91Ó°ÊÓ

Given that you are at the controls of a particle accelerator, sending a beam of$1.50×{10}^7$ m/s protons (mass m) at a gas target of an unknown element.

Your detector tells you that some protons bounce straight back after a collision with one of the nuclei of the unknown element.

All such protons rebound with a speed of $1.20×{10}^7\mathrm{m}/\mathrm{s}$.

Assume that the initial speed of the target nucleus is negligible and the collision is elastic.

Let A be the proton and B be the target nucleus.

Speed of incident beam$\mathrm{u}=1.5×{10}^7\mathrm{m}/\mathrm{s}\mathrm{m}/\mathrm{s}$ m/s

Rebound speed $\mathrm{v}=-1.2×{10}^7\mathrm{m}/\mathrm{s}$

Let${\mathrm{m}}_{\mathrm{A}}$ be mass of proton and equals m.

${\mathrm{m}}_{\mathrm{B}}$be mass of target nucleus say M

Now$\mathrm{v}=\left(\frac{{\mathrm{m}}_{\mathrm{A}}-{\mathrm{m}}_{\mathrm{B}}}{{\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}}\right)\mathrm{u}$

This implies${\mathrm{m}}_{\mathrm{B}}={\mathrm{m}}_{\mathrm{A}}\left(\frac{\mathrm{u}-\mathrm{v}}{\mathrm{u}+\mathrm{v}}\right)$

So

$$\begin{gathered} \mathrm{M}=\left(\frac{1.5×{10}^7\mathrm{kg}+1.2××{10}^7\mathrm{kg}}{1.5×{10}^7\mathrm{kg}-1.2××{10}^7\mathrm{kg}}\right)\mathrm{m} \\ =\frac{2.7}{0.3}\mathrm{m} \\ =9\mathrm{m} \end{gathered}$$

Hence, mass of one nucleus of unknown element is 9m

Let V be the speed of unknown nucleus just after collision.

$$\begin{gathered} \mathrm{V}=\frac{2\mathrm{m}}{\mathrm{m}+9\mathrm{m}}\left(1.5×{10}^7\mathrm{m}/\mathrm{s}\right) \\ =3×{10}^6\mathrm{m}/\mathrm{s} \end{gathered}$$

Speed of nucleus is$3×{10}^6$ m/s