Why does substitution F = ma in this problem not work?

[Nano-Passion]

We have a gravitational force on Planet X $$F=mγy^2$$ and we want to know the particle's final velocity. I know how to get the right answer, but I am wondering how come this doesn't work.

$$F = mγy^2$$
$$ma = mγy^2$$
$$∫a dt= γ∫y^2 dt$$
Integral from 0 to t, I take v_0 = 0 y_0=0
$$v = γy^3/3$$

 

[AlephZero]

You don't say what y is, but shouldn't a gravitational force be inversely proportional to some distance squared, not directly proportional?

 

[Nano-Passion]

You don't say what y is, but shouldn't a gravitational force be inversely proportional to some distance squared, not directly proportional?

Yes it should. But that isn't the case here. I'll quote the book:

"Near to the point where I am standing on the surface of Planet X, the graviational force on a mass m is a vertically down but has magnitude mγy^2, where γ is a constant and y is the mass's height above the horizontal ground...although most unusual [the gravitational force] is still conservative."

 

[Pengwuino]

You're integrating the position over the time variable. Your integral looks like

$\int y(t)^2 dt$

in reality and you don't know what the functional dependence of y(t) is. What you did was basically:

$\int y^2 dy$

which is very different from the first integral.

 

[Nano-Passion]

You're integrating the position over the time variable. Your integral looks like

$\int y(t)^2 dt$

in reality and you don't know what the functional dependence of y(t) is. What you did was basically:

$\int y^2 dy$

which is very different from the first integral.

Oh, thanks! Your right, I have absolutely no idea how y changes as a function of time, it all makes sense now.

 

[jhae2.718]

Here's a trick if $a = \ddot{y}$:
$$\begin{align*} m\ddot{y} &= m\gamma{}y^2 \\ \ddot{y} &= \gamma{}y^2 \\ \frac{\mathrm{d}\dot{y}}{\mathrm{d}t} &= \gamma{}y^2 \\ \frac{\mathrm{d}\dot{y}}{\mathrm{d}y}\frac{\mathrm{d}y}{\mathrm{d}t} &= \gamma{}y^2 \\ \frac{\mathrm{d}\dot{y}}{\mathrm{d}y}\dot{y} &= \gamma{}y^2 \\ \int_{\dot{y}_0}^{\dot{y}}\dot{y}\mathrm{d}\dot{y} &= \int_{y_0}^y\gamma{}y^2\mathrm{d}y \\ \frac{\dot{y}^2 - \dot{y}_0^2}{2} &= \gamma{}\frac{y^3 - y_0^3}{3} \\ \dot{y} &= \sqrt{\dot{y}_0^2 + \frac{2\gamma}{3}\left(y^3-y_0^3\right)} \end{align*}$$

 

[Nano-Passion]

Here's a trick if $a = \ddot{y}$:
$$\begin{align*} m\ddot{y} &= m\gamma{}y^2 \\ \ddot{y} &= \gamma{}y^2 \\ \frac{\mathrm{d}\dot{y}}{\mathrm{d}t} &= \gamma{}y^2 \\ \frac{\mathrm{d}\dot{y}}{\mathrm{d}y}\frac{\mathrm{d}y}{\mathrm{d}t} &= \gamma{}y^2 \\ \frac{\mathrm{d}\dot{y}}{\mathrm{d}y}\dot{y} &= \gamma{}y^2 \\ \int_{\dot{y}_0}^{\dot{y}}\dot{y}\mathrm{d}\dot{y} &= \int_{y_0}^y\gamma{}y^2\mathrm{d}y \\ \frac{\dot{y}^2 - \dot{y}_0^2}{2} &= \gamma{}\frac{y^3 - y_0^3}{3} \\ \dot{y} &= \sqrt{\dot{y}_0^2 + \frac{2\gamma}{3}\left(y^3-y_0^3\right)} \end{align*}$$

I like that trick! I recall a professor of mine showed it to me last year when I was trying to solve a problem. I know it is separation of variables and playing around with the math, but does this trick have a specific name?

 

[mfb]

The left part is basically the kinetic energy (and its time-derivative before the integration), so it is related to energy conservation.

$F=m\gamma y^2$ gives a potential of
$V(y)=\frac{m\gamma}{3} y^3$

The total kinetic energy $\frac{m}{2}\dot{y}^2 + V(y)$ is conserved, write it down, and you get the same result without integration.

 

[Nano-Passion]

The left part is basically the kinetic energy (and its time-derivative before the integration), so it is related to energy conservation.

$F=m\gamma y^2$ gives a potential of
$V(y)=\frac{m\gamma}{3} y^3$

The total kinetic energy $\frac{m}{2}\dot{y}^2 + V(y)$ is conserved, write it down, and you get the same result without integration.

Thanks, that is how the textbook tells you to solve it as well, that is to use conservation of energy. But I just wanted to try something different and was wondering why it didn't work.

I didn't realize that you can use ## for latex, thanks a lot!