Question

A 3.0-g bullet is fired horizontally into a 10-kg block of wood suspended by a rope from the ceiling. The block swings in an arc, rising 3 mm above its lowest position. The velocity of the bullet was:

Answer 1

i thought the formula was v=\[\sqrt{2gh}\] but i was wrong

Answer 2

also tried 0.5mv^2=mgh also wrong

Answer 3

I'm not sure if this is right but how about this....

If we assume all the kinetic energy of the bullet is transferred to the block, swings the block up by 3mm. We can calculate the change in potential energy in the block using

\[\Delta U = M_{block} \, g \, (h_{2}-h_{1})\]

where

\[h_{2} - h_{1} = 3mm\]

Then equate potential and kinetic energy to calculate the velocity, V

I roughly did the calculation and got something like 0.015m/s

What do you think? Assumptions valid?

Answer 4

nope
actually just found it is v= (total mass *impact velocity)/mass of bullet

Answer 5

i just don't see a formula... i know impact speed is v= sqrt{2gh}

Answer 6

i know that is right...but i don't see the general formula they would of refered to, to answer the question

Answer 7

I GET IT

Answer 8

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Answer 9

I AM INVINCIBILE!!!!!!!!!!!!!!!!!!

Answer 10

I WILL NOW FEED ON MY NEXT VICTIM