Curiosity Question 2: Stopping a mass on a spring

1. Mar 17, 2015

long758

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.

1. The problem statement, all variables and given/known data
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.

2. Relevant 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)

3. 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?

2. Mar 17, 2015

AlephNumbers

There are some pretty challenging problems in the Halliday & Resnick book. If you have the tenth edition try problems 33-37 on page 205. I would be interested in seeing someone else's solutions.

3. Mar 19, 2015

long758

Hmm... I am not totally sure how this relates at all to the question, but if you would like to comment on the original post, I would really appreciate it.

P.S. I don't have the tenth edition of H&R, but if you really would like, you can PM the questions and I could give them a shot. But this thread is for the problem above, not H&R