Question

A proton is traveling to the right at $2.0\times {10}^7\mathrm{m}/\mathrm{s}$. It has a head on perfectly elastic collision with a carbon atom. The mass of the carbon atom is $12$ times the mass of the proton. What are the speed and direction of each after the collision?

Short answer

The speed of the proton is $-1.69\times {10}^7\mathrm{m}/\mathrm{s}.$

The speed of the carbon atom is$3.1\times {10}^6\mathrm{m}/\mathrm{s}.$

Step-by-step solution

Given information

We have given,

speed of the proton =$2\times {10}^7\mathrm{m}/\mathrm{s}$

$\mathrm{Mass}\mathrm{of}\mathrm{carbon}\mathrm{atom}=2\times \mathrm{Mass}\mathrm{of}\mathrm{proton}$

We have to find the speed and direction of each one after the collision.

Simplify

From the elastic collision when one particle will be at rest, we know their final velocity will be,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{if}}=\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}{\mathrm{v}}_1 \\ {\mathrm{v}}_{2\mathrm{f}}=\frac{2{\mathrm{m}}_1}{{\mathrm{m}}_1+\mathrm{m}}{\mathrm{v}}_2 \end{gathered}$$

then,

After putting the value of

$$\begin{gathered} {\mathrm{m}}_1=\mathrm{mass}\mathrm{of}\mathrm{proton} \\ {\mathrm{m}}_2=\mathrm{mass}\mathrm{of}\mathrm{carbon}\mathrm{atom} \end{gathered}$$

$$\begin{gathered} {\mathrm{v}}_{\mathrm{if}}=\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}{\mathrm{v}}_1 \\ {\mathrm{v}}_{1\mathrm{f}}=\frac{{\mathrm{m}}_{\mathrm{p}}-12{\mathrm{m}}_{\mathrm{p}}}{{\mathrm{m}}_{\mathrm{p}}+12{\mathrm{m}}_{\mathrm{p}}}\left({\mathrm{v}}_1\right) \\ {\mathrm{v}}_{1\mathrm{f}}=\frac{-11{\mathrm{m}}_{\mathrm{p}}}{13{\mathrm{m}}_{\mathrm{p}}}\times (2\times {10}^7\mathrm{m}/\mathrm{s}) \\ {\mathrm{v}}_{1\mathrm{f}}=-1.69\times {10}^7\mathrm{m}/\mathrm{s} \end{gathered}$$

Similarly,

$$\begin{gathered} {\mathrm{v}}_{2\mathrm{f}}=\frac{2{\mathrm{m}}_1}{{\mathrm{m}}_1+{\mathrm{m}}_2}{\mathrm{v}}_1 \\ {\mathrm{v}}_{2\mathrm{f}}=\frac{2{\mathrm{m}}_{\mathrm{p}}}{13{\mathrm{m}}_{\mathrm{p}}}{\mathrm{v}}_1 \\ {\mathrm{v}}_{2\mathrm{f}}=\frac{2}{13}\times (2\times {10}^7\mathrm{m}/\mathrm{s}) \\ {\mathrm{v}}_{2\mathrm{f}}=3.1\times {10}^6\mathrm{m}/\mathrm{s} \end{gathered}$$