Consider the same situation as in Problem 54, but with a...

Consider the same situation as in Problem $54,$ but with a proton entering through the small hole at the same angle with a speed of $v=8.50 \times 10^{5} \mathrm{m} / \mathrm{s}$ (a) Can you ignore the force of gravity when solving this problem for the horizontal distance traveled by the proton? Why or why not? (b) How far will the proton travel, $\Delta x$ before it hits the metal plate?

Consider the same situation as in Problem $54,$ but with a proton entering through the small hole at the same angle with a speed of $v=8.50 \times 10^{5} \mathrm{m} / \mathrm{s}$
(a) Can you ignore the force of gravity when solving this problem for the horizontal distance traveled by the proton? Why or why not?
(b) How far will the proton travel, $\Delta x$ before it hits the metal plate?

Key Concepts

Projectile Motion Kinematics

Projectile motion kinematics is a fundamental concept where an object’s horizontal displacement is determined by its constant horizontal velocity and the time of flight, assuming that forces in the horizontal direction are absent or negligible. When vertical forces are either counteracted or occur over a very short timescale, the separation of horizontal and vertical dynamics allows one to calculate the horizontal distance traveled without accounting for the minute vertical effects, such as those introduced by gravity.

Negligibility of Gravitational Force

In many physics problems involving high-speed particles or situations where other forces dominate (such as magnetic or electric forces), the gravitational force can often be neglected because its magnitude is orders of magnitude smaller compared to the other forces acting on the body. This concept involves comparing the acceleration due to gravity with other accelerations in the system to justify simplifying the motion analysis by dropping gravity from the equations.

Transcript

00:01 Hi, in the given problem there is a sheet, a horizontal sheet which is having a hole in it.

00:10 A proton comes out of this hole moving in upward direction at an angle of 55 degree with the horizontal.

00:23 There is an electric field in a downward direction, vertically downward direction.

00:29 So here it should be positive upward, there should be negative.

00:33 Downward.

00:36 As this is the proton, so its mass is 1 .67 into 10 to the power minus 27 kilogram.

00:45 The charge carried by it is 1 .6 into 10 to 10 to minus 19 cullum.

00:51 It is coming with a speed v which is given as 8 .50 into 10 dash to the power 5 meter per second and the magnitude of electric field is 6 .50 into 10 to the power 3 newton per column.

01:17 Now in the first part of the problem we have to discuss whether the gravitational force acting on the proton should be considered or not if we have to find the total horizontal distance traveled by this proton before it hits the plate again.

01:37 So now first of all we will find the magnitude of these two forces for fg this is m into g means 1 .67 into 10 dash bar minus 27 into 9 .8 and this force here comes out to be 1 .64 into 10 dash the bar minus 26 newton and as far as electrostatic force is concerned that is given by eee means this is 1 .6 into 10 dash per minus 19 multiplied by electric field 6 .50 into 10 dash par 3.

02:11 So this electrostatic force comes out to be 1 .04 into 10 dash minus 15 newton means this gravitational force is negligibly small as compared with the electrostatic force.

02:25 Hence we may ignore gravitational force here in this problem which is the answer for the first part of the problem now in the second part we have to find the distance horizontal distance delta x traveled by the proton for which first of all we will find its vertical acceleration so using this force f e which is given as e into e then we can compare it with the using newton's second of motion m into so its acceleration will come out to be e e by m.

03:12 So now plugging in all known values here, this is 1 .6 into 10 dash par minus 19 into 6 .50 into 10 dash par 3.

03:22 And for mass, this is 1 .67 into 10 dash per minus 27.

03:27 So finally this acceleration comes out to be 6 .2 into 10 dash par 11 meter per second is now this velocity may be resolved into two components, the horizontal component, vx and vertical component, vy, and here it will be represented as vy, the initial vertical velocity.

03:53 So this vyi will be given as v into sign...

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