Solving Momentum and Energy Equations for Elastic Collision | Help Needed

[vorcil]

i used the conservation of momentum equation and conservation of energy equation

got m1.v1i + m2.v2i = m1.v1f + m2.v2f
and 1/2 m1 . v1i^2 + 1/2m2 . v2i^2 = 1/2 m1.v1f^2+ 1/2 m2.v2f^2

i know it's a simutaneous equation but i don't know how to solve it!

1.1 + 0 = 0.11 x + 0.3y
5.5 + 0 = 0.55x^2 + .15y^2

(this is for the elastic one)

http://img27.imageshack.us/img27/2373/masteringphysics.jpg

 

[vorcil]

i used the conservation of momentum equation and conservation of energy equation

got m1.v1i + m2.v2i = m1.v1f + m2.v2f
and 1/2 m1 . v1i^2 + 1/2m2 . v2i^2 = 1/2 m1.v1f^2+ 1/2 m2.v2f^2

i know it's a simutaneous equation but i don't know how to solve it!

1.1 + 0 = 0.11 x + 0.3y
5.5 + 0 = 0.55x^2 + .15y^2

(this is for the elastic one)

 

[vorcil]

anyone please?

 

[Doc Al]

i know it's a simutaneous equation but i don't know how to solve it!

1.1 + 0 = 0.11 x + 0.3y
5.5 + 0 = 0.55x^2 + .15y^2

One way is to solve for y (in terms of x) in equation 1 and substitute that into equation 2. Then you'll have an equation with one unknown, which you can solve.

 

[vorcil]

i got it wrong!
can someone show me how to do it please!

i solved Vf1 to be 3.153 ms^-1
vf2 to be 0.205 ms^-1

for the elastic collision

 

[Doc Al]

i got it wrong!

Show exactly what you did. Start by showing how you solved for y in terms of x in the first equation.

 

[vorcil]

Show exactly what you did. Start by showing how you solved for y in terms of x in the first equation.

1.1 + 0 = 0.11 x + 0.3y
5.5 + 0 = 0.55x^2 + .15y^2

5.5 = (0.55*x^2) + (0.15*((1.1-(0.11*x)-0.3)^2))
solving for x i got 3.15

y = 1.1 - (.11*x) -.3

y = .45

thanks for helping

 

[Doc Al]

1.1 + 0 = 0.11 x + 0.3y

...

y = 1.1 - (.11*x) -.3

Correct this.

 

[vorcil]

omg lol i don't believe it

dude i still got the wrong awnser

is it y = 1.1 / ((0.11x)+0.3)
y = (1.1 + (0.11x)) /0.3?

i don't know what the hell is wrong with me this morning, it's 3 am now

 

[Doc Al]

dude i still got the wrong awnser

is it y = 1.1 / ((0.11x)+0.3)
y = (1.1 + (0.11x)) /0.3?

Not quite (almost). Try again.

1.1 + 0 = 0.11 x + 0.3y

Solve this for y. Do it slowly, one step at a time.

 

[vorcil]

Not quite (almost). Try again.

Solve this for y. Do it slowly, one step at a time.

y = ( 1.1 - ( 0.11*X) ) / 0.3

i'm on my last try for this question, it's like high school all over again
just want to check with you before i go for my final attempt

5.5 = (0.55 * X^2) + (0.15 * ( ( ( 1.1 - (0.11*X ) ) / 0.3 ) ^2 ) )

 

[Doc Al]

y = ( 1.1 - ( 0.11*X) ) / 0.3

That's more like it.

 

[vorcil]

That's more like it.

I still managed to get it wrong,

V1f=2.85
V2f=2.62

2.85 was from the solved equation from above,
and v2f = (1.1-(.11*2.85))/0.3

 

[Doc Al]

I still managed to get it wrong,

V1f=2.85
V2f=2.62

2.85 was from the solved equation from above,
and v2f = (1.1-(.11*2.85))/0.3

When you solve a quadratic equation you get two solutions. Often, only one of the solutions will be physically realistic for a given problem. That's the case here. (Note that the solution that you chose has the first mass moving faster than the second, but in the same direction. How could that be possible unless the masses passed through each other?)

What's the other solution?