How do I find the percentage of energy lost duing a collision between two spheres going in opposite directions?

Jun 2017
276
6
Lima, Peru
The problem is as follows:

Two spheres are going in opposites directions as shown in the figure from below. The collision in progress is of inelastic nature. If the coefficient of restitution $COR=0.5$. Find the percentage of mechanical energy (with respect to the value of an instant before the collision) which is lost during the collision.



The alternatives given are as follows:

$\begin{array}{ll}
1.&50\,%\\
2.&55\,%\\
3.&60\,%\\
4.&65\,%\\
5.&75\,%\\
\end{array}$

I'm not sure exactly how should I use the information of the coefficient of restitution. Can someone help me here?.

I believe that for this situation I can use the conservation of mechanical energy as follows:

$m_1v_1+m_2(-v_2)=(m_1+m_2)v_3$

Then in order to find the kinetic energy will be found by finding $v_3$ but am I right with this?. Can someone help me?

If I replace in there the information given it would become into:

$2\times5+1\times\left(-10\right)= (2+1)v_3$

$v_3=0$

But this would give

$v_3=0$

How exactly can I find the energy lost if the speed in the end is zero?. Can someone help me with clearing out this inconsistency?.
 

skeeter

Math Team
Jul 2011
3,196
1,724
Texas
initial momentum of the system is $p_0 = -15 \dfrac{kg\, m}{s}$

momentum is conserved ...

$m_1v_{1f} + m_2v_{2f} = -15$

coefficient of restitution is the ratio $\dfrac{v_{1f}-v_{2f}}{v_{1i}-v_{2i}} = 0.5$

solve the system of equations for both final velocities


percentage of mechanical energy lost is $\left(\dfrac{E_f-E_i}{E_i}\right) \cdot 100$
 
  • Like
Reactions: Chemist116
Jun 2017
276
6
Lima, Peru
initial momentum of the system is $p_0 = -15 \dfrac{kg\, m}{s}$

momentum is conserved ...

$m_1v_{1f} + m_2v_{2f} = -15$

coefficient of restitution is the ratio $\dfrac{v_{1f}-v_{2f}}{v_{1i}-v_{2i}} = 0.5$

solve the system of equations for both final velocities


percentage of mechanical energy lost is $\left(\dfrac{E_f-E_i}{E_i}\right) \cdot 100$

Solving the system of equations would give me this:

$u_1+2\times u_2=-15$

$u_1-u_2=0.5 \times(5+10)$

$u_1=0$

$u_2=\frac{-15}{2}=-7.5$

Then the initial energy would be as follows:

$E_i=\frac{1}{2}(2)(5)^2+\frac{1}{2}(1)(-10)^2=75$

$E_f=\frac{1}{2}(2)(0)^2+\frac{1}{2}(1)(-7.5)^2=28.125$

So the percentage will be:

$\%= \frac{E_i-E_f}{E_i}\times 100 = \frac {75-28.125}{75} \times 100 = 62.5$

But this does not check with the alternatives. Where did I got it wrong?.
 
Jun 2017
276
6
Lima, Peru
There's a problem of a shell being fired (bullet) and exploding at the top and splitting in two fragments. Can you help me with that problem?. I'm stuck on it.
 

skeeter

Math Team
Jul 2011
3,196
1,724
Texas
...

$E_i=\frac{1}{2}({\color{red}1})(5)^2+\frac{1}{2}({\color{red}2})(-10)^2 \color{red}{\ne 75}$

$E_f=\frac{1}{2}({\color{red}1})(0)^2+\frac{1}{2}({\color{red}2})(-7.5)^2 \color{red}{\ne 28.125}$

... Where did I got it wrong?.
After working it out, I have a bit of heartburn with how this problem was set up ... $v_{2f} = -7.5$ indicates $m_2 = 2 \, kg$ has a final direction to the left and $v_{1f}=0$ indicates $m_1 = 1 \, kg$ comes to a stop. Reference the diagram ... see anything that doesn't make sense?
 
Jun 2017
276
6
Lima, Peru
After working it out, I have a bit of heartburn with how this problem was set up ... $v_{2f} = -7.5$ indicates $m_2 = 2 \, kg$ has a final direction to the left and $v_{1f}=0$ indicates $m_1 = 1 \, kg$ comes to a stop. Reference the diagram ... see anything that doesn't make sense?
Can you correct my calculations for the K.E for initial and final?.
 
Jun 2017
276
6
Lima, Peru
I thought I did ... in redin red
I'm sorry I got wrong the masses assigned for each object; it should have stated the object moving at $5\,\frac{m}{s}$ to have a $m=2\,kg$ and the one moving with $-10\,\frac{m}{s}$ to have $m=1\,kg$. This should clear up things.

But I want to know whether the rationale that I used for calculating the Kinetic energies before and after the collision are right. Can you help me with that?
 
Jun 2017
276
6
Lima, Peru
@skeeter I redid the problem and I got $75\%$, perhaps can you confirm if you arrived to the same answer by inserting the given quantitites?.
 

skeeter

Math Team
Jul 2011
3,196
1,724
Texas
Disagree with 75% ... try again