Question

Two protons are initially held at rest with a distance
d=0.020 cm between them.
(q[p]=1.60^(-19) C q[m]=1.67^(-27)kg)

a) when the protons are released from rest, will they
accelerate toward each other or away from each
other? why?

the protons accelerate away from each other because charges
with the same sign repel each other.

b) what is the distance between the two protons
when they reach their max speed?

d-> infinity as the protons reach their max speed

c) what is the initial kinetic energy of the system?

KE[i]=2(1/2mv^2)=m(0)^2 = 0

d) potential energy?

PE[i]=(k(q[p]^2))/d = 1.151^(-24) Nm/C

e) what is the final potential energy of the system?

PE[f]=0

f)use the law of conservation of energy to solve
for the final speed of the protons.

what is part f?

Answer 1