A proton and an alpha particle (q = +2.00e, m = 4.00u ) are fired directly toward each other from far away, each with an initial speed of 0.247c. What is their distance of closest approach, as measured between their centers? (Hint: There are two conserved quantities. Make use of both.)
To find the distance of closest approach between the proton and the alpha particle, we can use the conservation of momentum and the conservation of energy.
Let's denote the initial speed of both particles as v₀.
Conservation of momentum tells us that the initial total momentum of the system is equal to the final total momentum:
m₁v₁ + m₂v₂ = m₁v₁f + m₂v₂f
Where m₁ and m₂ are the masses of the particles, v₁ and v₂ are their initial velocities, and v₁f and v₂f are their final velocities.
In this case, m₁ is the mass of the proton (1.67e-27 kg) and v₁ is its initial velocity (0.247c). Similarly, m₂ is the mass of the alpha particle (6.64e-27 kg) and v₂ is its initial velocity (0.247c).
Since we are interested in the distance of closest approach, we can assume that at that point, they both momentarily come to rest (v₁f = v₂f = 0).
Using the conservation of momentum equation, we can solve for v₁f in terms of v₂:
m₁v₁ + m₂v₂ = m₁v₁f + m₂v₂f
m₁v₁f = m₁v₁ + m₂v₂
v₁f = (m₁v₁ + m₂v₂) / m₁
Now, let's consider the conservation of energy. The initial total energy of the system is equal to the final total energy:
(1/2)m₁v₁² + (1/2)m₂v₂² = (1/2)m₁v₁f² + (1/2)m₂v₂f²
Again, substituting v₁f and v₂f as 0:
(1/2)m₁v₁² + (1/2)m₂v₂² = (1/2)m₁v₁f² + (1/2)m₂v₂f²
(1/2)m₁v₁² + (1/2)m₂v₂² = (1/2)m₁((m₁v₁ + m₂v₂) / m₁)² + (1/2)m₂((m₁v₁ + m₂v₂) / m₁)²
(1/2)m₁v₁² + (1/2)m₂v₂² = (1/2)(v₁ + v₂)²(m₁ + m₂) / m₁
Now we have two equations. Solving them simultaneously will give us the distance of closest approach.
By substituting the values for the masses and velocities, we can calculate the distance of closest approach. Be sure to convert the given velocities to meters per second.