# Question #8effb

Mar 19, 2015

The speed of the proton at impact = $2 \times {10}^{6} \text{m/s}$

We use the idea of conservation of momentum.

The initial momentum of the proton = ${m}_{p} {v}_{p} = 1 \times {v}_{p}$

The initial momentum of the He atom is zero (its not moving).

The final momentum of the proton is zero since it is an elastic collision and has exchanged all its momentum with the He atom.

The final momentum of the He atom

= $4 \times 5 \times {10}^{5} = 2 \times {10}^{6} \text{amu m/s}$

So $1 \times {v}_{p} = 2 \times {10}^{6} \text{amu m/s}$

${v}_{p} = 2 \times {10}^{6} \text{m/s}$

Mar 20, 2015

Sorry for the formatting but I cannot see the symbols on this computer...

I found $v = 1.25 \cdot {10}^{6} \frac{m}{s}$.
You can use the Conservation of Momentum AND of Kinetic Energy:
Before ............................. After
o--> O ............................ <--o O-->

Cons. Momentum ($p = m v$:
${p}_{B} = {p}_{A}$
Cons. Kinetic E. ($K = \frac{1}{2} m {v}^{2}$)
${K}_{B} = {K}_{A}$

In numbers:
$m {v}_{B p r o t} + 0 = m {v}_{A p r o t} + 4 m \left(5 \cdot {10}^{5}\right)$
$\frac{1}{2} m {\left({v}_{B p r o t}\right)}^{2} + 0 = \frac{1}{2} m {\left({v}_{A p r o t}\right)}^{2} + \frac{1}{2} \left(4 m\right) {\left(5 \cdot {10}^{5}\right)}^{2}$
Simplifying:

${v}_{B p r o t} = {v}_{A p r o t} + 20 \cdot {10}^{5}$
${\left[{v}_{B p r o t}\right]}^{2} = {\left[{v}_{A p r o t}\right]}^{2} + 4 {\left[5 \cdot {10}^{5}\right]}^{2}$

I got from the second: ${v}_{A p r o t} = \sqrt{{\left({v}_{B p r o t}\right)}^{2} - 1 \cdot {10}^{12}}$ substituting in the first given as: ${v}_{B p r o t} - 20 \cdot {10}^{5} = {v}_{A p r o t}$
giving:
${v}_{B p r o t} - 20 \cdot {10}^{5} = \sqrt{{\left({v}_{B p r o t}\right)}^{2} - 1 \cdot {10}^{12}}$
and square both sides to get:
${v}_{B p r o t} = 1.25 \cdot {10}^{6} \frac{m}{s}$