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Imagine that a ball of mass $m$ is launched at a block, which also has mass $m$. Attached to the block, facing the ball, is a massless spring with a massless board at the end. Alternatively, we can assume that the block, spring, and board taken together have mass $m$. Assume there is no gravity and no dissipative forces. Suppose that the length of the spring and the spring constant are sufficient to stop the ball. When the velocity of the ball is zero, the spring becomes locked; it has some mechanism that physically prevents it from expanding or contracting, which is activated using an arbitrarily small amount of energy. After the interaction, how much potential energy is stored in the spring?

Setup

In solving this problem, an apparent paradox appears. Because the block-spring system and the ball have the same mass, and because the ball is at rest after the interaction, momentum conservation implies that the final velocity of the block should equal the initial velocity of the ball. This would imply that the final kinetic energy of the block is the same as the initial kinetic energy of the ball. But then there can be no potential energy stored in the spring, even though it has been compressed.

Obviously such a setup is impossible because of the massless spring and the lack of energy dissipation. However, these assumptions are fairly standard in physics, so one would not expect them to lead to a contradiction.

Also, I am aware of the principle that an idealized ratchet-like mechanism cannot exist because it could be used to violate the second law of thermodynamics. I understand that argument, but the problem here is not a violation of the second law, but rather a contradiction of energy and momentum conservation. Which of the premises of the problem is responsible for this contradiction, and why?

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    $\begingroup$ The speed will be half of the original ball, as now you have a mass of 2m moving with the same momentum, so the potential energy is $1/4mv_0^2$ $\endgroup$ – Wolphram jonny Oct 16 at 23:50
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    $\begingroup$ the spring locks when the ball is at rest in the lab frame $\endgroup$ – Maxwell Aifer Oct 16 at 23:53
  • $\begingroup$ locks where, in the floor and stays there? and the mass attached to the spring moves independently? $\endgroup$ – Wolphram jonny Oct 16 at 23:54
  • $\begingroup$ if the final kinetic energy of the block is equal to the initial kinetic energy of the ball, there is no energy left over for the spring. By "locks", it is meant that the length of the spring becomes fixed. One side remains attached to the block $\endgroup$ – Maxwell Aifer Oct 17 at 0:00
  • $\begingroup$ what trouble do you have applying conservation of angular momentum and energy? you will get the velocities of the two balls after the collision as a function of the compressed length. I do not think you can find this length by conservation laws, unless the initially moving mass stays attached to the system, but you can using the detailed dynamics. But in any case, energy is conserved. $\endgroup$ – Wolphram jonny Oct 17 at 0:28
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If the spring locks when the ball is at rest in the lab frame, then by the arguments you give it follows that the spring must not be compressed at all. This is indeed the case.

As the ball slows down, the block begins to speed up. Eventually they are traveling at the same speed, at which point the spring has reached its maximum compression. As the spring begins to expand, the block's velocity becomes greater than that of the ball. When the spring attains its uncompressed length, the ball comes to rest and the block is traveling with speed $v$.


This can be shown directly. Let $x(t)$ be the position of the ball and $y(t)$ be the position of the block, and let us consider left to be the positive direction in accordance with your figure. At time $t=0$ the ball makes contact with the spring. Let the initial position and velocity of the ball be $x(0)=0$ and $x'(0) = v$, and the initial position and velocity of the block be $y(0)=L$ and $y'(0) = 0$ where $L$ is the (unimportant) unstretched length of the spring.

The dynamics of the system are governed by the equations

$$m x'' = k(y-x-L)$$ $$m y'' = -k(y-x-L)$$

We can define the auxiliary variables $u = \frac{x+y}{2}$ and $w = \frac{x-y+L}{2}$ to obtain $$ u'' = 0 \implies u(t)= \frac{L+vt}{2}$$ $$w'' = -\frac{k}{m} w \implies w(t)= \frac{v}{2\omega}\sin(\omega t)$$ where $\omega=\sqrt{k/m}$ and I've applied the initial conditions stated above. We can invert these relations to find $x$ and $y$ to be $$x(t) = u+w-\frac{L}{2} = \frac{vt}{2}+\frac{v}{2\omega}\sin(\omega t)$$ $$y(t) = u-w+\frac{L}{2} = L + \frac{vt}{2} - \frac{v}{2\omega}\sin(\omega t)$$

