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Small oscillation after a perfectly inelastic collision

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A cymbal of mass M is stuck at one of the extremes of a spring of constant K. The system spring-cymbal is at equilibrium, in vertical position and stuck on the floor at the other spring’s extreme. Over the cymbal, and initially at rest and at a height h from the cymbal a block of mass m falls. The collision cymbal-block is perfectly inelastic.
a) Compute the new equilibrium position (now the system adds the block of mass m)
b) Compute the amplitude and angular frequency produced (after perfect inelastic collision)
c) At this section the block does not remain stuck to the system cymbal-spring. After the collision block-system the spring gets compressed and after returns to its equilibrium position and at some point the block separates from the system spring-massM. What is the maximum height the mass m reaches after doing so? (Compute the height from the point it separates from the mass M).

This is what I tried:

a) When you are asked for computing a new equilibrium position (in this case it happens after the perfectly inelastic collision) I guess all you have to do is state Newton's second law in equilibrium situation and solve for the distance at which the point particle is from the position where potential energy is zero.

b) Firstly I assumed this is a peak-to-peak amplitude. what I did was to equal the potential energy of the system to the potential energy of the system as it was not oscillating. This is why I am pretty sure I am wrong here.

c) I applied energy method

Thanks

asked Jun 27, 2018 in Physics Problems by Jorge Daniel (686 points)
edited Jul 3, 2018 by Jorge Daniel
I do not understand what you have written in the question at (c) : " ... suppose there’s no perfectly inelastic collision. After the collision... ". What does this mean? Please can you provide the original words. Or have you translated?

Does it mean the collision is not completely inelastic? If so, then the block and symbol do not stick together, so they cannot later separate.

The best interpretation which I can make is that after the collision the block moves at the same velocity as the cymbal, so the collision is totally inelastic, but it does not 'stick' to the cymbal, then it can later separate from it.
You interpreted it correctly Sr. At c) a block collides with the system spring-massM, which gets compressed and after that the restorage force makes the system go back to its initial position. That impulse propulses the mass m upwards, which separates from the system (at c) it is not a perfectly inelastic collision) and reaches a maximum height. I am having a look at your answer, I'll comment when doubts pop up
Sorry but this is still not clear. My interpretation is that there is only one collision in this problem, and it is perfectly inelastic. You said this is correct but later you say again that "at c) it is not a perfectly inelastic collision". Please can you explain what you mean.
I quote c) "Now supposing the block does not stick to the cymbal after the collision the spring, first gets compressed, after stretches (because of the restoring force) and at some point the block separates from the system spring-massM going upwards. Which is the maximum height reached by the block?"
At c) there is a collision block-system but it is not perfectly inelastic (i.e they do not remain stick)
I understand the quote. It does not say the collision is not perfectly inelastic. That is your interpretation and it is wrong. "Perfectly inelastic" does not mean the block and cymbal stick together, it means that the relative velocity of separation after the collision is zero.
Okey Sr. I let myself be taken in by the false assertion ‘A perfectly inelastic collision means that the block and cymbal do stick together’. Could you cast some light on your statement ‘It means that the relative velocity of separation after the collision is 0’ or post a link please?
The elasticity of the collision is defined by coefficient of restitution $e=\text{relative velocity of separation / relative velocity of approach}$. A totally inelastic collision has $e=0$ - ie relative velocity of separation is 0.

If the particles 'stick' together as if glued, then it follows that $e=0$. However, the converse is not necessarily true : if $e=0$ this does not guarantee that the particles have 'stuck' together.

Many sites do not make the distinction when they explain what a 'totally inelastic collision' means, because usually it is not important. Some say that the KE lost is used to form bonds, but this is not necessarily true. Whether the KE lost is used to heat the particles or deform them in some way or to form bonds is not something which $e=0$ can tell you.

Total inelasticity merely means the particles 'move together' after the collision. They do not necessarily 'stick together'. The extent to which they 'stick' in the sense that work must be done to separate them is a completely separate issue.

See https://www.quora.com/Will-two-objects-always-stick-together-after-an-inelastic-collision

1 Answer

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Best answer

a) Correct. The new equilibrium position occurs at distance $x=mg/k$ below the initial position (where the collision occurred). Alternatively it is $X=(M+m)g/k$ below the top of the unloaded spring (with no cymbal attached).


b) The new angular frequency of oscillations can be written down without any calculation : $\omega=\sqrt{\frac{k}{m+M}}$. Correct.

However, your calculation of amplitude is not correct.

Peak-to-trough is twice the amplitude - ie $2A$. Amplitude is the distance of a peak or trough from the new equilibrium position. Peaks and troughs occur when kinetic energy is zero - ie "when it is not oscillating" as you put it. See Energy Method below. You had the right idea, but your calculation neglects the fact that the cymbal is not released from rest - it has some initial kinetic energy due to the collision. Also your solution does not deal with elastic energy correctly.

CALCULATION OF AMPLITUDE OF OSCILLATIONS

1. Equation of Motion Method

First work out the speed of the cymbal immediately after the inelastic collision. From conservation of momentum this is $v=\frac{m}{m+M}u$ where $u=\sqrt{2gh}$ is the speed of $m$ immediately before the collision.

