Posted by Pooja on Thursday, November 27, 2008 at 1:43pm.
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)
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
(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.
What is MP?
MP = Moving Proton