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 carbon nucleus after the collision
To find the velocities of the proton and the oxygen nucleus after the collision, we can use the principle of conservation of linear momentum and the principle of conservation of kinetic energy.
1. Conservation of Linear Momentum:
The initial momentum of the system (proton + oxygen nucleus) before the collision is given by:
P_initial = m1 * v1_initial + m2 * v2_initial
where m1 is the mass of the proton, v1_initial is its initial velocity, m2 is the mass of the oxygen nucleus, and v2_initial is its initial velocity (which is zero since it is stationary).
The final momentum of the system after the collision is given by:
P_final = m1 * v1_final + m2 * v2_final
According to the principle of conservation of linear momentum, the initial momentum and the final momentum of the system should be equal:
P_initial = P_final
2. Conservation of Kinetic Energy:
The kinetic energy of the system before the collision is given by:
KE_initial = (1/2) * m1 * v1_initial^2 + (1/2) * m2 * v2_initial^2
The kinetic energy of the system after the collision is given by:
KE_final = (1/2) * m1 * v1_final^2 + (1/2) * m2 * v2_final^2
According to the principle of conservation of kinetic energy, the initial kinetic energy and the final kinetic energy of the system should be equal:
KE_initial = KE_final
Now, let's plug in the given values and solve for v1_final and v2_final.
Given:
- m1 (mass of the proton) = m
- v1_initial (initial velocity of the proton) = +250 m/s
- m2 (mass of the oxygen nucleus) = 16m
- v2_initial (initial velocity of the oxygen nucleus) = 0
From conservation of linear momentum:
m * v1_initial + 16m * 0 = m * v1_final + 16m * v2_final
m * v1_initial = m * v1_final + 16m * v2_final --- (Equation 1)
From conservation of kinetic energy:
(1/2) * m * (v1_initial)^2 + (1/2) * 16m * (v2_initial)^2 = (1/2) * m * (v1_final)^2 + (1/2) * 16m * (v2_final)^2
(1/2) * m * (v1_initial)^2 = (1/2) * m * (v1_final)^2 + (1/2) * 16m * (v2_final)^2 --- (Equation 2)
Simplifying Equation 1 and Equation 2:
Equation 1: m * v1_initial = m * v1_final + 16m * v2_final
v1_initial = v1_final + 16 * v2_final --- (Equation 3)
Equation 2: (1/2) * m * (v1_initial)^2 = (1/2) * m * (v1_final)^2 + (1/2) * 16m * (v2_final)^2
v1_initial^2 = v1_final^2 + 16 * v2_final^2 --- (Equation 4)
Now, square Equation 3:
(v1_initial)^2 = (v1_final + 16 * v2_final)^2
v1_initial^2 = v1_final^2 + 32 * v1_final * v2_final + 256 * v2_final^2 --- (Equation 5)
Substitute Equation 4 into Equation 5:
v1_initial^2 = v1_final^2 + 16 * v2_final^2 + 32 * v1_final * v2_final + 256 * v2_final^2
v1_initial^2 = 256 * v2_final^2 + 32 * v1_final * v2_final + v1_final^2 + 16 * v2_final^2
v1_initial^2 = 257 * v2_final^2 + 32 * v1_final * v2_final + v1_final^2 --- (Equation 6)
Substitute Equation 3 into Equation 6:
(v1_final + 16 * v2_final)^2 = 257 * v2_final^2 + 32 * (v1_final + 16 * v2_final) * v2_final + v1_final^2
(v1_final)^2 + 32 * v1_final * v2_final + 256 * v2_final^2 = 257 * v2_final^2 + 32 * v1_final * v2_final + v1_final^2
256 * v2_final^2 - 256 * v1_final * v2_final = 256 * v2_final^2
256 * v2_final * (v2_final - v1_final) = 0
Since v2_final cannot be zero (otherwise, the oxygen nucleus will remain stationary even after the collision), the only solution is:
v2_final - v1_final = 0
v2_final = v1_final --- (Equation 7)
Substitute Equation 7 into Equation 3:
v1_initial = v1_final + 16 * v1_final
v1_final = v1_initial / 17
Substitute the value of v1_final back into Equation 7:
v2_final = v1_final = v1_initial / 17
Therefore, the final velocities of the proton and the oxygen nucleus after the collision are:
v1_final = v1_initial / 17
v2_final = v1_initial / 17
Let's calculate the values based on the given velocity of the proton, v1_initial = +250 m/s:
v1_final = 250 m/s / 17 = 14.7 m/s (approx.)
v2_final = 14.7 m/s (approx.)
Hence, the velocities of the proton and the oxygen nucleus after the collision are approximately 14.7 m/s in the + direction for both.