A proton, moving with a velocity of vii, collides elastically with another proton that is initially at rest. Assuming that after the collision the speed of the initially moving proton is 1.50 times the speed of the proton initially at rest, find the following.
To solve this problem, we can apply the laws of conservation of momentum and kinetic energy. Here are the steps to find the following values:
1. Find the velocity of the initially moving proton (vf) after the collision.
2. Determine the velocity of the initially stationary proton (vf2) after the collision.
3. Calculate the magnitude and direction of the momentum of each proton before and after the collision.
4. Calculate the kinetic energy of each proton before and after the collision.
Step 1: Find vf (velocity of the initially moving proton after the collision)
Since the collision is elastic, we can use the conservation of kinetic energy:
(Kinetic energy before collision) = (Kinetic energy after collision)
(1/2)mvii^2 = (1/2)mvf^2 + (1/2)mvf2^2
where m is the mass of a proton and vii is the initial velocity.
Step 2: Determine vf2 (velocity of the initially stationary proton after the collision)
Since the law of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision:
(mvii) = (mvf) + (mvf2)
Step 3: Calculate the magnitude and direction of the momentum of each proton before and after the collision
The magnitude of the momentum for each proton is given by:
Momentum = mass x velocity
Step 4: Calculate the kinetic energy of each proton before and after the collision
The kinetic energy of each proton is given by:
Kinetic energy = (1/2) x mass x (velocity)^2
By following these steps, you can find all the requested values.
To find the unknowns in this problem, we can apply the conservation of momentum and the conservation of kinetic energy.
Given information:
Mass of both protons: m1 = m2 = m (since both are protons)
Initial velocity of the moving proton: vi = vii
Final velocity of the moving proton: vf = 1.50 * vi (1.50 times the initial velocity)
1. Find the final velocity of the initially resting proton after the collision.
Since this is an elastic collision, both momentum and kinetic energy are conserved. We can write the equations for conservation of momentum and conservation of kinetic energy as follows:
1) Conservation of momentum:
m1 * vi = m1 * vf1 + m2 * vf2
m * vi = m * vf1 + m * vf2
2) Conservation of kinetic energy:
(1/2) * m1 * vi^2 = (1/2) * m1 * vf1^2 + (1/2) * m2 * vf2^2
(1/2) * m * vi^2 = (1/2) * m * vf1^2 + (1/2) * m * vf2^2
Now we can solve these equations simultaneously to find vf1 and vf2.
Momentum equation:
m * vi = m * vf1 + m * vf2
Substituting the values:
m * vii = m * vf1 + m * vf2
m cancels out:
vii = vf1 + vf2
vf1 = vii - vf2
Now, substitute the expression for vf1 into the kinetic energy equation:
(1/2) * m * vi^2 = (1/2) * m * (vii – vf2)^2 + (1/2) * m * vf2^2
Simplify the equation:
vi^2 = (vii – vf2)^2 + vf2^2
Expand and rearrange:
vi^2 = vii^2 – 2 * vii * vf2 + vf2^2 + vf2^2
vi^2 = vii^2 + 2 * vf2^2 – 2 * vii * vf2
Now, substitute the value of vf1 from earlier:
vi^2 = vii^2 + 2 * vf2^2 – 2 * vii * vf2
Substitute vi = vii and multiply through by 2:
2 * vii^2 = 2 * vii^2 + 4 * vf2^2 – 4 * vii * vf2
Cancel out the common terms:
0 = 4 * vf2^2 – 4 * vii * vf2
Factor out 4vf2:
0 = 4vf2(vf2 – vii)
Setting this equation equal to zero gives us two possibilities:
1) 4 * vf2 = 0 -> vf2 = 0
2) vf2 - vii = 0 -> vf2 = vii
Since vf2 cannot be zero (as it represents the final velocity of one of the protons), we take the second possibility:
vf2 = vii
Therefore, the final velocity of the initially resting proton after the collision is vii.
2. Find the final velocity of the moving proton after the collision.
We have already determined that:
vf1 = vii
Therefore, the final velocity of the moving proton after the collision is also vii.
So, the final velocities of both protons after the collision are vii.