How Much Does the Spring Compress When Hit by a Rolling Ball?

[mrknowknow]

Homework Statement

A problem like this is on my final tomorrow and I can't seem to get past part C...Please Help!

A .321 kg, .27m radius thin shelled ball rolls (starting at rest) 3.82m from the top down a 35° 10.0m long incline without slipping. After the 3.82 m, the incline becomes frictionless for the rest of the board. After .8m of the frictionless unencumbered movement, the ball reaches a frictionless massless spring with a spring constant of 20 N/m.

a. How fast is the center of mass of the ball going after 3.82 (linear speed)
I solved for v using mgh=1/2mv^2 + 1/2(2/3MR^2)(v/r)^2

so v= 5.076 m/s

b. How fast is the ball rolling after the 3.82 m (angular speed)
v/r=5.076/.27 = 18.8 rad/s

c. How much does the spring compress?
d. what is the ball's maximum linear speed?
e) what is the balls angular speed when the spring is fully compressed?

Homework Equations

I tried using vf^2=vi^2+2(9.8sin35).8
to get the velocity before it hits the spring

The Attempt at a Solution

I'm using 1/2mv^2=1/2kx^2
and solving for x but I'm not getting the right answer!

 

[haruspex]

Did you forget that the ball is still descending as the spring compresses?

 

[mrknowknow]

so would it be 1/2mv^2=1/2kx^2+1/2mv^2

 

[haruspex]

so would it be 1/2mv^2=1/2kx^2+1/2mv^2

It could only be that if the two v's stand for different things, but i can't think what you have in mind.
What I meant was that the ball continues to lose PE after contacting the spring.

 

[mrknowknow]

I don't know how to set up the equation. A tutor at MIT on InstaEdu couldn't even figure it out. The answer is .211 m but I don't see how my professor got that.

 

[haruspex]

I don't know how to set up the equation. A tutor at MIT on InstaEdu couldn't even figure it out. The answer is .211 m but I don't see how my professor got that.

Suppose the max compression of the spring is x. Write out the change in PE and the change in KE from contacting spring to max compression.

 

[mrknowknow]

I'm thinking but nothing comes to mind!

 

[haruspex]

I'm thinking but nothing comes to mind!

You know the speed at contact with spring. You know the speed at max compression. So you know the change in KE.
If it compresses the spring a distance x, what is the PE stored in the spring?
All of the above you calculated and used before. The bit you forgot was:
If it compresses the spring a distance x, how much more PE did the ball lose while compressing the spring?

 

[mrknowknow]

From what you're saying I would think to the equation I posted in #4. I don't understand the last line, I'm stumped.

 

[haruspex]

From what you're saying I would think to the equation I posted in #4.

That equation may apply, I've no way of knowing since you still have not clarified what the two 'v's represent. Either way, it isn't the whole story, for the reason given below.

I don't understand the last line, I'm stumped.

The whole slope is 10m. After 4.62m it hits a spring. As it compresses the spring it is still going down the slope. if it travels a distance x while compressing the spring, how much gravitational PE does it lose in the process? I don't think I can make it any clearer.

 

[mrknowknow]

mg10sin35 - mg5.38sin35 ??

 

[haruspex]

mg10sin35 - mg5.38sin35 ??

" if it travels a distance x"