# Physics Elastic Collision Question

eo1989

## 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:

Homework Helper
"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

eo1989
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

Homework Helper
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

eo1989
can you please just get me going in the right direction i cant figure out how to solve it because of the two unknowns

Homework Helper
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

eo1989
is this right....

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

Last edited:
Homework Helper
is this right....

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

Yes. Now get rid of the coefficient of v1f: divide both sides by -0.06.

eo1989
i got this:

V1f = - 1.5 (V2f) + 4.225

....but, how do i plug this into the second equation?

eo1989
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?

Homework Helper

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

eo1989
.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?

Homework Helper
Why do you square 0.045? It has to be outside the parentheses.

ehild

eo1989
.2470 = .03((-1.5V2f) + 4.225)^2 + .045(V2f)^2

Do I distribute the .03 and .045, or am I missing something?

Homework Helper
You can distribute a sum: a(b+c) = ab+ac. (v2f)^2 is not a sum.

ehild

eo1989
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

Homework Helper
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

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

eo1989
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

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Homework Helper
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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eo1989
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

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

.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:
Homework Helper
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

0.045 +0.0675 = 0.1125, so the quadratic equation is

0.1125 V2f^2 - 0.3802 V2f + 0.2885=0

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

wrong.

$$v2f=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}$$

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

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

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

eo1989
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.