Set up:
There's a block of mass $M$ connected to a spring of negligible mass. This is set up horizontally, and there is no friction between the block and the surface. The block oscillates with simple harmonic motion, with an amplitude $A$.
Case 1:
A piece of clay of mass $M$ (the same mass as the block) is dropped on the block from a very small height when it is at its equilibrium point. The clay sticks to the block.
Finding the change in max velocity:
- $M \cdot v_1 = 2M \cdot v_2 $ (using conservation of momentum)
- $v_1 = 2\cdot v_2$
- $v_2 = \frac{v_1}2$
Energy change for this case (using the change in max velocity):
$ME_{1,total}=KE_{1,max}=\frac12 M v_{max}^2 = \frac12(M \cdot v_{max}^2)$
$ME_{2,total}=KE_{2, max}=\frac12 (2M) \cdot \left(\frac{v_{max}}{2}\right)^2 = \frac14\left(M \cdot v_{max}^2\right) $
$\frac{ME_1 }{ ME_2} = \frac12$
So the ME of the system decreases by $\frac12$.
Case 2:
Same as case 1 except the piece of clay goes on the block when it is furthest from its equilibrium point.
Energy change for this case:
The amplitudes before and after the clay is added are the same, so both systems have the same PE since the $k$ and $A$ are the same (in the equation $PE=\frac 12 \cdot k\cdot x^2$), and therefore the $ME$ of the systems are the same.
So my question is, where does the energy lost in the first case go? Having half the amount of energy after the clay is added seems like a big difference. I am thinking, with no horizontal velocity, the clay dropped at the equilibrium point will want to stay there (inertia) so energy is used up in adding the force to move it horizontally with the block, which is not necessary for case 2 since the clay is added while the block has no horizontal motion. Does the energy lost in the first case go to stress and possibly deformation on the material of the clay? Or does the energy lost simply go to slowing the block and not need to have another effect?
