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Homework Help: Projectile motion of protons

  1. Dec 18, 2005 #1
    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
     
  2. jcsd
  3. Dec 18, 2005 #2
    Check your calculations. I am getting ~ 5 and 85 degrees.

    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.
     
  4. Dec 18, 2005 #3
    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
     
  5. Dec 18, 2005 #4
    I used the electron mass. Sorry. You are right. I get ~37 and 53 deg.s now and t = 0.22 micro secs.
     
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