Calculating Final Speed and Kinetic Energy in an Elastic Collision

[BrainMan]

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

V_1i + V_1f = V_2f + V_2i
V_1i - V_2i = - (V_1f - V_2f)

The Attempt at a Solution

I plugged the numbers into the above formulas
3 + V_1f = -4 + V_2f
V_2f = 7 + V_1f

3 - (-4) = -(V_1f - V_2f)
7 = -V_1f + V_2f

substitute V_2f
7 = -V_1f + 7 + V_1f
I got 0 = 0 the answer should be V_1f = -2.60 ms, V_2f = 4.40 m/s

 

[Nathanael]

Homework Equations

V_1i + V_1f = V_2f + V_2i
V_1i - V_2i = - (V_1f - V_2f)

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]

 

[BrainMan]

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 1/2m₁v_1i² + 1/2m₂v_2i² = 1/2m₁v_1f² + 1/2m₂v_2f²
with some mathematical manipulation you can acquire the above formulas.

 

[Nathanael]

In this problem since it is perfectly elastic kinetic energy is conserved so 1/2m₁v_1i² + 1/2m₂v_2i² = 1/2m₁v_1f² + 1/2m₂v_2f²
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

 

[BrainMan]

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.

 

[Nathanael]

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

1/2m₁v_1i² + 1/2m₂v_2i² = 1/2m₁v_1f² + 1/2m₂v_2f²

into:

V_1i + V_1f = V_2f + V_2i
V_1i - V_2i = - (V_1f - V_2f)

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]

 

[Nathanael]

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.

 

[CAF123]

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.

 

[BrainMan]

No, those equations are correct for a perfectly elastic collision.

What was my mistake then?

 

[CAF123]

What was my mistake then?

I just edited my post. See above.

 

[BrainMan]

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(v_1f) + 2(v_2f)

equation 2
V_1i + V_1f = V_2f + V_2i
3 + v_1f = v_2f + 4
v_2f = 7 + v_1f

substitute v_2f
1 = 3(v_1f) + 2(7 + V_1F)
1= 3V_1F + 14 + 2V_1f
1- 3v_1f = 14 + 2v_1f
-3v_1f = 13 + 2v_1f
-v_1f = 13
v_1f = -13

the answer should be - 2.60 m/s

 

[CAF123]

equation 2
V_1i + V_1f = V_2f + V_2i
3 + v_1f = v_2f + 4
v_2f = 7 + v_1f

Small sign error there.

 

[BrainMan]

Small sign error there.

I am still getting the wrong answer

1 = 3v_1f + 2v_1f - 2
3/5 = v_1f

there must be something else I'm doing wrong.

 

[CAF123]

I am still getting the wrong answer

1 = 3v_1f + 2v_1f - 2
3/5 = v_1f

there must be something else I'm doing wrong.

From conservation of momentum, you correctly obtained the equation 1 = 3v_1f + 2v_2f.

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 = v_1f - v_2f

Solve these two equations for v_1f and v_2f.

 

[BrainMan]

From conservation of momentum, you correctly obtained the equation 1 = 3v_1f + 2v_2f.

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 = v_1f - v_2f

Solve these two equations for v_1f and v_2f.

I am having trouble deriving that formula. What I did was
V_1i + V_1f = V_2f + V_2i
3 - v_if = -4 -v_2f // the final velocities are negative because they will both switch directions.
-1(3 - v_if = -4 -v_2f)
-3 + v_if = 4 +v_2f
-7 + v_if = v_2f

I then tried to substitute this into the other equation.


1 = 3v_1f + 2v_2f
1 = 3v_1f + 2v_1f - 14
15 = 5v_1f
v_1f = 3
and I did not get the right answer

 

[SammyS]

From conservation of momentum, you correctly obtained the equation 1 = 3v_1f + 2v_2f.

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 = v_1f - v_2f

Solve these two equations for v_1f and v_2f.

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 v_2f .

 

[BrainMan]

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 v_2f .

What did I do wrong in my previous post?

 

[SammyS]

What did I do wrong in my previous post?

I am having trouble deriving that formula. What I did was
V_1i + V_1f = V_2f + V_2i
3 - v_1f = -4 -v_2f // 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 - v_1f = -4 -v_2f)
-3 + v_1f = 4 +v_2f
-7 + v_1f = v_2f

I then tried to substitute this into the other equation.


1 = 3v_1f + 2v_2f
1 = 3v_1f + 2v_1f - 14
15 = 5v_1f
v_1f = 3
and I did not get the right answer

 

[BrainMan]

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!