Protons are projected with an initial speed vi 955 mathrmkm...

Protons are projected with an initial speed $v_{i}=$ 9.55 $\mathrm{km} / \mathrm{s}$ from a field-free region through a plane and into a region where a uniform electric field $\mathbf{E}=-720 \hat{\mathrm{j}} \mathrm{N} / \mathrm{C}$ is present above the plane as shown in Figure $\mathrm{P} 23.48$ . The initial velocity vector of the protons makes an angle $\theta$ with the plane. The protons are to hit a target that lies at a horizontal distance of $R=1.27 \mathrm{mm}$ from the point where the protons cross the plane and enter the electric field. We wish to find the angle $\theta$ at which the protons must pass through the plane to strike the target. (a) What analysis model describes the horizontal motion of the protons above the plane? (b) What analysis model describes the vertical motion of the protons above the plane? (c) Argue that Equation 4.13 would be applicable to the protons in this situation. (d) Use Equation 4.13 to write an expression for $R$ in terms of $v_{i}, E,$ the charge and mass of the proton, and the angle $\theta .$ (e) Find the two possible values of the angle $\theta .$ (f) Find the time interval during which the proton is above the plane in Figure $\mathrm{P} 23.48$ for each of the two possible values of $\theta .$

Transcript

00:01 Okay, so since the horizontal direction velocity is the same at all points of chartatory, so therefore the analysis model we can use to describe the horizontal motion is the protons with constant velocity along horizontal direction.

00:16 For question b, the analysis model we can use is protons with constant acceleration.

00:24 Well, it's because there is only a electric field point along the y -axis.

00:30 Which is the vertical direction.

00:33 And for question c, while we know the x component of initial velocity, is equal to vi cosine theta, and vii, here is the y components of the initial velocity, which is equal to vi sine theta.

00:47 So we know the horizontal distance, r can be equal to vi square and sine 2 theta over g, which is equal to vi cosine cosine of times vi sine theta and then over g.

00:57 And then we'll have vxi times vii times 2 over g.

01:01 So the equation 4 .13 is applicable in this question.

01:07 For question d, well, we know horizontal distance can be equal to horizontal velocity times the total time, which is 2t, because the particle or the protons was moving from a low place to the highest place and then drop from highest place to the lowest place.

01:25 So the time of each segment is the same, and there are two such segments, which is 2t.

01:31 So therefore we'll have member t is also equal to the acceleration time, which is vertical initial velocity divided by the acceleration.

01:53 Okay? it's because when you reach the top points, the vertical velocity is zero since it stayed at the top.

02:03 So at the top, the velocity along the vertical direction is zero.

02:08 So therefore, we have t is equal to v .i.

02:10 Sine theta divide by a.

02:13 And then we can plug in back to the equation.

02:15 And we have r is equal to v .i.

02:20 Cosine theta times 2v .i.

02:26 Sine theta over 8.

02:34 And then therefore we have r is equal to 2 v .i square sine theta, cosine theta, over a.

02:49 Remember, 2 .5 .000.

02:53 C .s .0 .0 .0 .0.

02:54 0 .0 .0.

02:56 0.

02:57 Therefore, we have horizontal distance r is equal to 2 v .i.

03:10 I'm sorry, not 2 .5.

03:12 It's b .i.

03:13 Square.

03:15 Sin 2 theta over a.

03:21 And a, a.

03:24 And acceleration can be equal to force divided by the mass of protons.

03:28 Okay.

03:31 And the force in this case is the electric force, which is equal to qe and then over mp.

03:40 And we know q in this case is the charge on proton, and the charge on proton is just simply can be represented as e.

03:47 So we have ee as the electric force, and then divided by the mass of proton, we have acceleration.

03:52 So therefore, the a is equal to e over mp.

03:59 Therefore, we can plug in back to the equation.

04:03 And then we'll have horizontal distance to r is equal to vi square sine 2 theta over e e over mp.

04:27 And this will give us mp, vi square, sine 2 theta over ee.

04:49 Okay, so therefore this is the answer for question, for this question, which is question d...

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