INTRO:
A proton and an alpha particle are momentarily at rest at a distance r from each other. They then begin to move apart. Find the speed of the proton by the time the distance between the proton and the alpha particle doubles.
Both particles are positively charged. The charge and the mass of the proton are, respectively, e and m. The charge and the mass of the alpha particle are, respectively, 2e and 4m.
PROBLEM-SOLVING STRATEGY 25.1 Conservation of energy in charge interactions
MODEL: Check whether there are any dissipative forces that would prevent the mechanical energy from being conserved.
VISUALIZE: Draw a before-and-after pictorial representation. Define symbols that will be used in the problem, list known values, and identify what you are trying to find.
SOLVE: The mathematical representation is based on the law of conservation of mechanical energy:
Kf+qVf=Ki+qVi.
Is the electric potential given in the problem statement? If not, you'll need to use a known potential, such as that of a point charge, or calculate the potential using the procedure given in Problem-Solving Strategy 25.2.
Kf and Ki are the sums of kinetic energies of all moving particles.
Some problems may need additional conservation laws, such as conservation of charge or conservation of momentum.
ASSESS: Check that your result has the correct units and significant figures, is reasonable, and answers the question.
Model
The particles described in this problem interact under the effect of the electric force, which is a conservative force, so the system's mechanical energy is conserved.
Visualize
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PART A:
Which of the following quantities are unknown?
A. initial separation of the particles
B. final separation of the particles
C. initial speed of the proton
D. initial speed of the alpha particle
E. final speed of the proton
F. final speed of the alpha particle
G. mass of the proton
H. mass of the alpha particle
I. charge of the proton
J. charge of the alpha particle
Enter the letters of all the correct answers in alphabetical order. Do not use commas. For instance, if A, C, and D are unknowns, enter ACD.
SOLUTION:
The final separation between the particles is, essentially, known: It is twice the initial separation, or 2r. In this problem, then, there are really only two unknowns: the final speed of the proton, (vf)p, which is what you are trying to find, and the final speed of the alpha particle, (vf)α.Recall that a problem with two unknowns requires two equations to be solved. Here, the law of conservation of mechanical energy provides one of the equations. To find the second equation, think what other physical quantity besides energy is conserved, and translate that into a mathematical expression. But before you do that, it's helpful if you complete a before-and-after pictorial representation of the problem. Your drawing might look like this:
answer: BEF
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PART B:
Find the speed of the proton (vf)p by the time the distance between the particles doubles.Express your answer in terms of some or all of the quantities e, m, r, and ϵ0.
SOLUTION:
Using the law of conservation of mechanical energy, we know that the sum of the initial energies is equal to the sum of the final energies.
At first, when the two particles are right next to each other, there is no kinetic energy yet, but quite a bit of potential energy.
The potential energy can be written using U = qV -> Uelec = 1/[4π⋅ϵ0]⋅q1q2/dist
Therefore, since initial dist = r, q1 = e & q2 = 2e
The initial potential energy = Ui = 1/[4π⋅ϵ0]⋅2e⋅e/r = 1/[4π⋅ϵ0]⋅2e2/r
and sine the final dist = 2r,
the final potential energy =
Uf = 1/[4π⋅ϵ0]⋅2e⋅e/r = 1/[4π⋅ϵ0]⋅e2/r
What about the kinetic energy?
if the particles start at rest, then the initial kinetic energy =
Ki = 0
the total final kinetic energy will be equal to the final kinetic energy of the proton plus the final kinetic energy of the alpha. But how can we relate these two velocities to each other?
The only quantity that remains constant for the proton and alpha as they move apart is their magnitude of momentum.
momentum : P = m*v
conservation of momentum: ∑P1 = ∑P2∴ vα1⋅mα + vp1⋅mp = vα2⋅mα + vp2⋅mpsince both initial velocities are 0, 0 = vα2⋅mα + vp2⋅mp→vα⋅mα = vp⋅mp
→vα/vp = mp/mα = m/4m = 0.25
∴ vα = (0.25)vp
KE = 1/2 mv2∴ KE = 1/2mαvα2 + 1/2mpvp2 sub in for vα = (0.25)vp...KE = 1/2(4m)((0.25)vp)2 + 1/2(m)vp2
∴KE = (5/8)mvp2
plugging that into the final equation... 0 + 1/[4π⋅ϵ0]⋅2e2/r = (5/8)mvp2 +1/[4π⋅ϵ0]⋅e2/rthen solve for v!→ 1/[4π⋅ϵ0]⋅e2/r = (5/8)mvp2 → 2/[5π⋅ϵ0]⋅e2/r = mvp2 ⇒∴vp = SQRT{2⋅e2/[5r⋅m⋅π⋅ϵ0]}
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PART B:
The best way to check whether your result from Part B is correct is to check that it has the correct units. Which of the following expressions, where C stands for coulombs, N for newtons, kg for kilograms, and m for meters, represents the correct SI units for the expression found in part B?
SOLUTION:
just plug in for the units...
e ≡ C, r ≡ m, m ≡ kg, ϵ0 ≡ F/m = C2/Nm2
and all numbers have no units...
→ v = SQRT{C2/[m⋅kg⋅C2/Nm2]]}
The C's cancel out, one of the m's cancel out, and the N jumps up to the numerator...
⇒ units = √N⋅m/kg
