How Do Protons Hit a Target Under Electric Field Influence?

[an_mui]

Protons are projected with an initial speed of v0 = 9550 m/s into a region in which a uniform electric field E = 720N/C [down] is present. The protons are to hit a target that lies a horizontal distance of 1.27mm from the point at which the protons are launched.

a) find the two projection angles theta that will result in a hit
b) what is the total duration of flight for each of the two trajectories?

a) equations for projectile motion
(1) x = v0 cos (theta) t
(2) y = v0 sin (theta) t - 1/2gt^2

gelec = |field| / mass = |qfield x E| / m

x = (R, 0) where R = 1.27mm = 1.27 x 10^-3 m

solve equation 2 for t. t = 0 or t = 2v0sin theta / g (we use this one)

(3) R = 2Vo^2 sin (theta) cos (theta) / gelec
(4) sin (theta) cos (theta) = A = a number

square equation (4) : sin (theta)^2 cos (theta)^2 = A^2
sin (theta)^2 (1 - sin (theta)^2) = A^2
sin (theta)^4 - sin (theta)^2 + A^2 = 0

using quadratic equation ... Let x be sin (theta)^2

x^2 - x + A^2 = 0
1 +/- (1 - 4A^2)^1/2 <--- equation 5
---------------------
2

a) theta = inverse sin (sqrt x)
= 53 or 37 degrees.

I'd really appreciate it if someone could check my work / answer over for me. Thanks in advance.

b) t = (2vosintheta) / g
= 2.2 x 10^-7 s

 

[Gamma]

Check your calculations. I am getting ~ 5 and 85 degrees.

(3) R = 2Vo^2 sin (theta) cos (theta) / gelec

After this, you can simplyfy your calculation a lot by using the fact that,
2sin(theta)cos(theta)= sin(2theta)

So you have sin(2theta)= 2*A = a number and solve for theta.
You will get the second value for theta by knowing that sin(pi - theta)= sin (theta).

Hope this helps.

 

[an_mui]

hm . i still got 37 and 53 degrees using your method, i guess i might've calculated my A value wrong then.

this is what i got.

R = 2 Vo^2 sin (theta) cos (theta) / gelec

R = 1.27 x 10^-3
gelec = |Felec| / mass = |qfield x E| / m
= (1.6 x 10^-19)(720) / 1.67 x 10^-27
= 6.898 x 10^10
Vo = 9550

sin (theta) cos (theta) = (R x g elec) / 2Vo^2
= 0.48
= A

 

[Gamma]

I used the electron mass. Sorry. You are right. I get ~37 and 53 deg.s now and t = 0.22 micro secs.