Question

A moving electron accelerates at

Answer 1

Answer:

-4568.25 m/s (2 d.p.)

Explanation:

As we need to find the y-component of the initial velocity, consider the vertical and horizontal motion of the electron separately.

Trigonometry can be used to resolve the body's motion into its vertical and horizontal components:

• Horizontal component of u = u cos θ
• Vertical component of u = u sin θ

Vertical component of acceleration:

$\sf a_y=a \sin\theta = 5200 \sin(55^{\circ})\:\:\sf ms^{-2}$

Vertical component of final velocity:

$\sf v_y=v \sin \theta=6598 \sin (-20.5^{\circ})\:\:\sf ms^{-1}$

$\boxed{\begin{minipage}{9 cm}\underline{SUVAT}\\\\s = displacement in m (meters)\\u = initial velocity in ms ^{-1} (meters per second)\\v = final velocity in ms ^{-1} (meters per second)\\a = acceleration in ms ^{2} (meters per second per second)\\t = time in s (seconds)\\\end{minipage}}$

Therefore:

$\sf u=u_y\\ v=6598 \sin (-20.5^{\circ})\\a=5200 \sin(55^{\circ})\\ t=0.530$

To find the vertical component of the initial velocity (u):

$\begin{aligned}\textsf{Using }\:\:v&= u+at\\\\\implies \sf 6598 \sin (-20.5^{\circ}) & = \sf u_y+5200 \sin(55^{\circ})(0.530)\\\sf u_y & = \sf 6598 \sin (-20.5^{\circ})-5200 \sin(55^{\circ})(0.530)\\\sf \implies u_y& = \sf -4568.251336...m/s\end{aligned}$

Therefore, the y-component (vertical component) of the initial velocity is -4568.25 m/s (2 d.p.).