Answer to Question #271228 in Mechanics | Relativity for Bhumie

Question

A proton traveling at 107m/s Collides with a stationary particle and bounces back add 2×2 is collides with a stationary particle and bounces back at 2*10^6m/s. If the Particles moves forward at 3*10^6. Find the mass of the station iPad scree find the mass of the stationary Particle

Answer 1

by momentum conservation:

$m_1v_1-m_1v_2=m_2v$

where m₁, m₂ are masses of proton and stationary particle,

v₁, v₂ is speed of proton before and after collision

v is speed of stationary particle after collision

then:

$m_2=\frac{m_1v_1-m_1v_2}{v}=\frac{1.67\cdot10^{27}(10^7-2\cdot10^6)}{3\cdot10^6}=6.68\cdot10^{-27}$ kg