# Find penetration of bullet - classical mechanics

I am studying for my own education form a book and the last part of this question has an answer in the book which I disagree with. Can anyone confirm that it is I or the book that is in error?

Here's the question: I have proved the first three statements and agree with the book that.

$R_1=\frac{(u^2-v^2)m}{4a}$

$R_2=\frac{mv^2}{2a}$

emergence of bullet if $u>2v$

It is the final part, to determine x, that I have different from the book. Here is my work:

Let V = velocity of second bullet after emergence from Y, then through Y:

$u=3v$

$s=4a$

And using $F=ma$ as was done in the earlier parts of the question we have:

$acceleration = \frac{-v^2}{2a}$

Hence, using "$v^2=u^2+2as$" we have:

$V^2=9v^2-2*\frac{v^2}{2a}*4a$ therefore

$V=\sqrt{5}v$

Therefore though X we have (setting final velocity V to zero):

$u=\sqrt{5}v$

$V = 0$

and using "f=ma" we have:

$acceleration=\frac{-R_1}{m}$

Therefore

$Acceleration=\frac{v^2-u^2}{4a} = \frac{v^2-5v^2}{4a} = \frac{-v^2}{a}$

And so using "$v^2=u^2+2as$" we have:

$0^2=5v^2-\frac{2v^2x}{a}$

And so x = $\frac{5a}{2}$

But book answer is $\frac{5a}{4}$

Clearly my answer indicates that the bullet passes right through X whereas the book answer shows that the bullet is stopped inside of X.

Thanks for any help, Mitch.

• Maybe the book assumes that when bullet 2 hits bullet 1, you have to add their masses. – Trurl Apr 24 '18 at 14:26

## 1 Answer

Confusing that you used a for both acceleration and the thickness but you've used 4a for the thickness of X instead of 2a.

• Thanks so much. A stupid slip - I wasn't seeing it. Thanks again. – gnitsuk Apr 24 '18 at 15:41
• Sorry, maybe I'm still confused. Part 1 gave R1=(u^2-v^2)m/4a and because v^2=u^2+2as leads to a = (v^2-u^2)/2s and as s=2a then 2s=4a. That's where I get the 4a from. – gnitsuk Apr 24 '18 at 15:52
• But you state that s = 4a on line 8. And then you use that 4a to determine V which in turn determines other variables. Something isn't correct with that. I suggest you use a different symbol for acceleration so it isn't as confusing. – Phil H Apr 24 '18 at 16:08