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

#### Chemist116

Disagree with 75% ... try again
@skeeter I did it again and obtained the same answer. How you did it? Did you consider making the changes that I mentioned? I'm still stuck.

#### DarnItJimImAnEngineer

You would get 75 % loss if you took $1-\frac{final~relative~velocity^2}{initial~relative~velocity^2}$. Instead, try $\displaystyle 1-\frac{KE_{1,f}+KE_{2,f}}{KE_{1,i}+KE_{2,i}}$.

• Chemist116

#### skeeter

Math Team
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.
I didn't see this message earlier, so I reworked the problem swapping the values of the two masses ... and I still have heartburn with it.

initial momentum is (2 kg)(5 m/s) + (1 kg)(-10 m/s) = 0

therefore, final momentum is $2v_{1f} + v_{2f} = 0$

COR ... $\dfrac{v_{1f} - v_{2f}}{5 - (-10)} = \dfrac{1}{2} \implies v_{1f} - v_{2f} = 7.5$

solving the system yields $v_{1f} = 2.5$ and $v_{2f} = -5$

that says $m_1 = 2 \, kg$ ends up moving right after the collision and $m_2 = 1 \, kg$ moves left ... ???

Ignoring that "glitch" ...

$KE_f = \dfrac{1}{2}\bigg[2(2.5)^2 + 1(-5)^2] = 37.5 \, J$

$KE_i = \dfrac{1}{2}\bigg[2(5)^2 + 1(-10)^2] = 150 \, J$

$\dfrac{37.5 - 150}{150} = -0.75$ ... and you do get the 75% loss

#### skeeter

Math Team
Ok, I "dyslexically" applied the COR formula. Should be ...

$$\displaystyle \frac{v_{2f}-v_{1f}}{v_{1i}-v_{2i}} = \frac{v_{2f}-v_{1f}}{5-(-10)} = 0.5 \implies v_{2f}-v_{1f} = 7.5$$

Coupled with the final momentum equation, $2v_{1f} + v_{2f} = 0$, yields $v_{1f} = -2.5$ and $v_{2f} = 5$

... the final directions of motion for both masses now make sense.

#### Chemist116

Ok, I "dyslexically" applied the COR formula. Should be ...

$$\displaystyle \frac{v_{2f}-v_{1f}}{v_{1i}-v_{2i}} = \frac{v_{2f}-v_{1f}}{5-(-10)} = 0.5 \implies v_{2f}-v_{1f} = 7.5$$

Coupled with the final momentum equation, $2v_{1f} + v_{2f} = 0$, yields $v_{1f} = -2.5$ and $v_{2f} = 5$

... the final directions of motion for both masses now make sense.
Before the collision happens, we can tell that both objects have the same magnitude in momentum, so it is not easy to tell from that perspective which will be the final result without considering the $COR$ or $e$ for short.

At least for this problem I got it right as, like you, I was having big confusion regarding the direction, but what cleared up things was to use the COR equation as follows:

$COR=\frac{u_1-u_2}{v_2-v_1}$

Now why should this be the equation for the coefficient of restitution?. I'm assuming that it is because that's an absolute value so it has to go a minus in front of the expression to makes sense in the general case.

#### Chemist116

You would get 75 % loss if you took $1-\frac{final~relative~velocity^2}{initial~relative~velocity^2}$. Instead, try $\displaystyle 1-\frac{KE_{1,f}+KE_{2,f}}{KE_{1,i}+KE_{2,i}}$.
Doing what you mentioned would give me this:

$1-\frac{KE_{1,f}+KE_{2,f}}{KE_{1,i}+KE_{2,i}}$

$KE_{1,i}=\frac{1}{2} 2\times 5^2= 25\,J$

$KE_{2,f}=\frac{1}{2} 1\times \left(-10\right)^2=50\,J$

$KE_{1,f}= \frac{1}{2} 2\times \left(-2.5\right)^2= 6.25\,J$

$KE_{2,f}=\frac{1}{2} 1\times \left( 5\right)^2= 12.5\,J$

Therefore:

$1-\frac{KE_{1,f}+KE_{2,f}}{KE_{1,i}+KE_{2,i}}=1-\frac{6.25+12.5}{25+50}=1-\frac{18.75}{75}=0.75$

Which is exactly what I did in my notes book!. 