Next, the oscillation about the new equilibrium position can be described by the equation of motion $$\xi=A\sin(\omega t+\phi), \dot \xi=\omega A\cos(\omega t+\phi)$$ where $\phi$ is some unknown phase angle. Suppose the collision occurs at $t=0, \xi=x$ with $\dot \xi=v$. (It does not matter what time it occurs, we will only get a different phase angle $\phi$.) Then $$x=A\sin\phi, v=\omega A\cos\phi$$ $$A^2=x^2+(\frac{v}{\omega})^2=(\frac{mg}{k})^2+\frac{2ghm^2}{k(M+m)}$$

2. Energy Method

Let point U be the top of the spring when it has no load. I shall measure all potential and elastic energies from this point. Let P be the starting equilibrium position where the load is $M$, let O be the new equilibrium position when the load is $M+m$, and Q be the lowest point of the subsequent oscillations. The amplitude of oscillations is $OQ=A$. Distance $OP=x$ as already calculated. Other distances are $UP=UO-PO=X-x$ and $UQ=UO+OQ=X+A$.

At P there is kinetic energy of $\frac12 (m+M)v^2$. The spring is compressed by distance $UP=X-x$ so the elastic energy stored in the spring is $\frac12 k(X-x)^2$ where $X=(M+m)g/k$ as above. Gravitational PE at P relative to U is $-(m+M)g(X-x)=-kX(X-x)$.

At Q there is no KE, and gravitational PE is $-(M+m)g(X+A)=-kX(X+A)$. The elastic energy stored here is $\frac12 k(X+A)^2$.

By the conservation of energy, the total energy at P is the same as that at Q. Therefore $$\frac12 (m+M)v^2-kX(X-x)+\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2+2AX+A^2)-kX(X+A)$$ $$(m+M)v^2=k(A^2-x^2)$$ $$A^2=x^2+\frac{(m+M)v^2}{k}=x^2+(\frac{v}{\omega})^2$$ as found using Method 1.


(c) This part of the question is not clear. I assume that the block moves down with the cymbal as in (a) and (b). However, because it is not fixed to the cymbal it can separate from the cymbal when it rises from Q back above O. The block and cymbal do not 'stick together' because of the collision, they only 'move together' after it.

Separation occurs when the downward acceleration of the SHM becomes greater than $g$. Gravity is the only force holding the block in contact with the cymbal, so when gravity is no longer able to supply the required restoring force on the block, it leaves contact with the cymbal.

Suppose separation occurs at point R which has displacement $\xi$ above equilibrium position O. The downward acceleration at point R is $\omega^2 \xi=g$. So $$\xi=\frac{g}{\omega^2}=\frac{(m+M)g}{k}=X$$ This means that the lift-off point is always at the top of the spring when it has no load (R=U), whatever the values of $m, M, h, k$. This is a surprising result.

The explanation is that at this instant the only force acting on the cymbal and block is gravity, because the spring is no longer compressed or stretched so it exerts no force. Both cymbal and block are in 'free fall' so the force between them is zero. Just before this instant both cymbal and block are moving upwards but accelerating downwards at just less than $g$ because of a small upward push from the spring. After this instant the cymbal is being accelerated downwards at slightly more than $g$ because of a small pull from the spring. But the spring does not pull down on the block so the block is still accelerating downwards at $g$. There is relative acceleration, so the block and cymbal separate.

Note that if $A\lt X$ then the cymbal does not rebound to the relaxed position of the spring at U, then there is no lift-off. The condition for lift-off is that $A \gt X$. Substituting from the equations for $A^2$ and $X$ given above we get $$A^2 \gt X^2$$ $$\frac{m^2g^2}{k^2}+\frac{2ghm^2}{k(M+m)} \gt \frac{(m+M)^2g^2}{k^2}$$ $$1+\frac{2kh}{g(M+m)} \gt \frac{(m+M)^2}{m^2}$$ $$\frac{2kh}{g(M+m)} \gt \frac{(m+M)^2}{m^2}-1=\frac{m^2+2mM+M^2-m^2}{m^2}=\frac{M(2m+M)}{m^2}$$ $$h \gt \frac{M(M+m)(M+2m)g}{2km^2}$$

We can also find the maximum height $H$ above R=U reached by the mass $m$ after lift-off.

The total energy at P is the same as at U, so from above (Energy Method) we have $$\frac12 (m+M)v^2+\frac12 k(X-x)^2-kX(X-x)=\frac12 (m+M)V^2$$ $$\frac12 (m+M)(v^2-V^2)=k(X^2-xX)-\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2-x^2)$$ $$v^2-V^2=\frac{k}{(m+M)}(X-x)(X+x)$$ From earlier results we have $$v=\frac{m}{m+M}u=\mu u=\mu \sqrt{2gh}$$ $$v^2=2gh \mu^2$$ $$X-x=\frac{(m+M)g}{k}-\frac{mg}{k}=\frac{Mg}{k}$$ $$X+x=\frac{(m+M)g}{k}+\frac{mg}{k}=\frac{(m+M+m)g}{k}$$ The block rises to a height $H$ above U given by $2gH=V^2$. Making the above substitutions into the equation for $v^2-V^2$ we get $$2g(h\mu^2-H)=\frac{k}{m+M}\frac{Mg}{k} \frac{(m+M+m)g}{k}=\frac{Mg^2}{k}(1+\mu)$$ $$H=h\mu^2-\frac{Mg}{2k}(1+\mu)$$

