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For an introduction to the Curiosity Questions, please see the intro to Curiosity Question 1
Here is the first part of another problem I made up.
As I said in Question 1, I am posting this for
a) the enjoyment of everyone who reads them
b) work checking to see if they are riddled with mistakes or not
c) approach analysis, asking people to give me feedback on how I worked it.
Mass m oscillates on a spring of spring constant k, at an amplitude A. There is no kinetic friction, but static friction of Us (Us mg is less than kA)
A) Find the minimum speed with which a mass 0.25 m can be launched against m for it to stop after the elastic collision at point x.
B) If the 1/4 m is located 1 meter from the natural length of the spring, when should it be launched?
There are more parts but I will post them later.
I)##F=-kx##
II) ##F_{maxfriction}=Us mg##
III) KE is conserved in an elastic collision
##\frac{1}{2}(1/4m)v^2=\frac{1}{2}mv_m^2##
IV) Within the spring system, total energy is conserved and transferred between kinetic and potential
##\frac{1}{2}kA^2=\frac{1}{2}mv_{m max}^2=\frac{1}{2}mv_m^2+\frac{1}{2}kx^2##
V) ##v=dx/dt ##
or x/t for constant v
Unknowns:
x, vm, and v (I use v for the velocity of 1/4 m, and vm for the velocity of m on the spring)
For the mass to be stopped by the collision and to be hung up on the static friction, the kinetic energy of the 1/4 m must equal the kinetic energy of the mass on the spring at x.
For the KE and velocity of 1/4 m to be at minimum, the KE of the mass on the spring must be at its minimum, so x must be as close to A as possible, and as much energy as possible should be stored in the spring as potential energy. Thus, ##kx = Us mg##, so the maximum x = ##\frac{Us mg}{k}##.
Combining III and IV
Total energy = ##\frac{1}{2}kA^2=(\frac{1}{2}(\frac{1}{4}m)v^2) + \frac{1}{2}kx^2##
Combining above with I and II
##\frac{1}{2}kA^2=(\frac{1}{8}mv^2) + \frac{1}{2}k(\frac{Us mg}{k})^2##
Simplifying
##kA^2=(\frac{1}{4}mv^2) + \frac{(Us mg)^2}{k}##
Solving for v
##v=\sqrt{4\frac{kA^2-\frac{(Us mg)^2}{k}}{m}}##
or
##v=2\sqrt{\frac{kA^2}{m}-\frac{m(Us)^2(g)^2}{k}}##
B)
Now that I have the velocity, it is not hard to find the time at which the mass should be released when it is placed at 1 m away.
##d=1+x##
From above:
##x=\frac{Us mg}{k}##
(x also = ##\sqrt{A^2-\frac{mv^2}{k}}##--I checked)
So ##t=\frac{d}{v}=\frac{1+\frac{Us mg}{k}}{2\sqrt{\frac{kA^2}{m}-\frac{m(Us)^2(g)^2}{k}}} ##
That is about all the latex I can do right now, but in the next post, I will address
C) What is the distance between m and 1/4 m when 1/4 m is released in part B?
D) What speed of 1/4 m is needed if the collision is fully inelastic rather than elastic?
Here is the first part of another problem I made up.
As I said in Question 1, I am posting this for
a) the enjoyment of everyone who reads them
b) work checking to see if they are riddled with mistakes or not
c) approach analysis, asking people to give me feedback on how I worked it.
Homework Statement
Mass m oscillates on a spring of spring constant k, at an amplitude A. There is no kinetic friction, but static friction of Us (Us mg is less than kA)
A) Find the minimum speed with which a mass 0.25 m can be launched against m for it to stop after the elastic collision at point x.
B) If the 1/4 m is located 1 meter from the natural length of the spring, when should it be launched?
There are more parts but I will post them later.
Homework Equations
I)##F=-kx##
II) ##F_{maxfriction}=Us mg##
III) KE is conserved in an elastic collision
##\frac{1}{2}(1/4m)v^2=\frac{1}{2}mv_m^2##
IV) Within the spring system, total energy is conserved and transferred between kinetic and potential
##\frac{1}{2}kA^2=\frac{1}{2}mv_{m max}^2=\frac{1}{2}mv_m^2+\frac{1}{2}kx^2##
V) ##v=dx/dt ##
or x/t for constant v
Unknowns:
x, vm, and v (I use v for the velocity of 1/4 m, and vm for the velocity of m on the spring)
The Attempt at a Solution
For the mass to be stopped by the collision and to be hung up on the static friction, the kinetic energy of the 1/4 m must equal the kinetic energy of the mass on the spring at x.
For the KE and velocity of 1/4 m to be at minimum, the KE of the mass on the spring must be at its minimum, so x must be as close to A as possible, and as much energy as possible should be stored in the spring as potential energy. Thus, ##kx = Us mg##, so the maximum x = ##\frac{Us mg}{k}##.
Combining III and IV
Total energy = ##\frac{1}{2}kA^2=(\frac{1}{2}(\frac{1}{4}m)v^2) + \frac{1}{2}kx^2##
Combining above with I and II
##\frac{1}{2}kA^2=(\frac{1}{8}mv^2) + \frac{1}{2}k(\frac{Us mg}{k})^2##
Simplifying
##kA^2=(\frac{1}{4}mv^2) + \frac{(Us mg)^2}{k}##
Solving for v
##v=\sqrt{4\frac{kA^2-\frac{(Us mg)^2}{k}}{m}}##
or
##v=2\sqrt{\frac{kA^2}{m}-\frac{m(Us)^2(g)^2}{k}}##
B)
Now that I have the velocity, it is not hard to find the time at which the mass should be released when it is placed at 1 m away.
##d=1+x##
From above:
##x=\frac{Us mg}{k}##
(x also = ##\sqrt{A^2-\frac{mv^2}{k}}##--I checked)
So ##t=\frac{d}{v}=\frac{1+\frac{Us mg}{k}}{2\sqrt{\frac{kA^2}{m}-\frac{m(Us)^2(g)^2}{k}}} ##
That is about all the latex I can do right now, but in the next post, I will address
C) What is the distance between m and 1/4 m when 1/4 m is released in part B?
D) What speed of 1/4 m is needed if the collision is fully inelastic rather than elastic?
