Potential energy and momentum question

[Daniiel]

I included what i did under the pictures to make it easier to read.
but I am not sure, with the ski questoin, if v= root(2gd) that's somthing i pulled out of my textbook, and if d is 20 or the distance of the slope.
but if that's all right my next worry is that i shouldn't of solved it as a projectile, i found the range of the jump, found the time it took to reach the range, then halved the time to find when the skier was at max height. and got 4.4m or somthing

is there any other way to convert kinetic / potential energy to a velocity?
my tutor did somthing with the KE formula, he differentiated it, then v turned into a and he intergrated it, but I am not 100% what he did.

I'm pretty confident with the neutron question because it seems right having the particle with the smaller angle having a higher velocity.
But thanks to anyone who has a look


[PLAIN]http://img199.imageshack.us/img199/8657/eeebb.jpg

 

[kuruman]

For the skier problem, you can take a shortcut and write the energy conservation equation saying that the mechanical energy at the starting point is equal to the mechanical energy at maximum height. Note that at max height, the speed is just (1/2)mv_x². If you write down the equation, you will see that the masses cancel on each side, so the answer is independent of the skier's mass.

For the neutron problem you need to write the momentum conservation equations correctly in the x and in the y direction. You need to say that total momentum in the x-direction is the same before and after the collision and likewise for the y-direction. Momentum is a vector, so be sure to calculate its components correctly. Both of your equations in this problem are incorrect in that regard.

 

[Daniiel]

Pxi = Px1 + Px2
Pyi = 0 = mVy1 + mVy2
mVy1 = - mVy2
Vy1 = -Vy2

Vx1 = V1cos30, Vx2 = V2 cos 60

so

300m = m(V1 cos30) + m (V2 cos 60)

300 = (V1 cos 30) + (V2 cos 60)

is that right?
it mustn't be because when i try to get values for the separate v's i get 300

 

[kuruman]

Vy1 = -Vy2

What is this in terms of v₁ and v₂?

300m = m(V1 cos30) + m (V2 cos 60)

300 = (V1 cos 30) + (V2 cos 60)

is that right?

This is correct.

it mustn't be because when i try to get values for the separate v's i get 300

Can you show me exactly how you get 300 m/s?

 

[Daniiel]

I tried just then going like
In terms of the y components
V1 sin 30 = - V2 sin (-60)
and i got V1 = root(3) * V2

Then i tried subbing that into the equation for the x components and found V2 = 150
I've just been trying to do that again to find V1 with V2 = V1/root(3)
but it isn't working out

and with the other thing i did before and kept getting 300 i did

300 = (V1 cos 30) + (V2 cos 60)

changed them to exact values and rearranged to find V1
found V1= (-0.5V2 + 300)/ (root(3)/2) then pluged it into the formula and got

300 = -V2/2 + 300 + V2/2
i did it afew different ways but its all a mess

was doing the y component thing the right way?

could i do 300 = V1 + V2 and sub in V1= root3 V2

 

[kuruman]

You have the correct ingredients, but you don't seem able to assemble them properly. One step at a time. You have

V1 = root(3) * V2 ( Eq.1)

300 = (V1 cos 30) + (V2 cos 60) (Eq. 2)

What do you get if you take the value for V1 in equation (1) and put it in equation (2)?

 

[Daniiel]

are you sure that's right?
when i do that i get V2= 150 and v1 = 259 or somthing
if i plug it into
300 = v1 + v2 i get v1 = 190 and v2 = 110 which seems pretty realistic
because wouldn't the particle with the lower angle have a higher velocity
and also isn't equation 2 the equation for the x components? V1cos30 = Vx1 ?

 

[kuruman]

are you sure that's right?
when i do that i get V2= 150 and v1 = 259 or somthing

That's correct.

if i plug it into
300 = v1 + v2

Why do you think the above equation is correct? What is it an expression of?

i get v1 = 190 and v2 = 110 which seems pretty realistic
because wouldn't the particle with the lower angle have a higher velocity
and also isn't equation 2 the equation for the x components? V1cos30 = Vx1 ?

It ain't necessarily so. The numbers speak for themselves. If v₁ = sqrt(3)*v₂ = 1.732*v₂, which one is larger, v₁ or v₂?

 

[Daniiel]

If Pi = P1+P2

mVi = mV1 + mV2 m is the same
Vi = v1 + v2
300 = v1 + v2 ?
dont 150 and 259 have to add to give 300 for conservation of momentum to be kept?

 

[kuruman]

If Pi = P1+P2

mVi = mV1 + mV2 m is the same
Vi = v1 + v2
300 = v1 + v2 ?
dont 150 and 259 have to add to give 300 for conservation of momentum to be kept?

You are forgetting that the equation

v_i=v₁+v₂

is a vector equation. Because the angle between v₁ and v₂ is 90^o, when you draw the vector addition diagram, you get a right triangle where 300 is the hypotenuse and 150 and 259 are the right sides. Check if the Pythagorean theorem is obeyed.

 

[Daniiel]

ohhhh
that makes sense
thanks for all your help