Physics help please!

A 3.0kg block of wood sits on a table. A 3.0g bullet, fired horizontally at a speed of 500m/s , goes completely through the block, emerging at a speed of 250m/s .

What is the speed of the block immediately after the bullet exits?

Since energy is not conserved(heat generated in the mass) you cannot even remotely think of conserving energy.
Apply momentum conservation.

Momentum of the system Just before collision=0.003[-->3 grams] x 500 + 3 x 0
Momentum of the sys just after collision=0.003 x 250 + 3 x v

equate the 2 equations.
You have your answer.


Questions will come and go mate.
Try to analyze why YOU couldnt do this one,Where did you get stuck?
Did you even have a pen in your hand?
This is much more important.
Good luck;)
:D


@Menard:Must have been some parallel universe.:)

Here is a puzzle for you
an alleged commercial airliner
crashes into a wall - that is a steel framed skyscraper wall,
and penetrates the wall completely, not shedding any bit
of itself outside the wall and is observed to travel as fast
while penetrating the wall, as it did flying through air...

whats up with that?

from the character who was thrown out of Chemistry
for being a FREE RADICAL ...

This is a conservation of momentum problem. So our equation says that the momentum we start with is the momentum we end with.

p_initial, bullet=p_final, block+p_final, bullet

Since momentum is mass times velocity we get

0.003*500=3*v+0.003*250

Or more simply, (multiplying the whole equation by 1000/3)
500=1000 v+250

Then solve for v.