The ball comes to rest when $x'(t) = \frac{v}{2}(1+\cos(\omega t)) = 0 \implies \omega t = \pi$. However, at this time we have that $y-x = L$ and $y'= v$.

enter image description here enter image description here


An alternative is that the spring locks when the maximum compression is achieved, i.e. when $y'=x'$. This occurs when $\cos(\omega t)=0 \implies \omega t = \pi/2$. At this moment, the velocity of the ball and the block are both $v/2$, in accordance with the conservation of momentum in a completely inelastic collision.

enter image description here enter image description here

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  • $\begingroup$ You have the signs reversed on your first two equations $\endgroup$ – Wolphram jonny Oct 18 at 18:30
  • $\begingroup$ @Wolphramjonny I don’t think I do. $y-x-L<0$ means that the spring is compressed, so the force on the ball should be negative and the force on the block should be positive. $\endgroup$ – J. Murray Oct 18 at 18:40
  • $\begingroup$ My intuition saw the spring compress and the two masses continuing at v/2, so I'm happy with that. However, haven't we just kicked the can down the road in that we have twice the mass at half the velocity, but also a compressed spring, where did the energy to compress the spring come from? (I'm guessing something to do with the unrealistic conditions?). $\endgroup$ – Lamar Latrell Oct 18 at 18:42
  • $\begingroup$ you are right, sorry $\endgroup$ – Wolphram jonny Oct 18 at 18:43
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    $\begingroup$ @LamarLatrell Twice the mass at half the velocity means half the kinetic energy. The other half goes into the spring. $\endgroup$ – J. Murray Oct 18 at 18:51
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The problem is exactly how billiard balls work (in 1 dimension, it's a much easier game like that). The spring is replaced by the elasticity of the ball, which doesn't need to lock, because that's how billiards works: all the cue ball's E and p go to the target ball, leaving the cue ball at rest.

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If the spring locks when the deformation is maximum the collision is not ellastic. At that point the relative velocity of the two objects is zero so they continue to move with the same velocity. This is a model of non-elastic collision. The collisions are ellastic when the spring does not lock and restore its potential energy into kinetic. For actual balls the spring is played by the ellasticity of the material.

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  • $\begingroup$ I believe this is wrong, if $x=\sqrt{m/2k}v_0$, then you can conserve both, momentum and energy $\endgroup$ – Wolphram jonny Oct 17 at 0:43
  • $\begingroup$ Not if the spring is locked when compressed. See the detailed answer by J. Murray. $\endgroup$ – nasu Oct 18 at 5:11
  • $\begingroup$ You can conserve energy but not kinetic energy. $\endgroup$ – nasu Oct 18 at 17:18
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After the collision, the two blocks have the same velocity $v'$.

Conservation of momentum: $mv = 2mv'$, so $v' = v/2$.

Conservation of energy: $\frac{1}{2} m v^2 = \frac{1}{2}(2m)v'^2 + PE_{spring} = \frac{1}{4}mv^2 + PE_{spring}$

So, $PE_{spring} = \frac{1}{4}mv^2$. (Half the incident energy gets stored in the spring.)

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  • $\begingroup$ The blocks don't have the same velocity afterward. The spring locks when the ball is at rest in the lab frame. $\endgroup$ – Maxwell Aifer Oct 17 at 0:52
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    $\begingroup$ Hmm, maybe I'm not understanding your premise... I assumed you meant that the ball was locked in place on the block, with the spring compressed. The incident ball is only at rest in the lab frame when it has delivered all its KE to the block, at which point there is no PE stored in the spring, just as you said. Not sure what the contradiction is... the spring locks at its equilibrium position. $\endgroup$ – pwf Oct 17 at 0:56
  • $\begingroup$ The spring is locked at the moment the ball stops. The spring must be compressed at this moment, because the force acting on the ball must be nonzero (that is what caused it to stop), and this force can only come from the spring $\endgroup$ – Maxwell Aifer Oct 17 at 1:00
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    $\begingroup$ Actually I think you are right. It locks at equilibrium because the force goes to zero at the same instant that the ball stops. $\endgroup$ – Maxwell Aifer Oct 17 at 1:15

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