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October 1, 2014

October 1, 2014

Posted by **Pooja** on Thursday, November 27, 2008 at 1:43pm.

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

- Physics -
**Damon**, Thursday, November 27, 2008 at 2:03pmbefore 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)

- Typo -
**Damon**, Thursday, November 27, 2008 at 10:39pmspecial 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

- Typo -
- Physics -
**drwls**, Thursday, November 27, 2008 at 2:17pm(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.

- Physics -
**Frustrated**, Sunday, November 30, 2008 at 6:18pmWhat is MP?

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