A proton that has a mass m and is moving at 250 m/s in the + direction undergoes a head-on elastic collision with a stationary oxygen nucleus of mass 16m. Find the velocities of the proton and the oxygen nucleus after the collision
Conservation of momentum, and energy applies.
m*250= 32m*Vo+mVp
where Vo is the velocity of the oxygen molecule, and Vp is the velocity of the proton.
Vp= (250-32)Vo
Now, energy.
1/2 250^2=1/2 32 Vo^2+1/2 Vp^2
put the experession for Vp above into that, and solve for Vo
To find the velocities of the proton and the oxygen nucleus after the collision, we can apply the principles of conservation of momentum and conservation of kinetic energy.
1. Conservation of momentum:
In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision. Mathematically, we can express this principle as:
m1 * v1_initial + m2 * v2_initial = m1 * v1_final + m2 * v2_final
Here, m1 and v1_initial are the mass and initial velocity of the proton (m and 250 m/s, respectively), and m2 and v2_initial are the mass and initial velocity of the oxygen nucleus (16m and 0 m/s, respectively). We need to find v1_final and v2_final.
2. Conservation of kinetic energy:
In an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision. Mathematically, we can express this principle as:
(1/2) * m1 * v1_initial^2 + (1/2) * m2 * v2_initial^2 = (1/2) * m1 * v1_final^2 + (1/2) * m2 * v2_final^2
Substituting the given values, we get:
(1/2) * m * (250^2) + (1/2) * (16m) * (0^2) = (1/2) * m * (v1_final^2) + (1/2) * (16m) * (v2_final^2)
Now, we have two equations (momentum and kinetic energy conservation) with two unknowns (v1_final and v2_final). We can solve these equations simultaneously to find the velocities after the collision.
First, let's solve the momentum equation:
m * 250 + (16m) * 0 = m * v1_final + (16m) * v2_final
250m = m * v1_final + 16m * v2_final ----(1)
Next, let's solve the kinetic energy equation:
(1/2) * m * (250^2) + (1/2) * (16m) * (0^2) = (1/2) * m * (v1_final^2) + (1/2) * (16m) * (v2_final^2)
(1/2) * m * (250^2) = (1/2) * m * (v1_final^2) + (1/2) * (16m) * (v2_final^2)
62500 = v1_final^2 + 16 * v2_final^2 ----(2)
Now, we have two equations (equation 1 and equation 2) with two unknowns (v1_final and v2_final). We can solve this system of equations to find the velocities after the collision.