Question

A moving electron accelerates at

5200 m/s2 in a 55.0° direction.
After 0.530 s, it has a velocity of
6598 m/s in a -20.5° direction.
What is the y-component of the
initial velocity?

Answer 1

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

Step-by-step 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.).