A proton moves at 4.50 ×10^5 m / s in the horizontal direction. It enters a uniform vertical ele

A proton moves at 4.50 ×10^5 m / s in the horizontal...

A proton moves at $4.50 \times 10^{5} \mathrm{m} / \mathrm{s}$ in the horizontal direction. It enters a uniform vertical electric field with a magnitude of $9.60 \times 10^{3} \mathrm{N} / \mathrm{C}$ . Ignoring any gravitational effects, find $(\mathrm{a})$ the time interval required for the proton to travel 5.00 $\mathrm{cm}$ horizontally, (b) its vertical displacement during the time interval in which it travels 5.00 $\mathrm{cm}$ horizontally, and $(\mathrm{c})$ the horizontal and vertical components of its velocity after it has traveled 5.00 $\mathrm{cm}$ horizontally.

Key Concepts

Projectile Motion Decomposition

Projectile motion involves analyzing the movement of an object into horizontal and vertical components that are independent of each other. In scenarios where a particle moves under no gravitational influence in one direction, the horizontal component remains constant while the vertical component is subject to uniform acceleration due to the electric force.

Kinematics Equations for Uniformly Accelerated Motion

These equations describe the motion of objects under constant acceleration. They are used to determine parameters such as displacement, velocity, and time, both for the constant speed horizontal motion and the accelerated vertical motion, thereby enabling the analysis of the particle's overall trajectory.

Electric Force on a Charged Particle

This concept involves understanding that a charged particle in an electric field experiences a force proportional to the product of its charge and the electric field strength (F = qE). The direction of this force depends on the sign of the charge, and it governs the acceleration of the particle along the field's direction.

Newton's Second Law

Newton's Second Law, expressed as F = ma, is central in linking the force acting on an object to its acceleration. When a charged particle experiences an electric force, this law helps determine the resulting acceleration, thereby allowing one to analyze the particle's motion under the influence of the field.

Transcript

00:01 Okay, so the time interval will be equal to the horizontal distance divided by the horizontal speed.

00:05 We know horizontal distance is 5 centimeter versus 0 .05 meter, and the horizontal speed is given as 4 .50 times 10 to power 5 meter per second.

00:13 Therefore, tie interval t should be equal to 0 .05 meter over 4 .50 times 10 to the power 5 meter per second.

00:25 And this will give us the time interval is about 1 .11 times 10 to the power of negative 7 second, which is equal to 111 nanosecond.

00:37 Okay? and for question b was asking us the vertical distance, the proton travel, which is the heights, which is equal to 1 .580 square, since there's no initial velocity along the vertical direction.

00:49 Okay? and while we know force, it's equal to charge times electric field, which can be also equal to mass times the acceleration.

00:55 Since the electric force is the only force in this question, so it's equal to that force.

01:01 So therefore, if we do some arrangement here, we have a acceleration is equal to qe over m.

01:06 So we can substitute qe over m for a.

01:10 Therefore, we have h is equal to one half times qe over m times t squared.

01:14 Okay? we're going to charge for the proton is equal to 1 .6 times 10 to power of negative 19 colon.

01:22 Okay? electric fuel is given as 9 .60 times 10 to power 3, newton per koolog.

01:26 And the mass for a proton is 1 .67 times 10 to power negative 27 kilograms.

01:31 And the time, we'll just figure out from the last question, which is 1 .11 times 10 to the power of the next 7 second.

01:37 So therefore, we can plug in to the equation...

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