A point particle of mass [m] travels freely in the [x]-direction with uniform velocity [v_0]. At [x = 0 ], it enters a region between two plates oriented perpendicular to the [y-axis]; the plate spacing is [w], and the plate length in the [x]-direction is [L]. The particle enters on the mid-plane [y = 0]. While between the plates, it experiances a constant, spatially uniform force [F] in the [+y]-direction. After exiting the plates the particle again moves freely.

Ourt first task will be to obtain an expression for the [y]-coordinate of the point at which the particle exits the plates. We will assume that the plate spacing is wide enough that the particle never strikes either plate. But before we start, consider these possible solutions:

[ y = \frac{ F}{ m(v_0 + L)]

[ y = \frac{ FL}{ mv_0}]

[ y = \frac{ Fw^2}{ mv_0^2} ]

could any of them be correct? Why or why not? Remember, units and limiting behavior! In fact, from only those two considerations, you can write down the correct answer to within a factor of 2 without using any formulas at all. Want to git it a try? (This procedure is called dimensional analysis and physicists use it alll the time, especially when developing new theories.)

[\frac{ \frac{ kg * m}{ s^2}(m^2) }{kg( \frac{ m^2}{ s^2}} = m]

Therefore (3) is the answere.

Now go ahead and calculate the correct expression for [y].

[ y = \frac{ 1}{ 2} \frac{ F}{ m} \left( \frac{ L}{ v_0}\right)^2 ]

Next, Find an expression for the maximum value of the force [F] for which the particle passes through the force region without striking either plate.

[ F = mw \left( \frac{ v_0}{ L}\right) ]

For the conditions of part (c), find an expression for [ \tan \Theta ] where [ \Theta] is the angle of deflection at which the partical exits the force region. [Did you draw a sketch? Did you check your units?]

[ \tan \Theta = \frac{ F L}{ m v_0^2} = \frac{ w}{ L} ]