Calculating Final Speed and Kinetic Energy in an Elastic Collision

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Homework Help Overview

The discussion revolves around an elastic collision problem involving two balls of different masses and initial velocities. Participants are tasked with calculating the final speeds and kinetic energies of the balls after the collision using conservation laws.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of conservation of momentum and kinetic energy equations, questioning the validity of the equations used and the manipulations performed. There are attempts to derive relationships between the final velocities based on the conservation principles.

Discussion Status

The discussion is ongoing, with participants providing guidance on how to approach the problem and suggesting corrections to previous calculations. There is recognition of errors in sign and manipulation, and some participants are exploring different methods to derive the necessary equations.

Contextual Notes

Participants note potential misunderstandings regarding the equations derived from conservation laws and the implications of elastic collisions. There is a focus on ensuring the correct application of signs and the independence of equations used in the calculations.

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Homework Statement


A ball of mass = 3-kg moving to the right at 3 m/s collides elastically with a ball of mass m2 = 2 kg moving at 4 m/s to the left. Find (a) the final speed of each object and (b)
the kinetic energy of each ball before and after the collisions.


Homework Equations


V1i + V1f = V2f + V2i
V1i - V2i = - (V1f - V2f)


The Attempt at a Solution


I plugged the numbers into the above formulas
3 + V1f = -4 + V2f
V2f = 7 + V1f

3 - (-4) = -(V1f - V2f)
7 = -V1f + V2f

substitute V2f
7 = -V1f + 7 + V1f
I got 0 = 0 the answer should be V1f = -2.60 ms, V2f = 4.40 m/s
 
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BrainMan said:

Homework Equations


V1i + V1f = V2f + V2i
V1i - V2i = - (V1f - V2f)

Where'd you get those equations from? Conservation of momentum states m[itex]_{1}[/itex]V[itex]_{1i}[/itex]+m[itex]_{2}[/itex]V[itex]_{2i}[/itex]=m[itex]_{1}[/itex]V[itex]_{1f}[/itex]+m[itex]_{2}[/itex]V[itex]_{2f}[/itex]
 
Nathanael said:
Where'd you get those equations from? Conservation of momentum states m[itex]_{1}[/itex]V[itex]_{1i}[/itex]+m[itex]_{2}[/itex]V[itex]_{2i}[/itex]=m[itex]_{1}[/itex]V[itex]_{1f}[/itex]+m[itex]_{2}[/itex]V[itex]_{2f}[/itex]

In this problem since it is perfectly elastic kinetic energy is conserved so 1/2m1v1i2 + 1/2m2v2i2 = 1/2m1v1f2 + 1/2m2v2f2
with some mathematical manipulation you can acquire the above formulas.
 
BrainMan said:
In this problem since it is perfectly elastic kinetic energy is conserved so 1/2m1v1i2 + 1/2m2v2i2 = 1/2m1v1f2 + 1/2m2v2f2
with some mathematical manipulation you can acquire the above formulas.

Ah, I see now. Still you seem to have made a mistake with your manipulations since there's no possible way to manipulate out of the equation m1 and m2
 
Last edited:
Nathanael said:
Ah, I see now. Still you seem to have made a mistake with your manipulations since there's no possible way to manipulate out of the equation m1 and m2

I got these equations directly from my physics book.
 
BrainMan said:
I got these equations directly from my physics book.
Well don't trust your physics book. If it comes from conservation of KE then do the manipulations yourself.

And let me know if you figure it out, because I don't think it's possible to turn
BrainMan said:
1/2m1v1i2 + 1/2m2v2i2 = 1/2m1v1f2 + 1/2m2v2f2
into:
BrainMan said:
V1i + V1f = V2f + V2i
V1i - V2i = - (V1f - V2f)

The simplest I can make those equations is:
m[itex]_{1}(V_{1i}+V_1f)(V_{1i}-V_{1f})=m_{2}(V_{2f}+V_2i)(V_{2f}-V_{2i})[/itex]
 
Last edited:
But using either
m[itex]_{1}(V_{1i}+V_1f)(V_{1i}-V_{1f})=m_{2}(V_{2f}+V_2i)(V_{2f}-V_{2i})[/itex]
or
m[itex]_{1}[/itex]V[itex]_{1i}[/itex]+m[itex]_{2}[/itex]V[itex]_{2i}[/itex]=m[itex]_{1}[/itex]V[itex]_{1f}[/itex]+m[itex]_{2}[/itex]V[itex]_{2f}[/itex]
Should lead to the right answer (I think)

It's just that the equations you used (from your textboook) are wrong for this situation.
 
Nathanael said:
It's just that the equations you used (from your textboook) are wrong for this situation.
No, those equations are correct for a perfectly elastic collision.

Brainman, you got 0=0 because your 'two' equations are not independent. One is simply a rearrangement of the other. You can use the result you got in the OP, together with conservation of momentum. This will give the result because the first equation uses the fact that the collision was perfectly elastic, so provides additional information.
 
