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Suppose I have a body of mass $M$ connected to a spring (which is connected to a vertical wall) with a stiffness coefficient of $k$ on some frictionless surface. The body oscillates from point $C$ to point $B$ and $CB=d$. Its motion is harmonic. The total energy of such a system is simply $\frac{1}{2} k \left(\frac{d}{2} \right)^2=\frac{1}{8} k d^2$ (because the equilibrium point is at $d/2$ and $d/2$ is the amplitude). Now, suppose we drop vertically some plasticine with the same mass $M$ from some height $h$. After it hits the oscillating object, it just sticks to it.

My question is - why the total energy of the system doesn't change if the plasticine hits the object at point $C$ but it changes when it hits the object in the middle of $CB$ (it equals there to $\frac{1}{16} k d^2$)? Intuitively, I do understand that such a plastic collision contributes to the loss of energy, but I'm not sure how exactly it fits here and how the energy is lost. And even so, I do not understand how the position of the body influences the change in energy.


Proposed solution:

The total mechanical energy at point $d/2$ is $E_{k,max}=P_{total}=\frac{1}{8} k d^2$:

$E_k=\frac{M}{2}V^2\\ V=\sqrt{\frac{2E_k}{M}}=\sqrt{\frac{2P_{total}}{M}} $

Due to the conservation of the horizontal momentum we get:

$MV=2MV'\\ \frac{2E_k}{M}=4V'^2\\ V'=\sqrt{\frac{E_k}{2M}}=\sqrt{\frac{\frac{1}{8}kd^2}{2M}}=\sqrt{\frac{kd^2}{16M}}$

Therefore, the momentary kinetic energy (which is the total energy) after the hit is:

$$E_{k, after}=\frac{2MV'^2}{2}=M \frac{kd^2}{16M}=\frac{kd^2}{16}$$

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1 Answer 1

At point $C$, the total energy of the oscillator is completely due to the potential energy in the spring. The plastic adds mass to the system, but this does not affect the energy content. At any point between $C$ and $B$, the system will have some combination of potential and kinetic energy, and the kinetic energy will in general be affected by the collision with the plastic body.

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I do understand that the potential elastic energy of the spring transforms into the kinetic energy of the body and vice versa during the oscillations, but how does the collision affect the speed of the moving body? The plasticine is being thrown vertically, so there's no horizontal component of the velocity that can contribute to the change of kinetic energy. –  brmch8 Apr 14 '13 at 17:29
    
Here's another way to look at it: Since the plastic has no motion in the horizontal direction initially, the oscillator must do some work to bring it up to speed once it collides (the plastic sticks, so it must match the velocity of the oscillating body). At the end points the oscillator has no velocity, so no additional work is required. –  chase Apr 14 '13 at 18:03
    
So basically, when the plastic hits the body, it is willing to keep a zero horizontal velocity (because of inertia) so it is literally forcing the object to slow down a bit, until their kinetic energies are similar (thus the velocities are equal). I think that is the same as your explanation, and it seems fine for me, however it looks like the object "shares" its kinetic energy with the plastic, doesn't it? So there's no loss of energy. If the kinetic energy is lost then where it goes? And how can I explicitly calculate the total energy of the system at any given point of collision? –  brmch8 Apr 14 '13 at 18:31
    
Yes, you're correct that the two bodies "share" the kinetic energy. However, since the collision is inelastic, the oscillator must give up more energy than would simply be required to give the plastic body sufficient kinetic energy. I.e. some of the energy goes into heat/sound/deforming the object, instead of just imparting horizontal momentum. If you wanted to do an experiment to measure the energy loss, simply measure the amplitude of oscillation before and after the collision. –  chase Apr 14 '13 at 18:41
    
"the oscillator must give up more energy" - how can it give "more" energy? Energy must come from somewhere mustn't it? I do understand that during the collision some of the energy goes to heat and friction, but it is the energy of the plastic that is lost, am I wrong? Or is the energy of the moving object is also lost because of third Newton's law? And isn't the same amount of energy is needed to move both the plastic and the object when the object momentarily stops at the end points? It doesn't move there for a moment, but still you have a larger mass $2M$ to move, no? –  brmch8 Apr 14 '13 at 19:05

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