Physics Elastic Collision Question

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Homework Statement


a .060kg tennis ball, moving with a speed of 2.5 m/s, collides head-on with a .09kg ball initially moving away from it at a speed of 1.15m/s. Assuming a perfect collision, what are the speed and direction of each ball after the collision


Homework Equations



MaVa + MbVb = MaV’a + MbV’b (conservation of momentum)

½MaV²a +½ MbV²b = ½MaV’²a + ½MbV’²b (?)

The Attempt at a Solution


I plugged in the numbers into both equations and got....

1. (.06kg)(2.5m/s) + (.09kg)(1.15m/s) = (.06kg)(V1f) + (.09kg)(V2f)

and...

2. 1/2(.06kg)(2.5m/s)^2 + 1/2(.09kg)(1.15m/s)^2 = 1/2(.06kg)(V1f)^2 + 1/2(.09kg)(V2f)^2

i then started to solve the second equation...

.25 = 1/2(.06kg)(V1f)^2 + 1/2(.09kg)(V2f)^2

this is where I became confused....I got rid of the first part (1/2(.06kg)(V1f)^2) and just solved the second part [ 1/2(.09kg)(V2f)^2]

.25 = 1/2(.09kg)(V2f)^2

.25 = .05(V2f)^2

.25/.05 = (V2f)^2

5 = (V2f)^2

sqrt(5) = V

2.24 = V

I then went back and plugged 2.24 into the first equation ......and then rearranged the equation:

(.06kg)(2.5m/s) + (.09kg)(1.15m/s) = (.06kg)(V1f) + (.09kg)(V2f) =

- (.06kg)(V1f) = (.09kg) (2.23) - (.2523)

-.06kg (V1f) = -.0528

(-.0528) / -(.06) = V1f

v= .88

I know the answers are supposed to be .88 and 2.23 (back of the book), but it seems like I just randomly solved this (I don't get exactly how i got this), is this right?

I really need to pass physics (architecture major) but i don't understand it, any help would be greatly appreciated.
 
Last edited:

Answers and Replies

  • #2
ehild
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"this is where I became confused....I got rid of the first part (1/2(.06kg)(V1f)^2) and just solved the second part [ 1/2(.09kg)(V2f)^2]"

You can not do that. You can not just omit a term from an equation.

You have two equations with two unknowns.
Solve the first equation for v1f and substitute the expression you get in the second equation.


ehild
 
  • #3
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when you say to solve the first equation for V1f do you mean:

(.06kg)(2.5m/s) + (.09kg)(1.15m/s) = (.06kg)(V1f) + (.09kg)(V2f)

.2535 = (.06kg)(V1f) + (.09kg)(V2f)

.06kg(V1f) = .2535 + .09kg(V2f)


.06kg/.06(V1f) = .2535/.06 + .09kg/.06(V2f)

V1f = 4225 + 1.5 V2f
 
  • #4
ehild
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when you say to solve the first equation for V1f do you mean:

(.06kg)(2.5m/s) + (.09kg)(1.15m/s) = (.06kg)(V1f) + (.09kg)(V2f)

.2535 = (.06kg)(V1f) + (.09kg)(V2f)

It is good up to here, but the next steps are wrong:
.06kg(V1f) = .2535 + .09kg(V2f)


.06kg/.06(V1f) = .2535/.06 + .09kg/.06(V2f)

V1f = 4225 + 1.5 V2f

Check it.

ehild
 
  • #5
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can you please just get me going in the right direction i cant figure out how to solve it because of the two unknowns
 
  • #6
ehild
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Lets start from here: 0.2535=0.06*v1f+0.09*v2f

You want v1f be alone on one side of this equation. Subtract 0.09*v2f from both sides. Do it now.

ehild
 
  • #7
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is this right....

-.06(V1f) = .09 (V2f) - .2535
 
Last edited:
  • #8
ehild
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is this right....

-.06(V1f) = .09(V2f) - .2535

Yes. Now get rid of the coefficient of v1f: divide both sides by -0.06.
 
  • #9
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i got this:

V1f = - 1.5 (V2f) + 4.225

....but, how do i plug this into the second equation?
 
  • #10
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1/2(.06kg)(2.5m/s)^2 + 1/2(.09kg)(1.15m/s)^2 = 1/2(.06kg)[(-1.5)(V2f) +4.225)^2 + 1/2(.09kg)(V2f)^2

.25 = 1/2(.06kg)[(1.5)(V2f) + 4.225]^2 + 1/2(.09kg)(v2f)^2



....would this be right?
 
  • #11
ehild
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There should be a - sign in front of (1.5) (v2f).

Keep at least 4 significant digits on the left side of the equation.

