# Velocity of proton

1. Oct 7, 2011

### yardy_genius

can someone please demonstrate how a question such as this is solved.

A proton travelling with a velocity 6.0 × 105 m/s i
makes a glancing collision with a second proton
at rest. One of the protons is observed after the
collision to be moving with a velocity
V1 = 4.6 x 105 m/s at a 40 degree angle to the original
incoming proton’s direction.
i. Determine the velocity of the other proton
after the collision in unit vector notation
ii. Calculate the angle θ.

2. Oct 7, 2011

### WJSwanson

What kind of material is this covering? Does it incorporate the Coulomb force between the protons? (I suppose probably not, since the "glancing collision" would have some... complications, I believe.) Or is it just treating it like a particular perfectly elastic collision?

3. Oct 7, 2011

### yardy_genius

more on the basis of elastic collisions.

i know that the momentum in x direction is equal to the momentum in the y but i am not given a mass? but i suppose it cancels out.

here is a diagram.

[PLAIN]http://img444.imageshack.us/img444/9633/vectord.jpg [Broken]

Last edited by a moderator: May 5, 2017
4. Oct 7, 2011

### yardy_genius

ok this is what i did,

resolving x component

V1= 6.0X 10^5 cos40
= 4.5

V2= -4.6x 10^5 sin40

=-3

therefore velocity of other proton in unit vector form = (4.5 x10^5i -3 X 10^5j)

angle between them is tan (inv) -3/4.5 = 33 degrees.

is this correct?

5. Oct 8, 2011

### WJSwanson

Okay, so since this is just a problem involving a perfectly inelastic collision, I'll give you a couple of morsels for thought that should get you set. If it doesn't help at all, I'll be back in the morning. Anyway:

1. Think in terms of the law of conservation of momentum and that the masses are equal to each other: you can pick an arbitrary mass m and you should still get the right answer. Otherwise, IIRC the mass of a proton is just 1.67 * 10-27kg.

2. Remember that momentum is a vector quantity and consider the implication that has on what quantities are conserved by the law of conservation of momentum.

6. Oct 8, 2011

### yardy_genius

check my previous post to see if am on the right track

7. Oct 8, 2011

### WJSwanson

Yeah, sorry about that. You posted while I was typing lol. And my browser just crashed a moment ago so I lost the lengthy response I typed up. Anyway, checking the answer to make sure v0 = Ʃvf...

v0 = < 6.0 * 105m/s , 0 m/s >

So you've got...

v1 = < 4.6 * 105m/s , -3.0 * 105m/s >

v2 = < 4.5 * 105m/s , -3.0 * 105m/s >

unless I'm reading this horribly wrong.

So when we add the two vectors, we should get < 6.0 * 105m/s , 0 m/s >.

What we get instead is < 9.1 * 105m/s , -6.0 * 105m/s >

Remember that the velocity of that proton you're given after the collision is indeed given by

$v_{1}$ $= 4.5 * 10^{5}m/s < cos 40 , sin 40 >$

like you thought, and that your two final-velocity vectors have to add up to be equal to that initial velocity vector of

$v_{0} = 6.0 * 10^{5}m/s < 1 , 0 >$.