Velocity of impact

Electricity and Magnetism Level 5

Two identical metallic balls of mass $m$ and radius $R$ are initially far apart. One sphere is of charge $Q$ and the the other is uncharged (neutral). After they are released due to electrostatic attraction the two spheres come together and collide. If the speed $v$ of the balls after collision can be written as:

$v=\sqrt { \frac { (\ln(a)-b){ Q }^{ c } }{ (\ln(a))mR } }$

what is the value of $a+b+c$ ?

Details and Assumptions

The collision is perfectly elastic.

Image Credit: Flickr Russ B .

The answer is 7.

1 solution

Beakal Tiliksew
Sep 23, 2015

Setting up the images for this problem can be a headache....

From conservation of energy we have

$\large2(\frac{1}{2}mv^{2})+\frac{Q^{2}}{c_{o}}=\frac{Q^{2}}{c_{\infty}}$

where $c$ is the effective capacitance where $Q_{1}=Q$ and $Q_{2}=0$ .

$c_{\infty}=R$ and $c_{0}=2Rln2$ to find this values you can read upon here and here .Thus solving for velocity we get:

$v=\sqrt { \frac { (\ln(4)-1){ Q }^{ 2 } }{ (\ln(4))mR } }$

You can also get the full solution here

Nice problem.

Mardokay Mosazghi - 5 years, 6 months ago

uhmmm you should've stated that this problem uses cgs units

Mikhael Glen Lataza - 5 years, 5 months ago

Velocity of impact

Image Credit: Flickr Russ B .

The answer is 7.

1 solution

0 pending reports