Confused about a conservation of energy problem

AI Thread Summary
The discussion revolves around a conservation of energy problem involving a central collision between two bodies. The user calculated the final velocity of body 2 to be 16.67 m/s, resulting in a kinetic energy of approximately 9726.11 J. However, the book states that the energy transferred is only 24.3 J, which the user finds confusing. Participants in the discussion suggest that the book's answer may be incorrect and emphasize that the assumption of body 1 being at rest post-collision is unjustified. The conversation highlights the complexities of energy transfer in elastic collisions, particularly when mass differences are significant.
BikGer2
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Homework Statement
Determine the energy which body of mass m1=50 kg transfers to body of mass m2=70kg, if body 1 is moving towards body 2 with constant velocity of v1 = 20 m/s while body 2 is at rest. The collision is perfectly elastic and frontal (assuming this means the collision is central).
Relevant Equations
\begin{align}
m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2' \nonumber \\
\frac{m_1v_1^2}{2} + \frac{m_2v_2^2}{2} = \frac{m_1v_1'^2}{2} + \frac{m_2v_2'^2}{2} \nonumber
\end{align}
Hi,

I assumed I was supposed to find the amount of kinetic energy body 2 receives after contact, assuming the collision is central, body 1 will be at rest after the collision.

I started by using the equation for conservation of momentum:

\begin{align}
m_1v_1 = m_1v_1' + m_2v_2' \\
50 * 20 = 50v_1' + 70 v_2' \nonumber \end{align}

Then kinetic energy:
\begin{align}
m_1v_1^2 = m_1v_1'^2 + m_2v_2'^2 \\
50 * 400 = 50v_1'^2 + 70v_2'^2 \nonumber \end{align}

From the first equation, $$v_1' = \frac{1000 - 70v_2'}{50} = 20 - 1.4v2'$$

Which was then substituted into the kinetic energy equation:
\begin{align}
20000 = 50(20 - 1.4v_2')^2 + 70v_2'^2 \nonumber \\
20000 = 50(400 - 56v_2' + 1.96v_2'^2) + 70v_2'^2 \nonumber \end{align}
After dividing everything by 50 and sorting everything out:
\begin{align}
3.36v_2'2 - 56v_2' = 0 \nonumber \end{align}
Solving the quadratic yields v2' to be 16.67 m/s. (Other solution of quadratic is 0)

If I plug that into the kinetic energy formula, I get the energy of body 2 to be Ek = 9726.1115 J.

What I find confusing is that the solution of the problem in the book says ΔE = 24.3J which to me, doesn't make sense.

The problem asks for the amount of energy transferred from body 1 to body 2, shouldn't that amount be roughly the same, the collision is elastic.

Any help would be much appreciated.
 
Last edited:
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Hello @BikGer2 ,
:welcome:##\qquad##!​
BikGer2 said:
assuming the collision is central, body 1 will be at rest after the collision.
This assumption is not justified. What made you think that ? Did you check it ?

My compliments for your clear post and the effort to ##TeX## it ! :smile:

I get the same result you find (and with ##v_1' = \frac{1000 - 70v_2'}{50} = 20 - 1.4v2' \quad \Rightarrow\quad v_1'= - 3.33 ## m/s !)

So the book answer must be in error. It happens...##\ ##
 
Last edited:
BvU said:
This assumption is not justified. What made you think that ? Did you check it ?
I was not entirely sure, but I assumed saying the collision is 'frontal' was an implication the collision is central, though I have not tried solving this as an offset collision. I assumed the latter:
Screenshot 2024-01-10 at 10.54.22.png

BvU said:
I get the same result you find (and with ##v_1' = \frac{1000 - 70v_2'}{50} = 20 - 1.4v2' \quad \Rightarrow\quad v_1'= - 3.33 ## m/s !)

So the book answer must be in error. It happens...
It could be the book, since the solution states delta energy to be only 24.3 J, a weirdly small value, and the delta means it's a change in energy, which was not asked in the problem statement.
 
BikGer2 said:
aying the collision is 'frontal' was an implication the collision is central
It's the best you can do. Frontal is more a traffic term than a physics term

Bottom line: ##E_{\text kin}## of ##m_2## changes from 0 to 9.72 kJ
(hence the ##\Delta E##)

Physics: only 97% of the kinetic energy is transferred. This percentage decreases when the mass difference increases (in the extreme: 0% for ##m_2\uparrow \infty##)

Given data are 1 or 2 significant figures, so 9726.1115 is overdoing it
(and I suspect even the 6 will change if you don't round off intermediate results....)

##\ ##
 
You get the given answer if you replace 20m/s with 1m/s.
 
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