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dosvarog
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Hello everybody, this is my first time posting here, so if I make any errors, correct me please. :)
We have a body of mass 2 kg and it starts to move with constant velocity 5 m/s along a straight line. After 2 seconds a force that decreases with square of the distance starts to act on a body. Amount of the force at the moment it started to act is 10 N. After what time the speed of the body will be equal to half of its initial value? How far from the starting position body will be at that moment?
I hope everything is clear since this is rough translation from my language. :)
So, we have:
[itex]m = 2~kg[/itex] - mass of a body
[itex]v_0 = 5~m/s[/itex] - initial constant velocity
[itex]{F_x}_0 = 10~N[/itex] - amount of force that starts to act at position ##x_0##
[itex]{t_x}_0 = 2~s[/itex] - time that passes when body reaches position ##x_0##
##F \propto 1/x^2## - force that decreases with square of a distance
##x_0## - distance that body moves with constant velocity
Ok, now we can set the equation of motion. Since we have time and velocity with simple ##x_0~=~v_0*{t_x}_0## (because first 2 seconds body moves with constant velocity) we can calculate distance ##x_0## which is 10 m.
For force F we can write ##F = k/x^2##.
Given the initial amount of force and distance ##x_0## that we calculated we get that ##k = 1000~Nm^2##
Equation of motion would then be:
##\ddot{x}*m = -k/x^2##
Now the fun part. This is second-order nonlinear ordinary differential equation, which complicates things a bit.
So, basically I did this:
##\ddot{x} = -k/mx^2##
Then I made substitution ##\dot{x} = p(x)##, here p is function of x which means that ##\ddot{x} = p\dot{p}##.
That yields:
##p\frac{dp}{dx} = -a\frac{1}{x^2}## where ##a = k/m##
I separate the variables and integrate which yields:
##p^2/2 = a/x + C1##
Then I get constant C1 by plugging initial values:
##\frac{{v_0}^2}{2} - \frac{k}{mx_0} = C1 \Rightarrow C1 = -37.5##
Now I can calculate ##x_1##, a position where body has half of its initial velocity, that is 2.5 m/s:
##\frac{{\dot{x_1}}^2}{2} = \frac{k}{mx_1} - 37.5## where ##\dot{x_1} = \frac{\dot{x_0}}{2}##
which gives ##x_1 = 12.31~m## (note: ##x_1## is distance from ##x_0## because of initial conditions)
The problem is, how to get time? If I go and integrate this DE second time, I would get ##x(t)## and solve it, but the integral I have to solve is pretty nasty and I don't know how to solve it. Is there maybe some other way or some kind of catch that I'm missing here?
Homework Statement
We have a body of mass 2 kg and it starts to move with constant velocity 5 m/s along a straight line. After 2 seconds a force that decreases with square of the distance starts to act on a body. Amount of the force at the moment it started to act is 10 N. After what time the speed of the body will be equal to half of its initial value? How far from the starting position body will be at that moment?
I hope everything is clear since this is rough translation from my language. :)
So, we have:
[itex]m = 2~kg[/itex] - mass of a body
[itex]v_0 = 5~m/s[/itex] - initial constant velocity
[itex]{F_x}_0 = 10~N[/itex] - amount of force that starts to act at position ##x_0##
[itex]{t_x}_0 = 2~s[/itex] - time that passes when body reaches position ##x_0##
##F \propto 1/x^2## - force that decreases with square of a distance
##x_0## - distance that body moves with constant velocity
Homework Equations
Ok, now we can set the equation of motion. Since we have time and velocity with simple ##x_0~=~v_0*{t_x}_0## (because first 2 seconds body moves with constant velocity) we can calculate distance ##x_0## which is 10 m.
For force F we can write ##F = k/x^2##.
Given the initial amount of force and distance ##x_0## that we calculated we get that ##k = 1000~Nm^2##
Equation of motion would then be:
##\ddot{x}*m = -k/x^2##
The Attempt at a Solution
Now the fun part. This is second-order nonlinear ordinary differential equation, which complicates things a bit.
So, basically I did this:
##\ddot{x} = -k/mx^2##
Then I made substitution ##\dot{x} = p(x)##, here p is function of x which means that ##\ddot{x} = p\dot{p}##.
That yields:
##p\frac{dp}{dx} = -a\frac{1}{x^2}## where ##a = k/m##
I separate the variables and integrate which yields:
##p^2/2 = a/x + C1##
Then I get constant C1 by plugging initial values:
##\frac{{v_0}^2}{2} - \frac{k}{mx_0} = C1 \Rightarrow C1 = -37.5##
Now I can calculate ##x_1##, a position where body has half of its initial velocity, that is 2.5 m/s:
##\frac{{\dot{x_1}}^2}{2} = \frac{k}{mx_1} - 37.5## where ##\dot{x_1} = \frac{\dot{x_0}}{2}##
which gives ##x_1 = 12.31~m## (note: ##x_1## is distance from ##x_0## because of initial conditions)
The problem is, how to get time? If I go and integrate this DE second time, I would get ##x(t)## and solve it, but the integral I have to solve is pretty nasty and I don't know how to solve it. Is there maybe some other way or some kind of catch that I'm missing here?
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