artificially correct the sign. The only difference between going up and down is when you are going up, the frictional force points in the same direction as gravity, and when you going down it points in the opposite direction. Nice job solving this! I thought it was way too hard for someone "very new to physics" - but I see I was wrong.

 

[zoxee]

$mv \dfrac{dv}{dy}=ma=F$ regardless of whether you are going up or down, you don't need to artificially correct the sign. The only difference between going up and down is when you are going up, the frictional force points in the same direction as gravity, and when you going down it points in the opposite direction. Nice job solving this! I thought it was way too hard for someone "very new to physics" - but I see I was wrong.

Oh I see, I thought you had to change the sign originally. Guess not.

Thanks, I'm actually a mathematics major taking some calc based physics classes.

 

[Simon Bridge]

Well done.
I had a feeling you were a math major.
Physics can be a good way to learn calc - it concretizes the process.

The relation is F=ma
If there were no air friction, you'd have $$\frac{dv}{dt}=-mg: v(0)=v_0$$ ... for "up" = positive. (Use the chain rule to get it in terms of dv/dx)

There is no need to change the sign for downwards velocity because that is already what the function "v(t)" does all by itself. You needed separate math for up and down for the motion with air friction because the air friction is a piecewise function.

Which is another way of saying what Dick did :)

You could have done your work in the time domain and then done a bit of algebra to cancel off the times too... which, I suspect, is what most physics students would have done.

 

[Dick]

You could have done your work in the time domain and then done a bit of algebra to cancel off the times too... which, I suspect, is what most physics students would have done.

Yeah, trying to write the solutions as functions of time is the direct way. But I think that would have been REALLY nasty. That's why your hint not to do that is so good.