Question

Lottery balls are thrown into a hopper and are set into motion by air forced through the container. each ball has a mass of 0.0044 kg. at one instant in time, ball 1 is moving upward with a speed of 1.3 m/s, and ball 2 is moving downward with a speed of 1.66 m/s. if the kinetic energy of the three-ball system at this instant is 0.0268 j, what must be the speed of ball 3?

Answer 1

In this three body system total kinetic energy remains constant:
$E_{k} = constant =0.0268 J$

$E_{k} = E_{k1}+ E_{k2}+ E_{k3}$
If we assume that upwards movement is positive and downwards movement is negative the formula is:
$E_{k} = E_{k1}- E_{k2}+ E_{k3}$

Formula for kinetic energy is:
$E_{k} = \frac{m v^{2} }{2}$

Now we have:
$E_{k} = \frac{m v_{1} ^{2} }{2} -\frac{m v_{2} ^{2} }{2} +\frac{m v_{3} ^{2} }{2} \\ E_{k} = \frac{m}{2} (v_{1} ^{2}-v_{2} ^{2}+v_{3} ^{2}) \\ \\ 0.0268= \frac{0.0044}{2} (1.3 ^{2}-1.66^{2}+v_{3} ^{2}) \\ \\ 0.0536=0.0022(1.69-2.7556+v_{3} ^{2}) \\ \\ v_{3} ^{2}=25.43 \\ \\ v_{3} =+-5.04$

We got two solutions: positive and negative. The speed is 5.04 m/s and direction can be upwards and downwards.