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 need help with.
Assuming your part a is correct, now consider momenetum conservation.
If vi is called the x direction, then momentum is conserved
mvi=m(.94537)vi cosTheta11+m(.32599)vi*cosTheta2
and momentum in the perpendiuclar y directioni is conserved
0=m(.94537)vi*sinTheta1+m*.32599Vi*sinTheta2
solve for theta1, theta2,angles with respect to the x direction.
I do not know how to solve this. I tried.
well ur stupid
well too bad for u
To determine the direction of the velocity vectors after the collision, we need to consider the conservation of momentum and kinetic energy.
In an elastic collision, both momentum and kinetic energy are conserved. Let's denote the mass of each proton as m, the initial velocity of the moving proton as vi, and the final velocities of both protons as vf1 and vf2, respectively.
According to the conservation of momentum, the total momentum before the collision (which is zero since one proton is at rest) must be equal to the total momentum after the collision.
(0) = m × vf1 + m × vf2
Since the initially moving proton scatters toward the positive y-direction, the velocity component along the x-axis for that proton remains unchanged. Therefore, vf1 will have both x and y-components, while vf2 will only have an x-component.
Now, let's determine the y- and x-components of vf1 and vf2.
The y-component of the momentum is given by:
0 = m × vf1y + m × vf2y
Since the initially moving proton scatters toward the positive y-direction, vf1y must be positive, and vf2y must be negative.
Next, to find the x-components of vf1 and vf2, we use the conservation of kinetic energy. The initial kinetic energy is given by:
KE initial = 0.5 × m × vi^2
The final kinetic energy is given by:
KE final = 0.5 × m × vf1^2 + 0.5 × m × vf2^2
Since the collision is elastic, the kinetic energy before and after the collision must be equal:
KE initial = KE final
0.5 × m × vi^2 = 0.5 × m × vf1^2 + 0.5 × m × vf2^2
Simplifying this equation gives:
vi^2 = vf1^2 + vf2^2
Now we have two equations:
(1) 0 = m × vf1y + m × vf2y
(2) vi^2 = vf1^2 + vf2^2
From equation (1), we can see that the sum of the y-components of vf1 and vf2 must be zero. Therefore, vf1y = -vf2y.
Squaring both sides of this equation gives:
vf1y^2 = vf2y^2
Substituting this into equation (2), we have:
vi^2 = vf1^2 + vf1y^2
Since vf1y and vf2y are squared terms, we can rewrite equation (2) as:
vi^2 = vf1^2 + vf2^2 + 2vf1y^2
Substituting vf1y^2 = vf2y^2 and rearranging, we get:
vi^2 = vf1^2 + vf1y^2 + 2vf1y^2
Simplifying further, we have:
vi^2 = vf1^2 + 3vf1y^2
Now, let's substitute the expressions we found in part (a) for vf1 and vf1y:
vi^2 = (0.94537 × vi)^2 + 3(0.32599 × vi)^2
Simplifying this equation, we can solve for vf1:
vi^2 = 0.89496 × vi^2 + 0.32836 × vi^2
vi^2 - 0.89496 × vi^2 - 0.32836 × vi^2 = 0
0.32668 × vi^2 = 0
Since vi^2 cannot be zero (assuming a non-zero initial velocity), the only solution is that vf1 is zero.
Therefore, after the collision, the initially moving proton comes to a complete stop, and its velocity vector is zero degrees relative to the +x direction.
As for the initially at rest proton (vf2), since it only gains velocity along the x-axis, its velocity vector will have an angle of 0 degrees relative to the +x direction.
So, after the collision:
initially moving proton: 0 degrees relative to the +x direction
initially at rest proton: 0 degrees relative to the +x direction