Maximum compression distance of a spring

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Homework Help Overview

The problem involves two blocks colliding elastically, with one block attached to a spring. The original poster seeks to determine the velocities of the blocks after the collision and the maximum compression distance of the spring.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the conservation of momentum and kinetic energy in elastic collisions. There is uncertainty about the interpretation of "immediately after the collision" and its implications for the calculations. Some participants suggest that total mechanical energy is conserved, while others question the correct equations to use for the scenario.

Discussion Status

Participants are actively exploring different interpretations of the problem and discussing the relevant conservation laws. Guidance has been offered regarding the need to consider both momentum and energy conservation, but no consensus has been reached on the specific approach to finding the maximum compression distance of the spring.

Contextual Notes

There is ambiguity in the phrasing of the problem regarding the timing of the measurements post-collision, which affects the calculations. Participants are also navigating the distinction between kinetic energy and total mechanical energy in the context of the collision.

T Roth
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Homework Statement


Block #1 (m=2.00 kg) traveling to the right at 5.00 m/s along a level frictionless surface collides elastically with block #2 (m=3.00 kg) attached to a massless spring of spring constant 275 N/m. Assume that block #2 was at rest before the collision.
(a) calculate the celocity of each of the two blocks immediately after the collision
(b) calculate the maximum compression distance of the spring

The Attempt at a Solution


I found the answer to part a by using the equation m#1*V#1+m#2V#2=(m#1+m#2)V`
(2)(5)+(3)(0)=(2+3)V`
v`=2m/s
but I'm not sure how to go about part b
 
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T Roth said:
I found the answer to part a by using the equation m#1*V#1+m#2V#2=(m#1+m#2)V`
(2)(5)+(3)(0)=(2+3)V`
v`=2m/s
but I'm not sure how to go about part b
The collision is elastic, not inelastic. Use a different equation for momentum.

What else is conserved besides momentum?
 
T Roth said:

The Attempt at a Solution


I found the answer to part a by using the equation m#1*V#1+m#2V#2=(m#1+m#2)V`
(2)(5)+(3)(0)=(2+3)V`
v`=2m/s
On second thought, the question is a bit ambiguous when is says "immediately after the collision". If they mean at the point of maximum compression, then your method is correct. But if after the collision means after they separate, then you'd need a different equation. Given that they probably want you to use (a) to solve (b), I'd say you probably interpreted it as they wanted you to and have the correct answer. :wink:

To solve (b), answer my question: What else is conserved?
 
ok so i tried V#1-V#2=V`#2-V`1
and i got V`#2=V`#1+5
then i plugged that into m#1*V#1+m#2v#2=m#1V`#1+m#2V`#2 and that came out to be 4
is that correct? and I'm still not sure how to find the maximum compression distance of the spring
 
i believe kinetic energy is conserved
 
T Roth said:
ok so i tried V#1-V#2=V`#2-V`1
and i got V`#2=V`#1+5
then i plugged that into m#1*V#1+m#2v#2=m#1V`#1+m#2V`#2 and that came out to be 4
is that correct?
If you interpret question (a) as I first did (find the speeds after they separate), then that's correct. But you'll need to find the speed of the other mass as well.

As I said above, it's not clear to me which interpretation of "immediately after the collision" was intended.

and I'm still not sure how to find the maximum compression distance of the spring
Find the speed of the system when the spring is maximally compressed. Hint: You already did!

T Roth said:
i believe kinetic energy is conserved
Not kinetic energy, but total mechanical energy is conserved. Don't forget about the spring.
 

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