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A proton is projected into a uniform electric field that points vertically upward and has magnitude

e. the initial velocity of the proton has a magnitude v0 and is directed at an angle a below the horizontal. (a) find the maximum distance hmax that the proton descends vertically below its initial elevation. you can ignore gravitational forces. (b) after what horizontal distance d does the proton return to its original elevation? (c) sketch the trajectory of the proton. (d) find the numerical values of hmax and dife = 500 n>c,v0 = 4.00 * 105 m>s,anda = 30.0°.

Respuesta :

Part a)

here the proton is projected at some angle with the horizontal plate

so the two components of velocity is given as

[tex]v_x = v_0 cos\theta[/tex]

[tex]v_y = v_0 sin\theta[/tex]

now we will have acceleration due to electric force

[tex]a = \frac{eE}{m}[/tex]

now to find the maximum height we can use

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]0 - (v_0sin\theta)^2 = 2(-eE/m)h_{max}[/tex]

here we will now find the maximum height from equation above

[tex]h_{max} = \frac{mv_0^2sin^2\theta}{2eE}[/tex]

Part b)

total horizontal distance moved

[tex]d = v_x \times time[/tex]

now to find the time of whole motion we will have

[tex]v_f - v_i = at[/tex]

[tex]0 - v_0sin\theta = -\frac{eE}{m} \times \frac{T}{2}[/tex]

[tex]T = \frac{2mv_0sin\theta}{eE}[/tex]

now from above equation we have

[tex]d = (v_0 cos\theta)(\frac{2mv_0sin\theta}{eE})[/tex]

[tex]d = \frac{mv_0^2 sin2\theta}{eE}[/tex]

Part c)

now if the given values are

[tex]E = 500 N/C[/tex]

[tex]v_0 = 4.00 \times 10^5 m/s[/tex]

[tex]\theta = 30^0[/tex]

now we have

[tex]h_{max} = \frac{(1.67 \times 10^{-27})(2.00\times 10^5)^2sin^230}{2(1.6\times 10^{-19})(500)}[/tex]

[tex]h_{max} = 0.104 m[/tex]

[tex]d = \frac{(1.67\times 10^{-27})(2\times 10^5)^2sin60}{(1.6\times 10^{-19})500}[/tex]

[tex]d = 0.723 m[/tex]

Ver imagen aristocles

The maximum distance h_max that the proton descends vertically below its initial elevation is; h_max = m(v₀ sin θ)²/2eE

What are the velocity components?

A) The vertical and horizontal components of the initial velocity are;

v_y = v₀ sin θ

v_x = v₀ cos θ

From newton's second equation of motion, we know that;

v² - u² = 2ah

Since we are dealing with vertical motion of the projectile, then;

v = 0 and u = v₀ sin θ

Thus;

0² - (v₀ sin θ)² = 2ah

making h the subject gives;

h = -(v₀ sin θ)²/2a

where;

a = -eE/m

Thus;

h = -(v₀ sin θ)²/2a

h_max = m(v₀ sin θ)²/2eE

B) From newton's first equation of motion, we know that;

v - u = at

where;

u = v₀ sin θ

a = -eE/m

v = 0

t = T/2

Thus;

T = 2m(v₀ sin θ)/eE

Total horizontal distance is;

d = v_x * T

d = v₀ cos θ * 2m(v₀ sin θ)/eE

d = (mv₀² sin 2θ)/eE

Read more about Velocity Components at; https://brainly.com/question/14715903

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