A proton, moving with a velocity of vinitial i, collides elastically with another proton that is initially at rest. Assuming that the 2 protons have equal speeds after the collision, find (a) the speed of each proton after the collision in terms of vinitial and (b) the direction of the velocity vectors after the collision.
I found part (a) to be Vfinal= Vinitial(sqroott/2)
The keyword here is "elastically".
This means that
1. if they move with the same velocity after impact, they cannot move in the same direction.
2. Energy is conserved, so that
mv0²=mv1²+mv1²
from which you can conclude v1=v0/√2.
To solve this problem, we need to apply conservation of momentum and conservation of kinetic energy. Let's break it down step by step:
(a) Finding the final speed of each proton after the collision:
According to conservation of momentum, the total momentum before the collision must be equal to the total momentum after the collision. Since the second proton is initially at rest, the momentum before the collision is only due to the first proton and can be calculated as p_initial = m * v_initial, where m is the mass of each proton and v_initial is the initial velocity of the first proton.
After the collision, the two protons have equal speeds, let's call it v_final. The momentum of each proton can then be calculated as p_final = 2 * m * v_final since they have equal speeds.
Using conservation of momentum, we can equate the initial momentum with the final momentum:
m * v_initial = 2 * m * v_final
Simplifying the equation, we can cancel out the mass of the protons, resulting in:
v_initial = 2 * v_final
Dividing both sides by 2, we get the final speed of each proton after the collision:
v_final = v_initial / 2
Therefore, the speed of each proton after the collision is v_final = v_initial / 2.
(b) Finding the direction of the velocity vectors after the collision:
Since we are given that the collision is elastic, the kinetic energy of the system is conserved. This means that the total kinetic energy before the collision is equal to the total kinetic energy after the collision.
Before the collision, the first proton has a kinetic energy of KE_initial = (1/2) * m * v_initial^2, and the second proton has no kinetic energy since it is at rest.
After the collision, both protons have equal speeds v_final.
The kinetic energy after the collision for each proton can be calculated as KE_final = (1/2) * m * v_final^2.
Using conservation of kinetic energy, we can equate the initial kinetic energy with the final kinetic energy:
(1/2) * m * v_initial^2 = (1/2) * m * v_final^2
Cancelling out the mass and simplifying the equation, we get:
v_initial^2 = v_final^2
Taking the square root of both sides, we get:
v_initial = v_final
This implies that the direction of the velocity vectors after the collision is the same as the initial direction.
Therefore, after the collision, the speed of each proton is v_final = v_initial / 2, and the direction of the velocity vectors remains unchanged.