A 4.40g bullet moving with an initial speed of vi = 405 m/s is fired into and passes through a 1.20 kg block.
The block, initially at rest on a frictionless horizontal surface, is connected to a spring with a spring constant of 880 N/m. If the block moves x = 4.37 cm to the right after impact, calculate the speed at which the bullet emerges from the block.
2 answers
I don't see how this can be done without some crazy assumptions about energy lost. On the sliding, you have zero friction (no energy lost), but the block itself. Is one to assume no energy is lost in the block internally as a bullet goes through it? It can be worked with that assumption, but frankly, that is ridiculous.
I'm with Bob on this. The ridiculous answer goes something like:
1/2 (4.4e-3)405^2 - 1/2 880 (.0437^2) = 1/2 (4.4e-3) vf^2
But you've just torpedoed a whole generation of ballistic pendulum problems...
1/2 (4.4e-3)405^2 - 1/2 880 (.0437^2) = 1/2 (4.4e-3) vf^2
But you've just torpedoed a whole generation of ballistic pendulum problems...
1 answer