Confused about a conservation of energy problem

[BikGer2]

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.

 

[BvU]

Hello @BikGer2 ,

##\qquad##!​

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 !

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...##\ ##

 

[BikGer2]

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:

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.

 

[BvU]

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....)

##\ ##

 

[haruspex]

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