Question

An electron (mass m₁ = 9.11 x 10^-31 kg) and a proton (mass m₂ = 1.67 x10^-27 kg) attract each other via an electrical force. Suppose that an electron and a proton are released from rest with an initial separation d = 2.40 x 10^-6 m. When their separation has decreased to 0.510 x 10^-6 m, what is the ratio of (a) the electron's linear momentum magnitude to the proton's linear momentum magnitude, (b) the electron's speed to the proton's speed, and (c) the electron's kinetic energy to the proton's kinetic energy?

Answer 1

a)
Since there is no external force, the momentum is conserved.
If we consider the electron-proton as a single system, electrical force is an internal force.
Initial momentum = Final momentum
Initial momentum = 0 = Final momentum
Take ve and vp are the speeds of electron and proton respectively. me and mp are the masses of electron and proton.
Final momentum = me ve + mp vp = 0
me ve = - mp vp
Pe = - Pp
Magnitude of the ratio of the proton's linear momentum to that of electron, Pp/Pe = 1

b)
Pe = Pp (magnitude)
me ve = mp vp
ve/vp = mp/me
= [(1.67 x 10^-27)/(9.11 x 10^-31)]
= 1833.15

c)
Electron's kinetic energy, KEe = 1/2 me (ve)²
Protons's kinetice energy. KEp = 1/2 mp (vp)²
KEe/KEp = [(1/2 me (ve)²) / (1/2 mp (vp)²)]
= [me ve x ve] / [mp vp x vp]
= [Pe x ve] / [Pp x vp]
= [Pe/Pp] x [ve/vp]      
= ve/vp        (Pe = Pp)
= 1833.15