Homework Help: F=ma Problem

1. Dec 31, 2015

RubinLicht

1. The problem statement, all variables and given/known data
. A 2.0 kg object falls from rest a distance of 5.0 meters onto a 6.0 kg object that is supported by a vertical massless spring with spring constant k = 72 N/m. The two objects stick together after the collision, which results in the mass/spring system oscillating. What is the maximum magnitude of the displacement of the 6.0 kg object from its original location before it is struck by the falling object?
(A) 0.27 m
(B) 1.1 m ← CORRECT ANSWER
(C) 2.5 m ← what im getting
(D) 2.8 m
(E) 3.1 m

2. Relevant equations
Conservation of mechanical energy(spring, gravitational, and kinetic)
conservation of momentum for the inelastic collision

3. The attempt at a solution
I'm just copying the solution here because it's what I did, but I will point out where i did something different.

Conserve energy to find the speed of the 2.0 kg object at the instant before the collision:
m1gy = (2.0 kg)(10 m/s)(5.0 m) = 100 J

and this equals the kinetic energy, after plug and chug we get
v = 10.0 m/s

The collision is inelastic, so
vf = v1*m1/(m1+m2) = 2.5 m/s

Now conserve energy again
Kf = 1/2mtv2 = 1/2(8.0 kg)(2.5 m/s)2 = 25 J

This compresses the spring and changes the gravitational potential energy, so we must solve for x in the equation
1/2kx2 + mtgx = Kf
or
36x2 + 20x − 25 = 0 I used 8kg as the total mass instead of 2kg, why did they use 2kg?
which has solutions x = −1.15 m and x = 0.6 m.

solutions with mt=8kg: x = -2.5, 0.27.

I'm a bit reluctant to say this, but i think aapt might have made a mistake on the answer key. On top of my confidence that my solution is right, they already used mt=8kg in the previous step before apparently changing its value in the next step. Now, given I'm a mere high schooler, i would still like someone to double check my work so that i can ask aapt to make the small fix.

2. Dec 31, 2015

TSny

I agree that the solution as given in the key is set up in an odd way, even though I think it gives the correct numerical answer.

Clearly both masses move downward after the collision, so both masses decrease their gravitational PE. Also, the spring already has some potential energy before the collision.

I suggest setting up the energy conservation using the total mass and including the initial PE of the spring. You will need to take care to distinguish between the initial and final compression of the spring and the distance traveled by the masses.

3. Dec 31, 2015

RubinLicht

as in 1/2k(5/6-x)2+80x=25?
5/6 being the displacement from relaxed spring due to m2 alone.

4. Dec 31, 2015

RubinLicht

I searched the problem up and seems like other people are understanding the problem differently as well. I think I'll stop trying to figure out this problem. Thanks for the help.