It is not needed to use units in the equation. If you do, use them everywhere.

ehild
 
  • #12
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.2470 = 1/2(.06)[-(1.5)(V2f) + 4.225]^2 + 1/2(.09)(v2f)^2


.2470 = .03(-(1.5)(V2f) + 4.225)^2 + .045(V2f)^2


.2470 = .03((-1.5V2f) + 4.225)^2 + (.045V2f)^2


Is this right so far?
 
  • #13
ehild
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Why do you square 0.045? It has to be outside the parentheses.

ehild
 
  • #14
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.2470 = .03((-1.5V2f) + 4.225)^2 + .045(V2f)^2

Do I distribute the .03 and .045, or am I missing something?
 
  • #15
ehild
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You can distribute a sum: a(b+c) = ab+ac. (v2f)^2 is not a sum.



ehild
 
  • #16
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Oh, I see. So would I just leave the equation as is?


.2470 = .03((-1.5V2f) + 4.225)^2 + .045(V2f)^2

How do I go about answering when V2f is in there twice? Would I have to subtract one of them like so...

.2470 - .045(V2f)^2 = .03((-1.5V2f) + 4.225)^2

Sorry for all the questions I'm just really confused about this problem.
 
  • #17
ehild
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Oh, I see. So would I just leave the equation as is?


.2470 = .03((-1.5V2f) + 4.225)^2 + .045(V2f)^2

How do I go about answering when V2f is in there twice? Would I have to subtract one of them like so...

.2470 - .045(V2f)^2 = .03((-1.5V2f) + 4.225)^2

Sorry for all the questions I'm just really confused about this problem.

You can do that but it has not much sense.

First evaluate the square. Apply the identity: (a+b)^2=a^2+2ab+b^2

Next distribute 0.03.

Arrange the equation so one side is 0. Group like terms.

It is a quadratic equation. Use the quadratic formula to solve it.

You can find step-by step explanations for solving equations at

http://www.sosmath.com/algebra/solve/solve0/solve0.html.

ehild
 
  • #18
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this is what I have so far:


0 = 6.75(V1f)^2 + 53.55 + .045(V2f)^2 - .2470

Not to be so negative......but I think I did something wrong
 
Last edited:
  • #19
ehild
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It is not correct.

Evaluate the square by using the identity (a+b)^2=a^2+2ab+b^2: a=*-1.5 (v2f), b=4.225.

(-1.5(V2f) + 4.225)^2= 2.25 (v2f)^2-3(4.225) (v2f)+4.225^2=2.25 (v2f)^2-12.675 (v2f)+17.8506.

This expression is multiplied by 0.03. Distribute.

ehild
 
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  • #20
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After distributing I got:

.0675 V2f^2 - .3802 V2f + .5355


The original equation was:

.2470 = .03((-1.5V2f) + 4.225)^2 + .045(V2f)^2

I added .045 V2f^2 to .0675 V2f^2 and then subtracted .2470 from .5355

and got:

.1075 V2f^2 - .3802 V2f + .2885

Using the quadratic formula:

.3802 +/- √-.3802^2- 4 (.1075) (.2885) / 2(.1075)

adding:

.3802 + √.1445 - .1240 / .215

.3802 + √.0205 / .215

.3802 + .1431 / .215

.5233 / .215 = 2.43

subtracting:

.3802 - .1431 /.215

.2371 / .215 = 1.10


Did I do something wrong because my book says the answers should be .88 and 2.23
 
Last edited:
  • #21
ehild
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After distributing I got:

.0675 V2f^2 - .3802 V2f + .5355


The original equation was:

.2470 = .03((-1.5V2f) + 4.225)^2 + .045(V2f)^2

I added .045 V2f^2 to .0675 V2f^2 and then subtracted .2470 from .5355

and got:

.1075 V2f^2 - .3802 V2f + .2885

You made a mistake when adding .045 V2f^2 to .0675 V2f^2.

0.045 +0.0675 = 0.1125, so the quadratic equation is

0.1125 V2f^2 - 0.3802 V2f + 0.2885=0

Using the quadratic formula:

.3802 +/- √-.3802^2- 4 (.1075) (.2885) / 2(.1075)

wrong.

quadratic equation: ax^2+bx+c=0

quadratic formula:

[tex]

v2f=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}[/tex]

we have:
a=0.1125
b=-0.3802
c=0.2885


[tex]
v2f=\frac{0.3802\pm\sqrt{(-0.3802)^2-4 \cdot0.1125\cdot0.2885)}}{2\cdot0.1125}=\frac{0.3802\pm\sqrt{0.1445-0.1298}}{0.225}=\frac{0.3802\pm0.1212}{0.225}[/tex]

One root is 2.23, the other is 1.15, just the same as if the balls did not collided.
So v2f= 2.23 m/s. Find v1f.

ehild
 
  • #22
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Okay, I plugged the numbers in and sure enough it was .88. Thank you very much for helping me with this problem. I'll review this problem because I m sure it will be on my test Tuesday.Once again, thank you.
 

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