Electric potential energy question in EK

[Alongs50]

I have a quick algebraic question about this EK problem.

Theres a positive particle 25 cm from a 2nd positive particle (which is held stationary), and the first is released and accelerates away from the 2nd. When it's moved 25 cm it has reached a velocity of 10 m/s. What is the max velocity the 1st particle will reach?

a. 10
b. 14
c. 20
d. since the first particle will never escape the electric field of the 2nd, it will never stop accelerating and will reach an infinite velocity.

in the explanation they use U= kqq/r and say that the particle loses half potential energy to kinetic when r is doubled, so we should multiply K.E. by 2. Where I get lost is when they multiply KE by 2 "we must multiply the velocity by the square root of 2 or approx 1.4".

2 * KE = 1/2 m * v^2

are they taking the square root of both sides and thats where they got the 1.4? but if so why did they say the velocity is being multiplied by 1.4 and not the KE..

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[Arctangent]

The way I understand the explanation, you are basically starting with your normal kinetic energy equation:

KE = 1/2 m * v^2

The first point is that at 10m/s, half of your potential energy has been converted into kinetic, and you have half of the potential energy left for conversion. Thus, at this point in time:

KE_half = 1/2 m * (v_half)^2 ... where v_half is 10m/s

KE_half = 1/2 KE_max

By substitution:

KE_max = m * (v_half)^2

At this point (maximum kinetic energy) conceptually, you know that the original kinetic energy equation must hold:

KE_max = 1/2 m * (v_max)^2

By substitution:

v_half^2 = 1/2 v_max^2

100 = 1/2 v_max^2

v_max = ~14

Perhaps a more concise explanation is that given a kinetic energy value, and a corresponding velocity value, if you double kinetic energy, velocity must increase by the square root of 2. In the equation KE = 1/2 m * v^2, (1/2m) is a constant, so KE is proportional to the square root of v.

 

[ingramw1202]

I have a quick algebraic question about this EK problem.

Theres a positive particle 25 cm from a 2nd positive particle (which is held stationary), and the first is released and accelerates away from the 2nd. When it's moved 25 cm it has reached a velocity of 10 m/s. What is the max velocity the 1st particle will reach?

a. 10
b. 14
c. 20
d. since the first particle will never escape the electric field of the 2nd, it will never stop accelerating and will reach an infinite velocity.

in the explanation they use U= kqq/r and say that the particle loses half potential energy to kinetic when r is doubled, so we should multiply K.E. by 2. Where I get lost is when they multiply KE by 2 "we must multiply the velocity by the square root of 2 or approx 1.4".

2 * KE = 1/2 m * v^2

are they taking the square root of both sides and thats where they got the 1.4? but if so why did they say the velocity is being multiplied by 1.4 and not the KE..

I would do this problem using energy conservation. Initially, the particle has only potentially energy which is defined as U=kqq/r. When the particle is released, the potential energy is converted into kinetic energy which = KE=1/2mv^(2). So kinetic energy will reach a max when potential energy is at a minimum. Potential energy is at a minimum when the particle has moved an infinite distance away from the stationary particle, because the limit as r approaches infinity of PE=kqq/r = 0. So at infinity, the particle has a maximum kinetic energy, and as the particle moves further and further away from the stationary particle, its potential energy decreases as its kinetic energy increases and all of the potential energy is essentially been converted into kinetic energy. E initial = E final.

U=kqq/r=1/2mv^(2).
You are given the velocity of the particle at 25 cm from its original position, so the initial potential energy, which was once due to a distance of 25 cm, is now due to a distance of 50 cm. You can use this information, with the velocity they give you to write another equation.Since r is doubled, PE is halved. This means that half the PE has been converted to KE. Use this to write an equation for 1/2 KE max. ANd because total energy remains constant, then 1/2KEmax is half of the kinetic energy the particle will have at its maximum kinetic energy.
2* KEhalfmax=KEmax
2[1/2 (m)(10)^2]=1/2mvmax^2

2m(10)^2=mvamx^2
2(100)=vmax squared.
(200)^1/2 =vmax

 

[umdnjmed]

The kinetic energy is doubled but the 1/2m value in 1/2mv^2 is a constant. Therefore v^2 (not v) increases by 2. In order for V^2 to increase by 2, v must increase by 1.4 because then you will get (1.4V)^2 = 2V^2