An alpha particle collides with an oxygen nucleus, initially at rest. The alpha particle is scattered at an angle of 60.9° above its initial direction of motion, and the oxygen nucleus recoils at an angle of 51.9° on the opposite side of that initial direction. The final speed of the nucleus is 1.32 105 m/s. In atomic mass units, the mass of an alpha particle is 4.0 u. The mass of an oxygen nucleus is 16 u.

(a) Find the final speed of the alpha particle.
m/s
(b) Find the initial speed of the alpha particle.
m/s

Duplicate post; answered elsewhere

To solve this problem, we can use the principles of conservation of momentum and conservation of energy. Let's break down the problem step by step.

Step 1: Find the velocity components of the alpha particle and the oxygen nucleus after the collision.
We can use trigonometry to find the x and y components of the final velocities for both the alpha particle and the oxygen nucleus.

For the alpha particle, we know that it is scattered at an angle of 60.9° above its initial direction of motion. Let's call the x and y components of its final velocity Vax and Vay, respectively.
Vax = Vaf * cos(60.9°)
Vay = Vaf * sin(60.9°)

For the oxygen nucleus, we know that it recoils at an angle of 51.9° on the opposite side of the alpha particle's initial direction. Let's call the x and y components of its final velocity Vox and Voy, respectively.
Vox = Vof * cos(51.9°)
Voy = -Vof * sin(51.9°) (the negative sign is because it recoils in the opposite direction)

Step 2: Apply conservation of momentum in the x-direction.
Conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. Since the oxygen nucleus is initially at rest, the total momentum before the collision is zero. Therefore, the total momentum after the collision should also be zero in the x-direction.
malpha * Vax + moxygen * Vox = 0

Step 3: Find the final speed of the oxygen nucleus.
We are given the mass of the oxygen nucleus (moxygen = 16 u) and its final velocity in the x-direction (Vox). We can use this information to find the final speed of the oxygen nucleus.
Vof = sqrt(Vox^2 + Voy^2)
Vof = sqrt(Vox^2 + (-Vof * sin(51.9°))^2)
Solving this equation will give us the final speed of the oxygen nucleus.

Step 4: Apply conservation of energy.
Conservation of energy states that the total kinetic energy before the collision is equal to the total kinetic energy after the collision. Since the total energy before the collision is zero (the alpha particle is at rest), the total energy after the collision should be the sum of the kinetic energies of the alpha particle and the oxygen nucleus.
(1/2) * malpha * Vaf^2 + (1/2) * moxygen * Vof^2 = (1/2) * malpha * Vai^2

Step 5: Find the final speed of the alpha particle.
We are given the mass of the alpha particle (malpha = 4.0 u), the final speed of the oxygen nucleus (Vof), and the angles of scattering. We can use the conservation of energy equation from Step 4 to solve for the final speed of the alpha particle.
Solving the equation will give us the final speed of the alpha particle (Vaf).

Step 6: Find the initial speed of the alpha particle.
The initial speed of the alpha particle can be determined by using the final speed (Vaf) and the angles of scattering. We can use basic trigonometry to find the x and y components of the initial velocity and then combine them to find the magnitude of the initial velocity.

By following these steps, you should be able to find the final speed of the alpha particle (part a) and the initial speed of the alpha particle (part b).