Inelastic Collisions - Two Boxes One Spring

[Badger94]

Homework Statement

Mass 1 (M1=1 kg) slides on a frictionless surface at v0=1 m/s toward a mass (M2=2 kg) with a spring bumper of force constant k = 20 N/m. The two move together at vf at the instant that the spring is compressed to its maximum x. Find the loss in kinetic energy (KEi-KEf).

I truly don't know how to start this. Can I conserve momentum? Energy? Both? How do I incorporate the work done by gravity or the spring?

Thanks for your help!

 

[PeterDonis]

Badger94, in order for us to help you, you need to make at least an attempt to solve the problem yourself. Just saying "I don't know where to start" is not an attempt. Try solving the problem in a few different ways and see what you come up with; if you're still stuck, you'll be able to fill out the homework template completely and that will help us help you.

Please refer to the homework forum guidelines for more information (particularly item 4):

https://www.physicsforums.com/threads/guidelines-for-students-and-helpers.686781/

 

[Badger94]

alright fair enough.

I have KEi-KEf = Loss
and KEf = KEi - PEf (so essentially the final potential energy is equal to my loss in kinetic energy.

Then I get

1/2mv0^2 = 1/2(m1+m2)vf^2 - 1/2kxmax^2

But I have two unknowns in this equation, xmax and vf

should I also conserve momentum, solve for vf and plug that in for vf in my original equation?

 

[PeterDonis]

First, another good practice is to use LaTeX to format your equations; it really helps with readability. PF has a tutorial on LaTeX here:

https://www.physicsforums.com/threads/physics-forums-faq-and-howto.617567/#post-3977517

should I also conserve momentum, solve for vf and plug that in for vf in my original equation?

Have you tried that? What did you get? Did the result make sense?

 

[Badger94]

Maybe...

Okay so from the conservation of momentum I get

$v_f=(m_1v_0)/(m_1+m_2)$

and plugging that into the original equation

$1/2m_1v_0^2-1/2(m_1+m_2)((m_1v_0)/m_1+m_2))^2=1/2kx_{max}^2$

which I then chose to solve for
$x_{max}$

and got

$x_{max}^2=1/30$

And since I had previously determined that my loss in kinetic energy was my final potential energy I get

$KE_i-KE_f=(1/2)(20)(1/30)=1/3$

Which could be perfectly reasonable but springs are weird so I'm not confident in this answer.

 

[Badger94]

^^^^^ good heavens. Clearly not getting this LaTex thing either

 

[PeterDonis]

from the conservation of momentum I get

v_f^2= (m_1 v_0) / (m_1 + m_2)

This doesn't look right. Why is there a $v_f^2$ on the LHS?

 

[PeterDonis]

Clearly not getting this LaTex thing either

You did well for a first try. :) I used magic mentor powers to fix the two equations that weren't displaying right.

 

[Badger94]

My mistake, I just meant $v_f$

Fixed

 

[Badger94]

You did well for a first try. :) I used magic mentor powers to fix the two equations that weren't displaying right.

You are wonderful! Thank you

 

[PeterDonis]

My mistake, I just meant $v_f$

Ok, good. You should also check your second equation; I think the $v_0$ in the first term is not right either.

 

[Badger94]

Ok, good. You should also check your second equation; I think the $v_0$ in the first term is not right either.

Yes! Good catch. I promise on paper I didn't have these mistakes so I still am getting 1/3 as my answer

 

[PeterDonis]

I promise on paper I didn't have these mistakes so I still am getting 1/3 as my answer

Yes, 1/3 looks right as the final answer, but now the intermediate steps look right too. :)

As for whether the answer looks reasonable, you might check how it compares with the initial kinetic energy--i.e., what fraction of the initial kinetic energy is converted to potential energy in the spring, if this answer is correct? Does that fraction look reasonable?

 

[haruspex]

I am unable to make sense of the question. Is it stated correctly, word for word? If after collision the blocks move together then at the instant of maximum compression they are both stationary.

 

[PeterDonis]

at the instant of maximum compression they are both stationary.

No, they aren't. The spring can transmit force to the second mass well before it is at maximum compression.

 

[haruspex]

The spring can transmit force to the second mass well before it is at maximum compression.

Quite so, but I don't see how that resolves my objection.
Suppose the incoming mass is from the left. As long as the two masses are moving right, the spring is being compressed further. At the instant of maximum compression the masses will be stationary (transiently). Do I have the wrong model?

Here's the question that Badger94's algebra answers:

The maximum compression of the spring which subsequently occurs is x. Find the loss in kinetic energy (KEi-KEf) during the collision.​

Badger94, please check the wording of the question.

 

[Orodruin]

Quite so, but I don't see how that resolves my objection.
Suppose the incoming mass is from the left. As long as the two masses are moving right, the spring is being compressed further. At the instant of maximum compression the masses will be stationary (transiently). Do I have the wrong model?

If the right mass is moving faster to the right than the left one is, the spring is no longer being compressed. By conservation of momentum, both masses cannot be stationary at the same time (unless you transform the problem to the CoM frame).

 

[haruspex]

If the right mass is moving faster to the right than the left one is, the spring is no longer being compressed. By conservation of momentum, both masses cannot be stationary at the same time (unless you transform the problem to the CoM frame).

The penny has just dropped. I had the spring bumper on the far side of the second mass.
I plead jet lag.