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Old May 13th 2017, 11:10 AM   #1
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Collision problem

Hey, guys. I have problems with this exercise.
https://ibb.co/b4PiQk

I thought that the linear momentum would remain conserved and then i wrote the equation for the conservation of the total mechanical energy. But I don't know how to continue because actually I didnt understand what the problem asked me to find. Can you help me please?
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Old May 13th 2017, 11:38 AM   #2
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Hey Aledg...

The image you posted is rotated 90deg CCW. Text aligned vertical. Hard to read.
Here's all I have close to what you seek.

Box Bounces Between Springs | THERMO Spoken Here!

Good luck, JP
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Old May 13th 2017, 12:25 PM   #3
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Ok sorry
Here it is

https://ibb.co/nvbGBQ
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Old May 13th 2017, 01:00 PM   #4
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I don't look at unknown sites off forum, especially since the recent cyber attack.

However Thermo's reply seems to indicate that you have springs involved.

Why do you think momentum is conserved in this case?

Momentum is not conserved in elastic collisions.

https://www.google.co.uk/search?hl=e...73.VWvRINT2kBM
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Old May 13th 2017, 03:58 PM   #5
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Originally Posted by studiot View Post
I don't look at unknown sites off forum, especially since the recent cyber attack.
Which image hosting sites does physhelpforum say are safe?
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Old May 14th 2017, 07:46 AM   #6
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when the block achieves maximum compression of the spring, both the block and platform are at the same velocity, $v_1$ ...

momentum is conserved ...

$mv_0 = (m+M)v_1 \implies v_1 = \dfrac{mv_0}{m+M} = \dfrac{v_0}{41}$

all forces are conservative, so energy is conserved ...

$\dfrac{1}{2}mv_0^2 = \dfrac{1}{2}(m+M)v_1^2 + \dfrac{1}{2}k(x_{max})^2$

substituting given values for $m$, $M$, and $k$ ...

$x_{max} = \dfrac{v_0}{5\sqrt{82}}$


after the instant of maximum spring compression, momentum is conserved ...

$(m+M)\cdot \dfrac{v_0}{41} = mv_2 + Mv_3$

where $v_2$ is the cube's final velocity and $v_3$ is the platform's final velocity ... note both velocities are relative to an outside observer.

since energy is conserved, opening speed after the collision = closing speed before the collision ...

$v_3-v_2 = v_0 - 0$

$v_3 = v_0+v_2$

$(m+M)\cdot \dfrac{v_0}{41} = mv_2 + M(v_0+v_2)$

$\dfrac{v_0}{2} = 20.5v_2 + 20v_0$

$v_2 = -\dfrac{39v_0}{41} \implies v_3 = \dfrac{2v_0}{41}$
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