A proton and an alpha particle ( q= + 2e, m=4u ) are fired directly toward each other from far away, each with an initial speed of 0.01 c.What is their distance of closest approach, as measured between their centers?
This is not very different from your earlier problem with the three electrons and the proton in the middle if the triangle.
All they have in the beginning is kinetic energy, 1/2 m v^2
All they have when they stop, glaring at each other, is potential energy.
Again your potential energy is that integral of the force in from infinity k q1 q2/r
but i think when hey are closest their velocities are not zero since their masses are different. i tried using conservation of momentum also.
You are right. They do not stop. Momentum is conserved
(m1+m2) Vfinal = m1 v1 initial + m2 v2 initial
one v initial will be negative of course.
Then
final KE of system = (1/2) m Vfinal^2
That KE is smaller than the original sum of KEs
the difference is PE = k q1 q2/r
Holy bals this is hard
To find the distance of closest approach between a proton and an alpha particle, we can use the principles of conservation of momentum and conservation of kinetic energy.
Given:
Mass of the proton (m_p) = 1.673 × 10^-27 kg
Mass of the alpha particle (m_alpha) = 6.644 × 10^-27 kg
Charge of the alpha particle (q_alpha) = +2e = 2 × 1.602 × 10^-19 C
Initial speed of both particles (v_initial) = 0.01c
The first step is to convert the initial speed from a fraction of the speed of light to meters per second.
Light travels at a speed of approximately 3.00 × 10^8 m/s, so:
v_initial = 0.01 * 3.00 × 10^8 m/s
v_initial = 3.00 × 10^6 m/s
Now, let's calculate the initial momentum of each particle.
The momentum (p) of an object is given by the product of its mass and velocity.
For the proton:
p_proton_initial = m_proton * v_initial
For the alpha particle:
p_alpha_initial = m_alpha * v_initial
Next, since momentum is conserved in this interaction, we can equate the initial momentum of the proton to the final momentum of the system (proton + alpha particle) at the closest approach.
Therefore, p_proton_initial = p_proton_final + p_alpha_final
Similarly, the initial kinetic energy of the system is equal to the final kinetic energy at the closest approach.
So, the kinetic energy of the system is conserved during the interaction.
The initial kinetic energy of the proton is given by:
KE_proton_initial = (1/2) * m_proton * v_initial^2
Using the same logic, we can calculate the initial kinetic energy of the alpha particle as:
KE_alpha_initial = (1/2) * m_alpha * v_initial^2
At the closest approach, both particles will be momentarily at rest.
Therefore, the final kinetic energy of the system will be zero.
Now, since kinetic energy is conserved, we have:
KE_proton_initial + KE_alpha_initial = 0
Finally, we can solve for the distance of closest approach (r) using the equation for the potential energy associated with the electric force between the particles.
The potential energy (PE) is given by the equation:
PE = (q_proton * q_alpha) / (4πε₀r)
At the closest approach, the potential energy is at a minimum, which means it is equal to zero.
Therefore, we can set the equation equal to zero and solve for the distance of closest approach (r).
Given that the charge of the proton (q_proton) is e = 1.602 × 10^-19 C, and that ε₀ (epsilon naught) is the permittivity of free space (ε₀ = 8.85 × 10^-12 C²/(Nm²)).
0 = (q_proton * q_alpha) / (4πε₀r)
Substituting the known values, we can now solve for r, which will give us the distance of closest approach between the proton and the alpha particle.