Last edited:
CAF123 said:
No, those equations are correct for a perfectly elastic collision.

What was my mistake then?
 
  • #10
BrainMan said:
What was my mistake then?
I just edited my post. See above.
 
  • #11
CAF123 said:
No, those equations are correct for a perfectly elastic collision.

Brainman, you got 0=0 because your 'two' equations are not independent. One is simply a rearrangement of the other. You can use the result you got in the OP, together with conservation of momentum. This will give the result because the first equation uses the fact that the collision was perfectly elastic, so provides additional information.

OK I did that and I am still not getting the correct answer. I am pretty sure I did something wrong when I substituted. Here are my calculations.

equation 1
m[itex]_{1}[/itex]V[itex]_{1i}[/itex]+m[itex]_{2}[/itex]V[itex]_{2i}[/itex]=m[itex]_{1}[/itex]V[itex]_{1f}[/itex]+m[itex]_{2}[/itex]V[itex]_{2f}[/itex]
3(3) + 2(-4) = 3(v1f) + 2(v2f)

equation 2
V1i + V1f = V2f + V2i
3 + v1f = v2f + 4
v2f = 7 + v1f

substitute v2f
1 = 3(v1f) + 2(7 + V1F)
1= 3V1F + 14 + 2V1f
1- 3v1f = 14 + 2v1f
-3v1f = 13 + 2v1f
-v1f = 13
v1f = -13

the answer should be - 2.60 m/s
 
  • #12
BrainMan said:
equation 2
V1i + V1f = V2f + V2i
3 + v1f = v2f + 4
v2f = 7 + v1f
Small sign error there.
 
  • #13
CAF123 said:
Small sign error there.

I am still getting the wrong answer

1 = 3v1f + 2v1f - 2
3/5 = v1f

there must be something else I'm doing wrong.
 
  • #14
BrainMan said:
I am still getting the wrong answer

1 = 3v1f + 2v1f - 2
3/5 = v1f

there must be something else I'm doing wrong.
From conservation of momentum, you correctly obtained the equation 1 = 3v1f + 2v2f.

From the equation derived by manipulating the equations for conservation of kinetic energy and that of momentum (do you understand the derivation?) you obtain, by fixing the sign error in the previous post, -7 = v1f - v2f

Solve these two equations for v1f and v2f.
 
  • #15
CAF123 said:
From conservation of momentum, you correctly obtained the equation 1 = 3v1f + 2v2f.

From the equation derived by manipulating the equations for conservation of kinetic energy and that of momentum (do you understand the derivation?) you obtain, by fixing the sign error in the previous post, -7 = v1f - v2f

Solve these two equations for v1f and v2f.

I am having trouble deriving that formula. What I did was
V1i + V1f = V2f + V2i
3 - vif = -4 -v2f // the final velocities are negative because they will both switch directions.
-1(3 - vif = -4 -v2f)
-3 + vif = 4 +v2f
-7 + vif = v2f

I then tried to substitute this into the other equation.


1 = 3v1f + 2v2f
1 = 3v1f + 2v1f - 14
15 = 5v1f
v1f = 3
and I did not get the right answer
 
  • #16
CAF123 said:
From conservation of momentum, you correctly obtained the equation 1 = 3v1f + 2v2f.

From the equation derived by manipulating the equations for conservation of kinetic energy and that of momentum (do you understand the derivation?) you obtain, by fixing the sign error in the previous post, -7 = v1f - v2f

Solve these two equations for v1f and v2f.
Do what CAF123 says above.

One way to do this is by elimination: Multiply the 2nd equation by 2, then add the equations to eliminate v2f .
 
  • #17
SammyS said:
Do what CAF123 says above.

One way to do this is by elimination: Multiply the 2nd equation by 2, then add the equations to eliminate v2f .

What did I do wrong in my previous post?
 
  • #18
BrainMan said:
What did I do wrong in my previous post?

BrainMan said:
I am having trouble deriving that formula. What I did was
V1i + V1f = V2f + V2i
3 - v1f = -4 -v2f // the final velocities are negative because they will both switch directions.
Don't change signs on these velocities. You didn't on the other equation.

Algebra will take care of the signs! (That's one of Algebra's most endearing characteristics.)

-1(3 - v1f = -4 -v2f)
-3 + v1f = 4 +v2f
-7 + v1f = v2f

I then tried to substitute this into the other equation.


1 = 3v1f + 2v2f
1 = 3v1f + 2v1f - 14
15 = 5v1f
v1f = 3
and I did not get the right answer
 
Last edited:
  • #19
SammyS said:
Don't change signs on these velocities. You didn't on the other equation.

Algebra will take care of the signs! (That's one of Algebra's most endearing characteristics.)

OK I see what I did and now got the right answer. Thanks!
 

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