Question

You are at the controls of a particle accelerator, sending a beam of 3.60 x10^7 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 3.30 x 10^7 m/s. Assume that the initial speed of the target nucleus is negligible and the collision is elastic. (a) Find the mass of one nucleus of the unknown element. Express your answer in terms of the proton mass m. (b) What is the speed of the unknown nucleus immediately after such a collision?

Answer 1

Answer:

a) mass of unknown nucleus = 0.04245 mp, where mp is the proton mass

b) Speed of the unknown nucleus = (7.067 x 10^7) m/s

Explanation:

Considering the initial conditions, the observed collisions are ellastic, i.e, the total kinetic energy are conserved. The proton's mass will refer as $m_{p}$.

(a)

Total kinetic energy conservation  

$\frac{1}{2}m_{p}v_{p_0}^{2}+\frac{1}{2}m_{u}v_{u_o}^{2}=\frac{1}{2}m_{p}v_{p_f}^{2}+\frac{1}{2}m_{u}v_{u_f}^{2}$

where $v_{u_o}$ represents the initial velocity of the unknown element, $m_{u}$ the mass of the unknown element, and $v_{u_f}$ the final velocity of the unknown element

Linear momentum conservation

$m_{p}v_{p_0}+m_{u}v_{u_o}=m_{p}v_{p_f}+m_{u}v_{u_f}$

Using the initial speed of the target nucleus (unknown) is negligible, i.e,  its speed is zero. Thereby, using the relation of linear momentum conservation  given above, it is possible to find an expression of the final speed of the unknown nucleus in terms of its mass, which can be inserted in the relation of the kinetic energy conservation to obtain the value of the mass of the unknown elements, as follows;

$m_{u}v_{u_f}=m_{p}v_{p_0}-m_{p}v_{p_f}\\\\v_{u_f}=\frac{m_{p}(v_{p_0}-v_{p_f})}{m_{u}}$

Substituting this expression in the relation of total kinetic energy conservation,

$m_{p}(v^{2}_{p_0}-v^{2}_{p_f})={m_{u}}v^{2}_{u}_{f}$

Then,

$m_{p}(v^{2}_{p_0}-v^{2}_{p_f})={m_{u}}\frac{m^{2}_{p}(v_{p_0}-v_{p_f})^{2}}{m^{2}_{u}}\\\\m_{u}= \frac{m_{p}(v_{p_0}-v_{p_f})^{2}}{(v^{2}_{p_0}-v^{2}_{p_f})}$

Replacing the given data

$m_{u}= \frac{m_{p}(3.6x10^{7}-3.3x10^{7})^{2}}{((3.6x10^{7})^{2}-(3.3x10^{7})^{2})}$

Then,

$m_{u}=0.04245m_{p}$

(b) Using the relation of the final speed from linear momentum conservation and the above result, the speed of the unknown nucleus is calculated

$v_{u_f}=\frac{m_{p}(v_{p_0}-v_{p_f})}{m_{u}}\\\\v_{u_f}=\frac{m_{p}(3.6x10^{7}-3.3x10^{7})}{0.04245m_{p}}\\\\v_{u_f}=7.067x10^{7} m/s$