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