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

#### Chemist116

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
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 $\left(\dfrac{E_f-E_i}{E_i}\right) \cdot 100$

Chemist116

#### Chemist116

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?.

#### Chemist116

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
...

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

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

... 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?

#### Chemist116

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?.

#### skeeter

Math Team
Can you correct my calculations for the K.E for initial and final?.
I thought I did ... $\color{red}\text{in red}$

#### Chemist116

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?

#### Chemist116

@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
Disagree with 75% ... try again