Elastic collisions of a moving ball hitting a stationary ball

[doub]

Homework Statement

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.

Homework 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

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]

welcome to pf!

hi doub! welcome to pf!

A 2.0 kg ball moving with a speed of 3.0 m/s hits, elastically …

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 )

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

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! ​

pzzaaam!​

 

[doub]

so if we add are we left with;

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

?

 

[tiny-tim]

cos²(θ) = (v1² -v1²'cos30²)/v2'²

sin²(θ) = (-v1²'sin30²)/v2'²

actually, the RHS of that first equation should be (v1 -v1'cos30)²/v2'²

so if we add are we left with;

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

?

how did you get that?

where has the = sign gone?