A proton, moving with a velocity of vi, collides elastically with another proton that is initially at rest. Assuming that after the collision the speed of the initially moving proton is 2.90 times the speed of the proton initially at rest, find the following.
(a) the speed of each proton after the collision in terms of vi
I already figured out part a which the answer is:
initially moving proton: 0.94537 ✕ vi
initially at rest proton: 0.32599 ✕ vi
(b) the direction of the velocity vectors after the collision (assume that the initially moving proton scatters toward the positive y direction)
initially moving proton____° relative to the +x direction
initially at rest proton_____° relative to the +x direction
Part B is what I REALLY need help with.
I posted this question earlier, and another person said the following:
"Assuming your part a is correct, now consider momentum conservation.
If vi is called the x direction, then momentum is conserved
mvi=m(.94537)vicosTheta1+m(.32599)vi*cosTheta2
and momentum in the perpendicular y direction is conserved
0=m(.94537)vi*sinTheta1+m*.32599Vi*sinTheta2"
I cannot for the life of me solve these angles. I canceled out the m and v terms for both of them. Part A is correct. Please help!
how did you solve for part a?
To find the directions of the velocity vectors after the collision, we can utilize the concept of momentum conservation. Momentum is a vector quantity, so we need to consider the components of momentum in both the x and y directions separately.
Let's start with the x direction:
Using the conservation of momentum in the x direction, we have:
m_vi = m(0.94537 * vi) * cos(θ1) + m(0.32599 * vi) * cos(θ2)
Here, m_vi represents the initial momentum in the x direction, and θ1 and θ2 represent the angles of the velocity vectors of the moving proton and initially at rest proton, respectively, relative to the positive x direction.
Now, let's consider the y direction:
Using the conservation of momentum in the y direction, we have:
0 = m(0.94537 * vi) * sin(θ1) + m(0.32599 * vi) * sin(θ2)
Since the initially moving proton scatters towards the positive y direction, θ1 can be taken as 90 degrees. Substituting this in the above equation:
0 = m(0.94537 * vi) + m(0.32599 * vi) * sin(θ2)
To solve for θ2, we can rearrange and solve for sin(θ2):
0.32599 * vi * sin(θ2) = -0.94537 * vi
sin(θ2) = (-0.94537/0.32599)
Now, using the inverse sine function (sin^-1) in any scientific calculator, you can find the value of θ2. Note that there will be two possible values for θ2, as both sinθ and -sinθ will satisfy the equation.
Once you have the value(s) of θ2, you can find the angle relative to the +x direction by subtracting it from 180 degrees since the initial direction is towards the positive x direction.
Similarly, you can find the angle θ1 relative to the +x direction (which should be 90 degrees as mentioned earlier).
I hope this explanation helps you understand the approach to solve part B of the problem.