1. Ok... This one is bugging me...

(+) - Proton Ok, so we need to find the Velocity for W and V as shown in the image...
u=100m.s<sup>-1</sup>

Here is what I did:
if u = 100m.s<sup>-1</sup> with the angles 30Â° and 60Â° the collision is elastic because the angle between the two objects is 90Â°.

Wx/W = Cos(60)
Wx = W.Cos(60)

---------------
X-Dir
---------------
∑<sub>p before</sub> = ∑<sub>p after</sub>
m<sub>1</sub>V<sub>1</sub>+m<sub>2</sub>V<sub>2</sub> = m<sub>1</sub>V<sub>1</sub>+m<sub>2</sub>V<sub>2</sub>
(Ã·m) 100 = W.cos(60) + V.cos(30)
(1) 200 = W + √(3).V

---------------
Y-Dir
---------------
∑<sub>p before</sub> = ∑<sub>p after</sub>
m<sub>1</sub>V<sub>1</sub>+m<sub>2</sub>V<sub>2</sub> = m<sub>1</sub>V<sub>1</sub>+m<sub>2</sub>V<sub>2</sub>
(Ã·m) 0 = W.sin(60) + V.sin(30)
(2) 0 = √(3).W + V

---------------
(2) in (1)
---------------
200=W+√(3).(√(3)W)
200=2W
W=100m.s<sup>-1</sup>

But thats not true. That says that the 1st one stands still and the 2nd one moves. That means that this is 1Dimentional... but its not...

Thanks  2.

3. So you have a proton moving with initial velocity u=100ms<sup>-1</sup>, which collides elastically with a stationary particle, where after collision, the velocity of the particle is w, and the velocity of the proton is v? Is that what you're describing?

If so, you know that momentum is conserved before and after the collision, both horizontally and vertically. Therefore with u=<u<sub>x</sub>, u<sub>y</sub>>, v=<v<sub>x</sub>,v<sub>y</sub>>, and w=<w<sub>x</sub>,w<sub>y</sub>> being the three velocity vectors, and m<sub>1</sub> and m<sub>2</sub> being the masses of the proton and the other particle respectively, you have:

for x:

m<sub>1</sub>u<sub>x</sub>=m<sub>1</sub>v<sub>x</sub> + m<sub>2</sub>w<sub>x</sub>
100m<sub>1</sub>=m<sub>1</sub>vcos30 + m<sub>2</sub>wcos300
100m<sub>1</sub>=(1/2)*√3*v*m<sub>1</sub> + (1/2)*w*m<sub>2</sub> (1)

for y:

m<sub>1</sub>u<sub>y</sub>= m<sub>1</sub>v<sub>y</sub> + m<sub>2</sub>w<sub>y</sub>
0=m<sub>1</sub>v<sub>y</sub> + m<sub>2</sub>w<sub>y</sub>
0=m<sub>1</sub>vsin30 + m<sub>2</sub>wsin300
0=(1/2)*v*m<sub>1</sub> - (1/2)*√3*w*m<sub>2</sub> (2)

From equation (1) and (2) you might be able to go ahead and solve this by simultaneous equations, but it's worth asking what other information you have about the scenario (masses velocities etc) first. And you are trying to calculate w right??

When you are dividing your expressions by "m", then are you assuming that m<sub>1</sub> = m<sub>2</sub>? If not then you cant get rid off the m<sub>1</sub> AND m<sub>2</sub> terms by diving by ONE of them. If you write down the exact question, then we can help you more.  4. Sorry, didnt see you edited your post after the food...

Well it was kinda explained in class. So I mostly have it in my head.

Its basicaly
u is 100m.s-1
Then the 30Â° and 60Â° added up give 90 which makes it a perfectly elastic collision.
All the protons have the same mas
Because the chance a proton has to actually collide with an air molecule is so small we say that it is frictionless so we can use conservation of momentum.

Think basic. Nothing big.

So now i need to work out the x and y speed for V and W...

But I keep getting W=100... tried like 3 different ways already and now im just stuck...

Thanks  5. Originally Posted by Dark_Fire
(Ã·m) 0 = W.sin(60) + V.sin(30)

---------------
One of those has to be negative.  6. OK, if both particles are protons, then they will have the same mass and the collision will be elastic as in generally the case for sub-atomic particles.

So from my working above take (1) and (2) and if m<sub>1</sub>=m<sub>2</sub>, then we can just call this 'm'. Divide (1) and (2) by 'm' to obtain:

(1) / m ----> 100 = (1/2)*√3*v + (1/2)*w ------(3)
(2) / m ----> 0 = (1/2)*v - (1/2)*√3*w ----------(4)

Now we can solve the equations simultaneously by making the 'v' terms equal. To do this mulitply (3) by (1/2), and (4) by (1/2)*√3, to obtain:

(3)*(1/2) --------> 50 = (1/4)*√3*v + (1/4)*w -----(5)
(4)*(1/2)*√3 ---> 0 = (1/4)*√3*v - (3/4)*w --------(6)

Then subtract the equations:

(5) - (6) ----> 50 = (1/4)*w + (3/4)*w = w( 1/4 + 3/4 ) ----> w = 50 ms<sup>-1</sup>

You can then put this back into one of the equations to get a value for v, and then put both back into the original to find the mass.

Alternatively, the mass of a proton = 1.673Ã10<sup>-27</sup> kg, so you could use that too.   7. Wow. It actually worked. I don't quite understand why you multiplied with 1/2 and 1/2 root3...

Ill look at it again in the morning thou. But currently Im still confused. But good thing is your magic touch works. Thanks a million. I just like to know why instead of just use it. I like to learn. Im addicted to knowledge...  8. Glad to help The reason I multiplied one by (1/2) and one by (√3/2) is to make all the terms in the equation with a 'v' in them, equal....so that it's then possible to get rid of the 'v' terms when we subtract the equations, and are just left with something in terms of w, which is easily solved. In general you can cross-multiply co-efficients from the y-terms in each equation leaving you with equal terms to subtract. Have you done simultaneous equations? They almost always come in helpful with these momentum questions.  9. Jup. We have done them. And I can't thank you enough. Its funny How you can sometimes do the smallest thing wrong and still being in that mind set you'll keep doing it wrong as far as you go. :P i now realize How stupid I was. Hehe.

Thanks again.
Have a wonderfull week.  Bookmarks
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