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

Jun 2017
345
6
Lima, Peru
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
Sep 2015
2,885
1,609
USA
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.
 
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skeeter

Math Team
Jul 2011
3,276
1,769
Texas
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%
 
Jun 2017
345
6
Lima, Peru
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
Jul 2011
3,276
1,769
Texas
0.586... is sufficiently close to 59%, wouldn't you say?
 
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