Maximum compression distance of a spring

AI Thread Summary
The discussion revolves around calculating the velocities of two blocks after an elastic collision and determining the maximum compression of a spring. The first block, weighing 2.00 kg, collides with a stationary 3.00 kg block attached to a spring, and the initial calculations indicate a post-collision velocity of 2 m/s for the combined system. Participants clarify that both momentum and total mechanical energy are conserved during the process, not just kinetic energy. There is some ambiguity regarding the interpretation of "immediately after the collision," which affects the calculations for maximum spring compression. Ultimately, the correct approach involves using the conservation of momentum and energy principles to find the maximum compression distance.
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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