Elastic and Inelastic: Finding Maximum Spring Compression

[1831]

[SOLVED] Elastic and Inelastic

A 100 g block on a frictionless table is firmly attached to one end of a spring with k = 29 N/m. The other end of the spring is anchored to the wall. A 25 g ball is thrown horizontally toward the block with a speed of 5.9 m/s.

Q1: If the collision is perfectly elastic, what is the maximum compression of the spring?

Q2: in a perfectly inelastic collision, what is the maximum compression of the spring?

For Q1 I tried to use:
(1/2)mvi^2+(1/2)kx^2 = (1/2)mvf^2 + (1/2)kxf^2

and got change in x was .44 m, but it wasn't right...

I did the same thing for the other part, and it was still wrong...please help.

Oh, and I found the velocity of the ball after collision:

Perfectly Elastic: vf=3.5m/s
Perfectly Inelastic: vf=1.2m/s

These were right.

 

[G01]

Hi 1831,

First things first, please post homework questions like these in the homework help section of the site next time. This forum is more for general questions than homework type questions.Alright,

Since you found the final speed in each case correctly. You must know the difference between an inelastic and elastic collision.

HINT: Due to this difference between the two types of collisions, the speed of the block attached to the spring will be different. Can you find this speed in both cases? Can you use this speed to solve for what your looking for?

 

[1831]

Ok, I will place it in the HW section last time...I really thought I did. I'm sorry.

I do know the difference between inelastic and elastic. For the inelastic collision, the ball sticks and moves with the block and spring. I used Vf= (m1v1i + m2v2i)/(m1+m2) =1.2 m/s to find the velocity of the ball+block system.

And for perfectly elastic, I used Vf1= ((m1 - m2)/(m1 + m2)) * (vi1) to get the velocity of the ball after the collision.

However, I do not understand what I need to do from here to find the change in compression of the spring.

I want to use the formula: (1/2)mvi^2 + (1/2)kxi^2 = (1/2)mvf + (1/2)kxf^2

but I guess I'm using it incorrectly.

 

[G01]

You intuition is correct. That formula, conservation of energy, is indeed what you have to use, but I think your using the wrong speeds, at least in one case.

HINT:

What is the mass and speed of the object on the end of the spring in each case?

It is this mass and speed that you want to use in the conservation of energy formula.

 

[1831]

For the inelastic equation, the mass on the final side is mball+mblock as is the velocity final. (1.2 m/s)
And for the elastic equation, the final velocity of the ball (3.5) and the vf of the block should be .4m/s...right?

 

[1831]

So for the perfectly Inelastic equation for max compression I got:

m=mass of ball
M=mass of block
vi=initial v of ball
vf=final v of ball+block

.5*m*vi^2 = .5*(m+M)*vf^2 + .5*k*x^2

Evaluated:
.5(.25)(5.9^2) = (.5)(.25+1)(1.2^2) + (.5)(29)(x^2)
x=.48m...but this is incorrect, so i didn't try the one for perfectly elastic.

 

[1831]

ok, i figured it out...thanks guys!

 

[G01]

Cool!