F4 Phy__急*

Ok. I understand your question. i am sorry to say that the answer suggested by the guy above is wrong. The correct answer is shown as follows:

** Step One
Note that the mass of the bullet is 2g, i.e. 0.002kg.
Initial K.E. of the bullet = 0.5 x mass x (speed^2) = 0.5 x (0.002) x (100^2) = 10J

K.E. of the bullet passing through a piece of paper = 0.5 x (0.002) x (80^2) = 6.4J

** Step Two
Change of K.E. of the bullet = 10-6.4 = 3.6J
(NOTE: this amount of energy is lost to the paper, as a form of heat energy or whatever other possible forms

** Step Three
Since the initial K.E. is 10J and each paper causes 3.6J energy loss for the bullet, the number of paper that the bullet can pass through = 10 / 3.6 = 2.78

** Step Four
Hence, we choose the smallest integer, i.e. TWO.
The bullet can pass through TWO pieces of paper and the remaining energy of bullet is equal to 10 - 3.6 - 3.6 = 2.8J which is smaller than 3.6J and does not allow the bullet passing through the third paper.

*** PS: What's the problem of the previous calculation demonstrated by the guy above?

Note that in the KE formula, there is a speed^2 term. The squared term makes the calculation entirely different. Also, the momentum change is not a linear change. You can see this fact from the simple calculation below:

@ v =100m/s, momentum of the bullet = 0.2kgm/s
@ v =80m/s, momentum of the bullet = 0.16kgm/s
@ v =52.9m/s, momentum of the bullet = 0.11kgm/s

Thank you for reading. I hope this helps

p=mv
p=2*100=200
afterpassing through one paper
the loss of p
=2*(100-80)=40
so the bullet can at most pass throngh 200/40=5