Solve for the total Kinetic Energy change with unknowns

[Balti]

Homework Statement

Dave and Paula are outside, playing upon frictionless ice provided by their physics teacher. Paula is stationary when Dave, sliding with velocity v suddenly crashes into her. They move off together. Dave has twice the mass of Paula. What percentage of the initial kinetic energy remains after the collision? Give your answer to two significant figures.

Relevant Equations

$E_k=\frac 1 2 (m)(v^2)$
I suppose

Firstly I tried defining into an equation to make the whole thing more 'tangible'.

$m_1= Paula's~Weight⋅2 = m_p⋅2$

$m_2= \frac {Dave's~weight}{2} = \frac {m_d}{2}$

Before impact
$E_k1= \frac 1 2 (m_p⋅2)(0^2)$
$E_k1= \frac 1 2 (2m_p)$
$E_k1= m_p$

After Impact
$E_k2= \frac 1 2 (\frac {m_d} {2}) (v^2)$
$E_k2= ({m_d}) (v^2)$
And I guess
$E_k2=({m_d}) (v^2) ({m_p})$

And as I'm told we find the change of kinetic energy by
$ΔE_k={E_k2}-{E_k1}$
And then I get stuck. Provided, it is 2 am and I am dead exhausted.I don't want an answer, I just want a direction to go in. I was never really good with having a few variables and you figure it out with formula manipulation. I don't know, just not really my thing.

Any help?

 

[haruspex]

$m_1= Paula's~Weight⋅2 = m_p⋅2$

$m_2= \frac {Dave's~weight}{2} = \frac {m_d}{2}$

What are m₁ and m₂, if not the respective masses?

 

[gleem]

Keep you notation simple so you know to whom each energy or mass or velocity refers to. Use d for Dave and p for Paula and pd for their combined parameters no need for using 1 and 2.

Right off bat you got things confused; read the problem carefully.

Paula's mass = m_p;
Dave's mass = m_d = 2m_p
v_p = 0

Thus E_p =0

m_pd = ?
v_pd = ?

 

[PeroK]

Homework Statement: Dave and Paula are outside, playing upon frictionless ice provided by their physics teacher. Paula is stationary when Dave, sliding with velocity v suddenly crashes into her. They move off together. Dave has twice the mass of Paula. What percentage of the initial kinetic energy remains after the collision? Give your answer to two significant figures.
Homework Equations: $E_k=\frac 1 2 (m)(v^2)$
I suppose

Firstly I tried defining into an equation to make the whole thing more 'tangible'.

$m_1= Paula's~Weight⋅2 = m_p⋅2$

$m_2= \frac {Dave's~weight}{2} = \frac {m_d}{2}$

Before impact
$E_k1= \frac 1 2 (m_p⋅2)(0^2)$
$E_k1= \frac 1 2 (2m_p)$
$E_k1= m_p$

After Impact
$E_k2= \frac 1 2 (\frac {m_d} {2}) (v^2)$
$E_k2= ({m_d}) (v^2)$
And I guess
$E_k2=({m_d}) (v^2) ({m_p})$

And as I'm told we find the change of kinetic energy by
$ΔE_k={E_k2}-{E_k1}$
And then I get stuck. Provided, it is 2 am and I am dead exhausted.I don't want an answer, I just want a direction to go in. I was never really good with having a few variables and you figure it out with formula manipulation. I don't know, just not really my thing.

Any help?

1) As others have said, sort out your notation.

2) Energy is not conserved.

3) What is conserved?

 

[Mister T]

I don't want an answer, I just want a direction to go in

Then pick some numbers and work it out that way first. That is, choose values for Paula's mass, Dave's mass, and the speed $v$. See if you can get an answer that way.

 

[Balti]

Ah, I see. You god damn geniuses.

