Block move down the curved hill and hit attached block spring with diagram

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SUMMARY

The discussion revolves around a physics problem involving a 100-gram block moving at 2 m/s down a curved hill, colliding elastically with a stationary 400-gram block attached to a spring with a spring constant of 500 N/m. The participants utilized the Work-Energy theorem and Conservation of Energy principles to solve for the maximum compression of the spring, which was calculated to be 0.08966 meters, and the maximum height the 100-gram block reaches post-collision, determined to be 1.154 meters. The conversation emphasizes the importance of showing work and understanding momentum conservation in elastic collisions.

PREREQUISITES
  • Understanding of the Work-Energy theorem
  • Knowledge of Conservation of Momentum in elastic collisions
  • Familiarity with basic kinematics and dynamics
  • Ability to apply Hooke's Law for spring compression
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  • Study the principles of elastic collisions and their equations
  • Learn about the Work-Energy theorem in detail
  • Explore Hooke's Law and its applications in spring mechanics
  • Practice solving problems involving conservation of energy and momentum
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for practical examples of energy conservation and momentum in collisions.

cecico
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A 100grams block is moving at 2m/s down the curved hill. The block slides along the smooth surface and collides elastically with the 400gram block. The 400gram block is initially at rest and is attached to an ideal spring with spring constant of 500N/m
a)Find the maximum distance the spring compresses
b)Find the maximum height the 100gram block reaches after the collision.

I added a diagram... help me T.T
 

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oh~ height is 3.0m...
 
Welcome to PF!

Hi cecico! Welcome to PF! :smile:
cecico said:
A 100grams block is moving at 2m/s down the curved hill …

I don't understand … it will be accelerating … at which point is its speed 2m/s? :confused:
 
Someone else asked this exact same question on another thread, and that poster made the same mistake you did-- you have not attempted the problem. You must show that you have worked on the problem before asking for help.
 
DavidWhitbeck said:
Someone else asked this exact same question on another thread, and that poster made the same mistake you did-- you have not attempted the problem. You must show that you have worked on the problem before asking for help.

Not someone else.

Saladsamurai said:
In order to receive help with this problem I suggest that you

a) Provide a diagram
b) Show the work you have tried thus far. Forum rules.

To start you off. I would ask my self what theory(s) I would like to apply to solve this problem. What theories have you learned that involve finding velocities and displacements?

Work-Energy theorem and Conservation of Energy come to mind. Which would be easier in this case?

tiny-tim said:
Hi cecico! Welcome to PF! :smile:I don't understand … it will be accelerating … at which point is its speed 2m/s? :confused:

And this too. Where is its velocity 2 m/s ?
 
Last edited:
Oh you're right SaladSamurai! That poster must be impatient.
 
SaladSamurai your right... I have to use Work-Energy theorem and Conservation of Energy and the speed is 2m/s at point A and the surface is smooth...
 
cecico said:
… the speed is 2m/s at point A …

Hi cecico! :smile:

ok … then what is the speed of the 100grams block just before it hits the other block? :smile:
 
according to my calculation I got 7.925m/s...after this i have no idea...
 
  • #10
cecico said:
according to my calculation I got 7.925m/s...after this i have no idea...

That's right. :smile:

Now pretend that there's no spring, and work out the speeds of the two blocks immediately after the collision. :smile:
 
  • #11
immediately after the collision I got 1.585m/s... is this right?
 
  • #12
cecico said:
immediately after the collision I got 1.585m/s... is this right?

uuh? you should get two speeds … one for each block! :rolleyes:

(and if you show your working, that'll save me time checking! :smile:)
 
  • #13
ok...now i really don't get it... what equation do i have use??
momentum equation? Mv+Mv=Mv+Mv but there's two unknown...
how do i calculate this? -0-
 
  • #14
cecico said:
ok...now i really don't get it... what equation do i have use??
momentum equation? Mv+Mv=Mv+Mv but there's two unknown...
how do i calculate this? -0-

Hi cecico! :smile:
DON'T PANIC!

(i can tell you're panicking because the number of question marks is increasing faster than exponentially!)

In collisions, momentum is always conserved.

And you're told that this collision is elastic, and that means that energy is also conserved.

Two equations … two unknowns … physicist's heaven! :smile:
 
  • #15
alright for 100gram I got -4.755m/s and for 400gram I got 3.17m/s...it seems wrong..T.T
 
  • #16
Well, let's assume it's right.

You now have an initial velocity for the 400gram block.

So … what is the maximum distance the spring compresses? :smile:
 
  • #17
I got 0.08966m...
 
  • #18
Well, that's the answer to part a) then! :smile:

And the maximum height that the 100gram block goes to?
 
  • #19
I used 1/2mv2=mgh that I got 1.154m...
 
  • #20
Hi cecico! :smile:

Well, that's the whole question then, isn't it?

Or is something still worrying you about it? :smile:

( … if aok now, click on "Thread Tools" to mark the thread [SOLVED] … :smile: )
 
  • #21
this is it~~~omg thanks for your help really... I was stuck with this prob from yesterday.. T.T now I got it... thanks bro~~
 

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