# How to find the percentage of kinetic energy from the collision of three spheres in tandem?

#### Chemist116

The problem is as follows:

Three spheres identical in size, shape and mass are situated over a surface as indicated in the picture. Initially, $2$ and $3$ are at rest. Sphere labeled $1$ makes a collision with $2$ and as a result $2$ makes a collision with $3$. Find the percentage of kinetic energy which $3$ receives with respect from the initial. Assume that in all collisions the coefficient of restitution (COR) is $0.75$.

The alternatives are as follows:

$\begin{array}{ll} 1.&29\%\\ 2.&39\%\\ 3.&49\%\\ 4.&59\%\\ \end{array}$

For this specific situation, what I attempted was to tackle the problem using the conservation of momentum as follows:

$p_i=p_f$

$mv_1=mu_1+mu_2$

$v_1=u_1+u_2$

Since $e=0.75$

$\frac{1}{4}=\frac{u_{1}-u_{2}}{v_{2}-v_{1}}$

But from this part I end up with many equations; how exactly can I find the percentage which is being asked?

#### romsek

Math Team
You have two collisions to analyze.

In the first collision we have two equations arising from conservation of momentum, and the coefficient of restitution.

$m v_{1i}+0 = m v_{1f} + m v_{2f}$
$\dfrac{v_{2f}-v_{1f}}{v_{1i}-0} = \dfrac{3}{4}$

This can be solved for $v_{2f}$ which then becomes $v_{2i}$ for the 2nd collision

The second collision is analyzed exactly the same way but with the initial velocity for ball #2 as calculated above.

Solving this second set of equations will give you a ratio of the velocity of ball #3 to the velocity of ball #1.

This must be squared to get the ratio of kinetic energies.

Do all this and you'll find the answer is one of the choices.

topsquark

#### skeeter

Math Team
You have two collisions to analyze.

In the first collision we have two equations arising from conservation of momentum, and the coefficient of restitution.

$m v_{1i}+0 = m v_{1f} + m v_{2f}$
$\dfrac{v_{2f}-v_{1f}}{v_{1i}-0} = \dfrac{3}{4}$

This can be solved for $v_{2f}$ which then becomes $v_{2i}$ for the 2nd collision

The second collision is analyzed exactly the same way but with the initial velocity for ball #2 as calculated above.

Solving this second set of equations will give you a ratio of the velocity of ball #3 to the velocity of ball #1.

This must be squared to get the ratio of kinetic energies.

Do all this and you'll find the answer is one of the choices.
What he said $\uparrow \uparrow \uparrow$ ... I get 59%

#### Chemist116

What he said $\uparrow \uparrow \uparrow$ ... I get 59%
Since nobody did explicitly mentioned I'll do on my own:

$\frac{u_{1}-u_2}{v_2-v_1}=\frac{3}{4}$

$v_1=u_1+u_2$

Then:

$\frac{u_1-u_2}{0-v_1}=\frac{3}{4}$

$4u_1-4u_2=-3v_1$

$u_1=v_1-u_2$

$4v_1-4u_2-4u_2=-3v_1$

$u_2=\frac{7v_1}{8}$

Then for the following collision is rinse and repeat:

$u_2=v_2$

$\frac{u'_2-u_3}{0-v_2}=\frac{3}{4}$

$\frac{u'_2-u_3}{0-\frac{7v_1}{8}}=\frac{3}{4}$

$\frac{7v_1}{8}=u'_2+u_3$

$\frac{7v_1}{8}-u_3=u'_2$

$4u'_2-4u_3=-\frac{21v_1}{8}$

$4u'_2=4u_3-\frac{21v_1}{8}$

Multiplying by $4$ on both sides to: $\frac{7v_1}{8}-u_3=u'_2$

$\frac{28v_1}{8}-4u_3=4u'_2$

Inserting this value into: $4u'_2=4u_3-\frac{21v_1}{8}$

$\frac{28v_1}{8}-4u_3=4u_3-\frac{21v_1}{8}$

$\frac{49v_1}{8}=8u_3$

$u_3=\frac{49v_1}{64}$

This must be the velocity for the third sphere after the tandem collision:

Therefore the relationship in the kinetic energy to the first must be:

$\frac{K.E_3}{K.E_1}=\frac{\frac{1}{2}m\left(\frac{49v_1}{64}\right)^2}{\frac{1}{2}mv_1^2}=\frac{49^2}{64^2}$

Finally, this is approximately to:

$\frac{K.E_3}{K.E_1}\approx 0.5861816406$

which times $100$ is $58.62\%$ which isn't exactly $59\%$ but should it be considered this? Or did somebody obtained a closer answer to $59\%$?

The answers sheet mentions the answer is $59\%$. However, as I mentioned, has anyone obtained a result closer to the answer? @skeeter Did you obtained the same?

#### skeeter

Math Team
0.586... is sufficiently close to 59%, wouldn't you say?

topsquark