The reason I ask this is that I've come across questions like this before that seem unsolvable. But are really not, they sometimes give you actual answers. So I'm just wary about picking random values. I have to go learn conservation of energy and the like. I know for a fact that Kinetic Energy is not conserved, because the collision is inelastic. Should I then use $E_k=\frac {p^2}{2m}$ formula to try to solve it?

 

[Mister T]

So I'm just wary about picking random values.

Okay, so now you have a choice. Do something that you were advised to do even though you're wary of it, or just ignore the advice.

 

[neilparker62]

Ah, I see. You god damn geniuses.

The reason I ask this is that I've come across questions like this before that seem unsolvable. But are really not, they sometimes give you actual answers. So I'm just wary about picking random values. I have to go learn conservation of energy and the like. I know for a fact that Kinetic Energy is not conserved, because the collision is inelastic. Should I then use $E_k=\frac {p^2}{2m}$ formula to try to solve it?

Good idea. Since momentum is conserved it will be the same in the expressions for your KE before and KE after.

 

[haruspex]

wary about picking random values.

I share your caution. Much better is to insert unknowns, as you did in post #1, but the way you inserted them made no sense.
If you write Paula's mass as m_p and Dave's as m_d then you have no need to invent m₁ and m₂ as well. Just write the equation relating m_p and m_d that represents the statement "Dave has twice the mass of Paula".

 

[Balti]

Good idea. Since momentum is conserved it will be the same in the expressions for your KE before and KE after.

I am new to maths entirely (as in I spent June-July learning all of Algebra), so would I do $E_k1 = E_k2$? Or just stick to what I did up in my original post?

 

[haruspex]

I am new to maths entirely (as in I spent June-July learning all of Algebra), so would I do $E_k1 = E_k2$? Or just stick to what I did up in my original post?

No, energy is not conserved. You are asked to find how much is lost.
What you have to use here is conservation of momentum. Can you write expressions for the total momentum before and after collision?

 

[Balti]

No, energy is not conserved. You are asked to find how much is lost.
What you have to use here is conservation of momentum. Can you write expressions for the total momentum before and after collision?

Yep. That's what I did and I ended up with an actual answer to my surprise. It's fascinating how you can have so little information and still deduct something from a system.

First, I realized that the end result is a compound of the two $E_k$, so I wrote out.
$\frac 1 2 {(2m)}{(v^2)} = \frac 1 2 {(m)}{(v^2)} + \frac 1 2 {(2m)}{(v^2)}$
Then the rest is simple maths, and I ended up with the amount of Kinetic Energy retained to be 66.67%

Thank you all so much

 

[SammyS]

Yep. That's what I did and I ended up with an actual answer to my surprise. It's fascinating how you can have so little information and still deduct something from a system.

First, I realized that the end result is a compound of the two $E_k$, so I wrote out.
$\frac 1 2 {(2m)}{(v^2)} = \frac 1 2 {(m)}{(v^2)} + \frac 1 2 {(2m)}{(v^2)}$
Then the rest is simple maths, and I ended up with the amount of Kinetic Energy retained to be 66.67%

Thank you all so much

Which speed does the variable, $v$, represent?

From a strictly mathematical viewpoint, that is an inconsistent equation. The left-hand side is clearly not equal to the right-hand side (unless $v=0$ or $m=0$).

If you mean that the left-hand is the initial K.E. and the right-hand side is the final K.E., as separate statements, with no equality involved between the two sides, then this also leads to an incorrect result. Why? Because that gives 3/2 as the ratio of final K.E. to initial K.E. It does not give 2/3 as the ratio.

Yes, the initial K.E. is given by $\dfrac 1 2 {(2m)}{(v_{Dave}^2)}$, where $2m$ is Dave's mass and $v_{Dave}$ is his initial speed.

And, the final K.E. is given by $\dfrac 1 2 {(m)}{(v_{Both}^2)} + \dfrac 1 2 {(2m)}{(v_{Both}^2)}$.

Use conservation of momentum to find the relationship between $v_{Dave}$ and $v_{Both}$.