Elastic collisions

doub

1. The problem statement, all variables and given/known data

A 2.0 kg ball moving with a speed of 3.0 m/s hits, elastically, an identical stationary ball. If the first ball moves away with an angle 30 degrees to the original path, determine the speed of the first ball after the collision, and the speed and direction of the second ball after the collision.

2. Relevant equations

v_1 = V_1cos30 + v_2cos(theta) for the movement in the "x" direction

and

0 = v_1cos30 + v_2cos(theta) for movement in the "y" directions


3. The attempt at a solution

Played with this for hours but to me it does not seem like there is enough information. I feel like I am missing at velocity for after the collision.

Thanks

tiny-tim

hi doub! welcome to pf!
an elastic collision is defined as one in which energy is conserved (as well as momentum, which is always conserved in collisions)

doub

Right,

this is the best answer I got however I do not feel anywhere near confident.

3.0 m/s = v_1'cos30 + v_2'cos(theta) = 2.598 m/s + v_2'cos(theta) = 3.0 m/s - 2.598 m/s
so v_2'cos(theta) = 0.402 m/s in the "x" direction

0 = v_1'sin30 + v_2'sin(theta) = 1.299 m/s + v_2'sin(theta). so v_2'sin(theta) = -1.299 m/s

tan^-1 = v_2'cos(theta)/v_2'sin(theta) = -1.299/0402 = -72 degrees

and using sqrt(v_2'sin(theta)^2 + v_2'cos(theta)^2 = 1.36 m/s

am I anywhere in the ballpark at least?

tiny-tim

hi doub!

(try using the X₂ button just above the Reply box )
no, v₁' isn't 3.0, it's unknown!

try the momentum equations again (and you'll need an energy equation also)

doub

yeah I'm totally lost now

tiny-tim

start again, with v₁' v₂' and θ as your three variables

(you have three equations: x, y, and energy, so that should be solvable )

show us what you get

doub

Ok,

The equations I have gotten are

x --> v_1 = v_1'cos30 + v_2'cos(theta)

y --> 0 = v_1'sin30 + v_2'sin(theta)

Energy --> v_1^2 = v_1'^2 + v_2'^2

tiny-tim

fine so far

now fiddle about with the first two equations so that cosθ and sinθ are on their own, then use cos²θ + sin²θ = 1 to eliminate θ

doub

So,

cos(theta) = v₁ - (v₁'cos30)/v₂'

and

sin(theta) = (-v_a'sin30)/v₂'

where do the sin²theta come from?

tiny-tim

uhh?

square both equations!

doub

I am just not seeing this...

cos²(θ) = (v1² -v1²'cos30²)/v2'²

sin²(θ) = (-v1²'sin30²)/v2'²

thanks very much for helping btw

tiny-tim

ok, now add …

θ will miraculously disappear!

doub

so if we add are we left with;

(v1² -v1²'cos30²) + (-v1²'sin30²) / 2v2'²

?

tiny-tim

actually, the RHS of that first equation should be (v1 -v1'cos30)²/v2'²
how did you get that?

where has the = sign gone?