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