answered Jun 28, 2018 by sammy gerbil (28,806 points)
edited Jul 24, 2018 by sammy gerbil
The new equilibrium position (i.e after the collision) is $x= \frac{(m+M)g}{k}$ right?
That depends on your definition of $x$. If $x$ is the total compression of the spring from its unloaded length then this formula is correct. If $x$ is the extra distance below the equilibrium position for the cymbal of mass $M$ on its own, then $x=mg/k$.
Does Unloaded length  mean the length of the spring totally unwrapped?
By "unloaded length" I mean the length without any load $M$ or $m$. The natural length or relaxed length of the spring.
After checking out, if the origin of coordinates is located at the top of the spring (ie where is the cymbal) wouldn’t the equilibrium position be x=-mg/k?
Yes. That is stated in the first sentence of my answer. Or are you quibbling about the sign?
Yes I am quibbling over the sign. Then we can state the new equilibrium position occurs at distance x=mg/k  below the initial position (implicitly we are saying it is negative), got it.
Sr then $\frac12 (m+M)v^2+\frac12 k(x)^2+gy(M+m)=\frac12 k(x+y)^2$ is correct as well as $\frac12 (m+M)v^2+kX(A+x)+\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2+2AX+A^2)$ The only difference is that in the first case we state $x=mg/k$ and in the second case we also take into consideration the definition $X=(M+m)g/k$ right? Note $y= A + x$
Sorry but I don't want to check everything line by line. If there is something you do not understand or something you disagree with, I am happy to explain.
I agree, it could be tedious for you. I am trying to understand your energy approach. Particularly I am confused by the values of the elastic energy at P $\frac12 k(X-x)^2$ and at Q $\frac12 k(X+A)^2$ (I do not understand why X-x and X+A respectively). The first time you answered without using the value of the equilibrium position with the unloaded length $X=(M+m)g/k$ If you do not mind to add it again... It was this if I do not recall badly $\frac12 (m+M)v^2+\frac12 k(x)^2+gy(M+m)=\frac12 k(x+y)^2$ and defining $y= A + x$
If you prefer not to add it I am happy if you just explain what I asked you. Thank you
I have edited my answer so that potential energies (gravitational and elastic) are now measured from the top of the unloaded spring. I have also explained how the various distances relate to $x, X, A$.
I was rereading this exercise and I realised I did not really understand why $\omega^2 \xi=g$. I tried to obtain it from both Newton's second law and deriving twice the equation of motion $\xi=A\sin(\omega t+\phi)$ but I did not get $\omega^2 \xi=g$. May you please cast some light on it?
In SHM the acceleration is $\ddot \xi=\omega^2 \xi$ where $\xi=A\sin(\omega t+\phi)$ is displacement. The only force providing downward acceleration on the mass $m$ is gravity (it is not attached to the cymbal). So if SHM acceleration exceeds $g$ then mass $m$ leaves contact with the cymbal, because the cymbal is accelerating faster. The cymbal is pulled down by gravity **and** the tension in the spring.
Okey but the issue here is that I get $\ddot \xi=- \omega^2 \xi$, -ve sign because of the derivative of cosine function
Yes that's right : $\ddot \xi=-\omega^2 \xi$. Here $\xi$ is measured upwards so $\ddot \xi$ is downwards, same direction as $g$. I was referring to the magnitudes : $\ddot \xi=\omega^2 |\xi|=g$.

"...Suppose separation occurs at point R which has displacement ξ above equilibrium position O. The downward acceleration at point R is $ω^2ξ=g$."
Sr beforehand I would like to say sorry as it may seem to you I am making a nuisance of myself but once I redid this exercise doubts started popping up so ...
I am pretty sure $h \gt \frac{M(M+m)(M+2m)g}{2km^2}$ comes from isolating h from this equation $(\frac{mg}{k})^2+\frac{2ghm^2}{k(M+m)} > ((M+m)g/k)^2$ but I do not get the exact result. Could you add one step more of the required algebra?
Besides, I do not get this equation as you have it neither $V^{2} = 2gh\mu^2-\frac{mg^2}{k}(\frac{1}{\mu}-\mu)$ You have made changes as , for instance, where should be $\frac{k}{M+m}$ you typed $\frac{mg^2}{k}$ instead. Could you add one step more of the required algebra as well?
I have updated my answer. The result for $H$ is different, so I might have made a mistake this time or last.
Now everything is clear except for $V^2=\omega^2 (A^2-X^2)=v^2-\omega^2 (X^2-x^2)=\omega^2(X-x)(X+x)$ Why $v^2-\omega^2 (X^2-x^2)=\omega^2(X-x)(X+x)$?
Typing error : there should be a $v^2$ term at the front.
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