# nuclear physics [closed]

1 vote
92 views

In the above question how they have written the ratio in first step , and what is the use of it .

closed with the note: Not enough effort shown. Please at least show your thought process.
closed Dec 16, 2016
The equation for fractional energy loss is probably derived in your book.
I've reopend this one to allow you to edit and improve it.
@Einstein : Isn't it possible to edit the question after it has been closed? This feature is necessary.
The feature to edit after a question has been closed has been added, so I have closed the question as before.

1 vote

Initial KE of particle 1 is $\frac12m_1u_1^2$. Final KE is $\frac12m_1v_1^2$. Fractional KE retained after collision is $(v_1/u_1)^2$.

Loss of momentum by particle 1 equals gain of momentum by particle 2 :
$m_1(u_1-v_1)=m_2(v_2-u_2)$
$u_1-v_1=\mu(v_2-u_2)$ ... (i)
where $\mu=m_2/m_1$.
Law of Restitution with $e=1$ for elastic collisions :
$v_2-v_1= -e(u_2-u_1) = u_1-u_2$
$u_1+v_1=v_2+u_2$ ... (ii)

Combining (i) and (ii) :
$2v_1=(1-\mu)v_2+(1+\mu)u_2$
$2u_1=(1+\mu)v_2+(1-\mu)u_2$.

Assuming that the target is initially at rest $(u_2=0)$ then
$v_1/u_1=(1-\mu)/(1+\mu)$.

In this case $\mu=m_2/m_1=2$ so
$(v_1/u_1)^2=(-1/3)^2=\frac19$.

After $n$ collisions the KE of particle 1 is $(\frac19)^n$ of its initial KE :
$(\frac19)^n=\frac{1}{6561}$
$9^n=6561=9^4$.

The number of collisions required is 4.

answered Dec 13, 2016 by (28,876 points)
selected Dec 14, 2016 by koolman