Two protons at rest and separated by 7.00 mm are released simultaneously. What i

Question

Two protons at rest and separated by 7.00 mm are released simultaneously. What is the speed of either at the instant when the two are 11.00 mm apart?

I initially tried to solve as if acceleration would be contant:
F=(kqQ)/(r^2) found electrostatic force
a=F/m plugged in for force and mass of a proton solved for a
X-Xo=Vot + 1/2at^2 => 2-0=(0)t +1/2at^2 solved for t
V=Vo +at solved for V

I used the formula U=(kq^2)/(d) for d=7E-3 then 11E-3
U1-U2 should be the energy spent or converted, I'm not sure how to get to the velocity from here.

Explanation / Answer

Initially both are at rest

we have

Ua = 0 Ub =0

Now initial Potential Energy = (Kxqxq)/r

r= 7mm = 7 x 10^-3 m

q = 1.602 x 10^-19 C

Initial Potential energy = 3.3 x 10-26 Joules

FInal DIstance = 11 mm

Final Potential Energy =2.1 x 10-26 Joules

Now final speed of either of the protons will be the same as both have the same mass and the same charge Therefore the force acting on each of them is also the same

Therefore Va =Vb = V m/s

From Conservation of energy,

Initial Energy = Final Energy

0 +  3.3 x 10-26 = =2.1 x 10-26 + 2 x 0.5 mV2

1.2 x 10-26 = 1.6726 x 10-27 V2

Velocity V = 2.68 m/s

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