91Ó°ÊÓ

Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause this motion. In our problem, the electron's motion horizontally and vertically is described using kinematic principles. To determine the time it takes for the electron to move a given horizontal distance, we use the basic formula for speed: \ \ \ \[ t = \frac{d}{v} \] Here, \( t \) is the time, \( d \) is the distance covered, and \( v \) is the speed. Given that the electron moves horizontally with a constant speed of \( 1.2 \times 10^{7} \text{ m/s} \) over a distance of \( 30 \text{ mm} = 30 \times 10^{-3} \text{ m} \), we substitute these values into the formula to find the time \( t \). This gives us: \ \ \ \[ t = \frac{30 \times 10^{-3} \text{ m}}{1.2 \times 10^{7} \text{ m/s}} = 2.5 \times 10^{-9} \text{ s} \] Understanding kinematics helps us lay the foundation for solving motion problems before diving into the forces causing the motion.

Newton's second law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration, expressed as: \ \ \ \[ F = ma \] In this problem, the vertical motion of the electron is influenced by a constant vertical force of \( 4.5 \times 10^{-16} \text{ N} \). To find the vertical acceleration of the electron, we rearrange the above formula to solve for acceleration \( a \): \ \ \ \[ a = \frac{F}{m} \] Here, \( F \) is the force and \( m \) is the mass of the electron \( 9.11 \times 10^{-31} \text{ kg} \). Substituting these values into the equation provides: \ \ \ \[ a = \frac{4.5 \times 10^{-16} \text{ N}}{9.11 \times 10^{-31} \text{ kg}} \ a \rightarrow 4.94 \times 10^{14} \text{ m/s}^2 \] This significant acceleration demonstrates Newton's second law, showing that a small force can cause a large acceleration on a particle with a very small mass like the electron.

Constant acceleration implies that an object's velocity changes by the same amount every second. In our problem, after finding the vertical acceleration of the electron, we can compute the vertical distance \( y \) it travels while under this constant acceleration. Using the kinematic equation that relates distance, acceleration, and time: \ \ \ \[ y = \frac{1}{2}at^2 \] Where: \ \ \( y \) is the vertical distance, \( a \) is the constant acceleration \( 4.94 \times 10^{14} \text{ m/s}^2 \), and \( t \) is the time interval \( 2.5 \times 10^{-9} \text{ s} \). Substituting these values, we get: \ \ \ \[ y = \frac{1}{2} \times 4.94 \times 10^{14} \times (2.5 \times 10^{-9})^2 \text{ m} \ \rightarrow y = \frac{1}{2} \times 4.94 \times 10^{14} \times 6.25 \times 10^{-18} \ y \rightarrow 1.54 \times 10^{-3} \text{ m} \] Thus, under constant vertical acceleration, the electron is deflected by approximately \( 1.54 \text{ mm} \) after moving \( 30 \text{ mm} \) horizontally. Understanding the concept of constant acceleration helps in predicting motion behavior accurately over time.