Hi guys,

Im a bit stumped on how to solve an online physics homework problem and it's really bothering me. I have no idea how to come up with the angles or even attempt the problem and neither do my friends. We'd all really appreciate it if someone could help us.
Many Thanks,
Pooja
A proton, moving with a velocity of viihatbold, collides elastically with another proton that is initially at rest. Assuming that after the collision the speed of the initially moving proton is 2.00 times the speed of the initially at rest proton, find the following.
(a) the speed of each proton after the collision in terms of vi
initially moving proton multiplied by vi
initially at rest proton multiplied by vi

(b) the direction of the velocity vectors after the collision
initially moving proton __° relative to the +x direction
initially at rest proton __° relative to the +x direction

(a) They tell you that is is an elastic collision. Let's call the velocities

V1i = vi = speed of initially moving proton
V2i = 0 (initially at rest proton)
Vif (final speed of initially moving proton)
V2f = (final speed of initially moving proton)
They tell you that V2f/V2f = 2

From energy conservation
Mp vi^2 = Mp V2f^2 + Mp V2f^2/4
= (5/4)Mp Vf2
V1f = sqrt(4/5) * vi = 0.8944 vi
V2f = 0.4472 vi

(b) Now you have to consider the momentum equation to determine the angles the protons leave. Assume +x is the initial direction of motion.

Let the angles be A1 and A2
Total momentum in the y direction must remain zero
V1f sin A1 = -V2f sin A2
sin A2/sin A1 = -V1f/V2f = -2
Mp V1f cos A1 + Mp V2f cos A2 = Mp vi
Vif cos A1 + (Vif/2) cos A2 = vi
0.8944 cos A1 + 0.4472 cos A2 = 1

You have two equations in the two unknown angles A1 and A2. The remaining steps involve some trig which I will leave up to you. If it gets too messy, solve graphically. Remember A1 and A2 must be on opposite sides of the x axis.

special 2 times constraint

W1x^2+W1y^2 = 4 (W2x^2+W2y^2)

must be 4 not two because it is the speed twice, therefore the square of the speed 4 times

What is MP?

Hi Pooja,

I can help you with this physics problem. To solve it, we can use the principles of conservation of momentum and conservation of kinetic energy.

Let's start with part (a) to find the speed of each proton after the collision in terms of vi (the initial velocity of the moving proton).

According to the conservation of momentum, the total momentum before the collision should be equal to the total momentum after the collision. Since one proton is initially at rest, its momentum is zero. Therefore, only the moving proton's momentum contributes to the total momentum before the collision.

Momentum before collision = Momentum after collision

mv_initial = m1 * v1_final + m2 * v2_final

where m is the mass of each proton and v_initial is the initial velocity of the moving proton.

The problem statement tells us that the speed of the moving proton after the collision is 2.00 times the speed of the initially at rest proton.

v1_final = 2 * v2_final

To solve for the final velocities, we need to express everything in terms of vi.

From conservation of momentum:

m * vi = m1 * v1_final + m2 * v2_final

From the information given:

v1_final = 2 * v2_final

Substituting this into the conservation of momentum equation:

m * vi = m1 * (2 * v2_final) + m2 * v2_final

We can also express m1 and m2 in terms of the proton mass (m). Since both protons are the same, m1 = m2 = m.

m * vi = m * (2 * v2_final) + m * v2_final

Canceling out the mass (m) from both sides:

vi = 2 * v2_final + v2_final

Combining like terms:

vi = 3 * v2_final

Simplifying, we get:

v2_final = vi / 3

Since v1_final is twice the speed of v2_final:

v1_final = 2 * (vi / 3) = 2vi / 3

Therefore, the speed of the moving proton after the collision is 2vi / 3, and the speed of the initially at rest proton after the collision is vi / 3.

Moving on to part (b) to find the direction of the velocity vectors after the collision.

To find the direction, we need to determine the angle each proton's final velocity vector makes with the +x direction.

We know that the moving proton after the collision has a velocity of 2vi / 3, but we need to find the angle.

Let's consider the moving proton's final velocity vector as V1. We'll call the angle it makes with the +x direction as α1.

Using the trigonometric definition of the angle, we have:

tan(α1) = (y-component of V1) / (x-component of V1)

Since V1 has a magnitude of 2vi / 3, we can write:

V1x = (2vi / 3) * cos(α1)
V1y = (2vi / 3) * sin(α1)

Now, we can write the tangent of α1:

tan(α1) = V1y / V1x

tan(α1) = (2vi / 3) * sin(α1) / (2vi / 3) * cos(α1)

tan(α1) = sin(α1) / cos(α1)

This gives us:

tan(α1) = tan(α1)

This equation is satisfied for any angle α1. Therefore, we can conclude that the angle of the moving proton's velocity vector after the collision is not determined by the given information. It could have any angle relative to the +x direction.

Similarly, for the initially at rest proton, since its speed after the collision is vi / 3, its final velocity vector will have the same angle α2 as α1. Again, this means that the angle of the initially at rest proton's velocity vector after the collision is not determined by the given information. It could have any angle relative to the +x direction.

I hope this explanation helps you understand the problem and how to approach it. Let me know if you have any further questions!

before collision

V1x is x velocity of proton 1 (your Vihat)
V1y = 0 is y velocity
V2x = V2y = 0 = velocities of proton 2

after collision
W1x = x velocity of proton 1
W1y = y velocity of proton 1
W2x = x velocity of proton 2
W2y = y velocity of proton 2

Now try

now constraints (m = mass of proton)
momentum same before and after
m V1x = m W1x + m W2x
m V1y = 0 = m W1y + m W2y
energy(ke) the same before and after
.5 m V1x^2 = .5m (W1x^2+W1y^2 +W2x^2+W2y^2)
special 2 times constraint
W1x^2+W1y^2 = 2 (W2x^2+W2y^2)

MP = Moving Proton