# Physics: 2D Motion

**Problem: **Older televisions display a picture using a device called a cathode ray tube, where electrons are emitted at high speed and collide with a phosphorescent surface, causing light to be emitted. The paths of the electrons are altered by magnetic fields. Consider one such electron that is emitted with an initial velocity of 2.20 10^{7} m/s in the horizontal direction when magnetic forces deflect the electron with a vertically upward acceleration of 5.00 10^{15} m/s^{2}. The phosphorescent screen is a horizontal distance of 7.4 cm away from the point where the electron is emitted.

**A) ***How much time does the electron take to travel from the emission point to the screen?*

We’ll be using the following formula: Xf = Xi + VixT + 1/2Ax(T^2)

For this first part, we set Xf = 0.074 m and Vix = 2.20E7 m/s. We can set Ax and Xi equal to 0 as their values aren’t given. This gives us:

0.074 m = 0 + 2.20E7 m/s * T + (^{1}⁄_{2})(0)(T^2) = T (2.20E7 m/s)

Solving for T, we get T = 3.364E-9 s

**Answer: **T = 3.364E-9 s

**B) ***How far does the electron travel vertically before it hits the screen?*

For this second part, we will use the same formula, just substitute Y values in for X. By using the same time T from the previous problem, we can find the distance the particle traveled vertically before hitting the screen at Xf.

Yf = Yi + ViyT + ^{1}⁄_{2} Ay(T^2) = 0 + 0T + (^{1}⁄_{2})(5.00E15 m/s^2)(3.364E-9 s)^2 = 0.028 m -> 2.8 cm

**Answer: **Yf = 2.8 cm