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- Homework Statement
- A block is performing SHM with an amplitude ##A## and angular frequency ##\omega##, if an identical block is dropped on the original block while it passes through its mean position find the common velocity of the two blocks, assume friction to be sufficient.

- Relevant Equations
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at the mean position of the original block, its velocity is ##V = \omega A##

once the new block is dropped we can conserve momentum in the horizontal direction $$m\omega A = (2m)v$$ $$v = \frac{\omega A}{2}$$

where ##v## is the common velocity of the blocks.

but if instead of conserving linear momentum I try to conserve energy to find the common velocity

$$ (1/2)\omega^2 m A^2 = (1/2)(2m)v^2$$

$$v = \frac {\omega A}{ \sqrt {2}}$$

why is conserving energy to find the common velocity of the blocks wrong here? is it because the friction has done work on the upper block to increase its velocity from zero? but instantaneously the block has no displacement so ##W_{friction} = 0##

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