# Spring and maximum compression time

1. May 4, 2010

### smhippe

1. The problem statement, all variables and given/known data
A mass m1 = 8 kg is at rest on a frictionless horizontal surface and connected to a
wall by a spring with spring constant k = 70 N/m as shown in the figure. A second mass
m2 = 5 kg is moving to the right at vo = 17 m/s. The two masses collide and stick
together. How long will it take after the collision to reach the maximum compression of
the spring?

2. Relevant equations

3. The attempt at a solution
I just want to make sure I did this right...
So solve for the maximum compression distance using energy conservation. The spring has no energy and there is no energy change for potential gravity. So, 1/2*m*v^2=1/2*k*x^2; solving for x I got 5.75. Next we know F=-kx so ma=-kx; solving for a I got a=-30.96. Putting this into kinematics gives us .55s

2. May 5, 2010

### aim1732

By kinematics, do you mean eqns. of uniform acceleration?

3. May 5, 2010

### sArGe99

What was your "v" in 1/2 mv^2??
You need to apply Conservation of Linear Momentum principle in the horizontal direction (as there are no external forces in this direction) to obtain the velocity of the combined mass after collision.
Then obtain the maximum compression "x" at which the combined mass will be at rest, by using Conservation of Mechanical Energy principle (since there are no non-conservative forces here).

Acceleration given by a = (k*x)/m is time-variant as x depends on time.
So, put a = (v.dv)/dx = kx and integrate both sides. Determine the constant by putting x=0, v= velocity of the combined mass obtained. Now you will get v = a function of x. Put v = dx/dt and integrate to obtain x in terms of time t.
This is a direct relation between compression and the time taken for it.

4. May 5, 2010

### aim1732

Oh I should have seen that before - conservation of energy here in an inelastic collision is wrong.

5. May 5, 2010

### sArGe99

We're only concerned with what happens after the inelastic collision. Conservation of energy is valid then.

6. May 5, 2010

### aim1732

Yes but I only meant the conservation statement the OP wrote was wrong.

7. May 5, 2010

### smhippe

Okay so I can plug it into momentum conservation laws to find what the velocity of the combined blocks is. But, does that mean that I can still use energy conservation to figure out how far the spring is compressed after the collision? (I mean 1/2*m*(v^2)=1/2*k*(x^2))

8. May 5, 2010

### sArGe99

Yes. That's correct.

9. May 5, 2010

### smhippe

Try #2
So from momentum conservation i got the velocity of both blocks to be 10.46 just before they hit the spring. Using that I got the distance that the spring compressed by .5*m*(v^2)=.5*k*(x^2) and got x to be 4.5m. Then using that I used F=-kx=ma and got a=-24.23. From there I just used a simple equation (v)f=(v)i+a*t. I got time to be .43s. Sound right to anyone else?

10. May 7, 2010

### aim1732

Acceleration as pointed earlier is not constant so the eqn. you wrote can not be used.