Physics Elastic Collision Question

In summary: Thank you so much! I got the right answers! In summary, using the equations of conservation of momentum and conservation of kinetic energy, the speeds and directions of two colliding balls can be calculated. After plugging in the given values and solving for the unknowns, the final speeds of the balls were found to be 0.88 m/s and 2.23 m/s.
  • #1
eo1989
12
0

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:
Physics news on Phys.org
  • #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]"

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
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
eo1989 said:
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
can you please just get me going in the right direction i can't figure out how to solve it because of the two unknowns
 
  • #6
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
is this right...

-.06(V1f) = .09 (V2f) - .2535
 
Last edited:
  • #8
eo1989 said:
is this right...

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

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

V1f = - 1.5 (V2f) + 4.225

...but, how do i plug this into the second equation?
 
  • #10
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
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
.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
Why do you square 0.045? It has to be outside the parentheses.

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

Do I distribute the .03 and .045, or am I missing something?
 
  • #15
You can distribute a sum: a(b+c) = ab+ac. (v2f)^2 is not a sum.
ehild
 
  • #16
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
eo1989 said:
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
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
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
 
Last edited:
  • #20
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
eo1989 said:
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
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.
 

1. What is an elastic collision in physics?

An elastic collision is a type of collision between two particles in which there is no loss of kinetic energy. This means that the total kinetic energy of the system before and after the collision remains the same. In other words, the particles bounce off each other without any loss of energy.

2. How is momentum conserved in an elastic collision?

Momentum is conserved in an elastic collision because the total momentum of the system before and after the collision remains the same. This means that the sum of the individual momentums of the particles before the collision is equal to the sum of the individual momentums after the collision.

3. What is the difference between an elastic collision and an inelastic collision?

In an elastic collision, the total kinetic energy of the system is conserved and there is no loss of energy. In an inelastic collision, some of the kinetic energy is lost and converted into other forms of energy, such as heat or sound. This means that the total kinetic energy of the system after the collision is less than before the collision.

4. How is the coefficient of restitution related to an elastic collision?

The coefficient of restitution is a measure of the elasticity of a collision. It is defined as the ratio of the relative velocity of separation between the two particles after the collision to the relative velocity of approach before the collision. In an elastic collision, the coefficient of restitution is equal to 1.

5. Can an elastic collision occur between objects of different masses?

Yes, an elastic collision can occur between objects of different masses. The conservation of momentum and kinetic energy still apply, regardless of the masses of the objects involved. However, the change in velocity of each object will be different based on their respective masses.

Similar threads

  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
999
  • Introductory Physics Homework Help
Replies
2
Views
980
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
892
  • Introductory Physics Homework Help
Replies
17
Views
5K
  • Introductory Physics Homework Help
Replies
20
Views
882
  • Introductory Physics Homework Help
Replies
17
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
2K
Back
Top