# Question #dffe3

Feb 19, 2017

It is given that collision between the protons is elastic. We know that for such collisions both momentum and kinetic energy are conserved.
The total system kinetic energy before the collision equals to the total system kinetic energy after the collision.
Also, in a perfectly elastic collision coefficient of restitution is $1$

Let two protons be designated as $A \mathmr{and} B$. Let the proton $A$ of mass ${m}_{p}$ be traveling along $+ x$ axis, initially with velocity ${\vec{u}}_{A}$.

Momentum of system before collision $= {\vec{p}}_{A} + 0 = {\vec{p}}_{A}$
Momentum of system after collision $= {\vec{p}}_{A}^{'} + {\vec{p}}_{B}^{'}$

By Law of conservation of momentum we have
${\vec{p}}_{A} = {\vec{p}}_{A}^{'} + {\vec{p}}_{B}^{'}$

Since it is proton-proton collision, mass of both is same. As such the above expression reduces in terms of velocities ${\vec{v}}_{A} \mathmr{and} {\vec{v}}_{B}$.
${\vec{u}}_{A} = {\vec{v}}_{A} + {\vec{v}}_{B}$ .....(1)

From Law of Conservation of Energy we get
$\frac{1}{2} {m}_{p} {u}^{2} = \frac{1}{2} {m}_{p} | {\vec{v}}_{A} {|}^{2} + \frac{1}{2} {m}_{p} | {\vec{v}}_{B} {|}^{2}$
$\implies {u}^{2} = | {\vec{v}}_{A} {|}^{2} + | {\vec{v}}_{B} {|}^{2}$ .....(2)

As mass of both protons is identical kinetic energy is shared equally between the two after the collision.
$\frac{1}{2} {m}_{p} | {\vec{v}}_{A} {|}^{2} = \frac{1}{2} {m}_{p} | {\vec{v}}_{B} {|}^{2}$
$\implies | {\vec{v}}_{A} | = | {\vec{v}}_{B} | = v$

Inserting in(2) we get
${u}^{2} = {v}^{2} + {v}^{2}$
$\implies 2 {v}^{2} = {u}^{2}$
$\implies v = \sqrt{{u}^{2} / 2}$

Inserting given value
$\implies v = \sqrt{{\left(317\right)}^{2} / 2}$
$v = 224.15 m {s}^{-} 1$