At1=0, a proton is projected in the positive x-direction into a region of a uniform electric field of E=-6x1051. The proton travels 7.00 cm as it comes to rest. Determine The time interval over which the proton comes to rest
in the simplest steps
To determine the time interval over which the proton comes to rest, we can use the equation for the acceleration of a charged particle in an electric field:
a = qE / m
where a is the acceleration, q is the charge of the proton, E is the electric field, and m is the mass of the proton.
First, we need to determine the acceleration of the proton. The charge of a proton is q = 1.602 x 10^-19 C, and the mass of a proton is m = 1.6726 x 10^-27 kg. Plugging these values into the equation, we get:
a = (1.602 x 10^-19 C)(-6 x 10^5 N/C) / (1.6726 x 10^-27 kg)
≈ -5.72 x 10^16 m/s^2 (negative because the acceleration is in the opposite direction of the electric field)
Next, we need to determine the initial velocity of the proton. We are given that At1 = 0, which means the proton starts from rest. Therefore, the initial velocity is v0 = 0 m/s.
We can use the kinematic equation to find the time interval, t, over which the proton comes to rest:
v = v0 + at
Since v0 = 0 m/s and a = -5.72 x 10^16 m/s^2, the equation becomes:
v = -5.72 x 10^16 m/s^2 * t
We are given that the proton travels a distance of 7.00 cm = 0.07 m. We can use another kinematic equation to relate the acceleration, initial velocity, time, and displacement:
s = v0t + 0.5at^2
Since v0 = 0 m/s and s = 0.07 m, the equation becomes:
0.07 m = 0 + 0.5(-5.72 x 10^16 m/s^2)t^2
Simplifying, we get:
0.07 m = -2.86 x 10^16 m/s^2 * t^2
Dividing both sides by -2.86 x 10^16 m/s^2, we get:
t^2 = 0.07 m / -2.86 x 10^16 m/s^2
≈ -2.44 x 10^-19 s^2
Taking the square root of both sides, we get:
t = √(-2.44 x 10^-19 s^2)
≈ ± 4.94 x 10^-10 s
Since time cannot be negative, we only consider the positive value as the time interval over which the proton comes to rest:
t ≈ 4.94 x 10^-10 s
Therefore, the time interval over which the proton comes to rest is approximately 4.94 x 10^-10 seconds.
To determine the time interval over which the proton comes to rest, we can use the equation of motion:
v = v0 + at
where:
v = final velocity (0 m/s, since the proton comes to rest)
v0 = initial velocity (unknown)
a = acceleration (from the electric field)
t = time interval (unknown)
We can rearrange the equation to solve for t:
t = (v - v0) / a
Since the proton comes to rest, v = 0. Thus, the equation becomes:
t = -v0 / a
Now, let's find v0.
We know that the initial velocity (v0) is given by:
v0 = at1 + v1
where:
a = acceleration (from the electric field)
t1 = time interval (unknown)
v1 = initial velocity at t1 (unknown)
From the given information, at t = t1, the proton is projected in the positive x-direction, so the only force acting on it is the electric field force, given by:
F = qE
where:
F = force (unknown)
q = charge of proton (1.60 x 10^-19 C)
E = electric field (-6 x 10^51 N/C)
From Newton's second law (F = ma), we know that:
ma = qE
Simplifying, we find:
a = (qE) / m
where:
m = mass of proton (1.67 x 10^-27 kg)
Now, substitute the values into the equation to find a:
a = (1.60 x 10^-19 C * -6 x 10^51 N/C) / (1.67 x 10^-27 kg)
We can calculate the acceleration (a) using the given values. After calculating a, we can substitute it along with v = 0 into the equation t = -v0 / a to find the time interval over which the proton comes to rest.
To determine the time interval over which the proton comes to rest, we can use the equation of motion for a particle under constant acceleration:
v^2 = u^2 + 2aΔx
where v is the final velocity, u is the initial velocity, a is the acceleration, and Δx is the displacement.
In this case, the proton starts with an initial velocity of At1=0, meaning u = 0. The acceleration can be calculated using the equation F = qE, where F is the force experienced by the proton, q is the charge of the proton (e = 1.6 × 10^-19 C), and E is the electric field strength.
Since F = ma, we have:
qE = ma
Rearranging the equation, we get:
a = qE/m
Substituting the known values, q = 1.6 × 10^-19 C, E = -6 × 10^5 N/C (note the negative sign represents the direction of the electric field), and m = 1.67 × 10^-27 kg (mass of the proton), we can calculate the acceleration:
a = (1.6 × 10^-19 C)(-6 × 10^5 N/C) / (1.67 × 10^-27 kg)
Now we can substitute the values of u = 0, a, and Δx = 7.00 cm = 0.07 m into the equation:
v^2 = u^2 + 2aΔx
(0)^2 = (0)^2 + 2a(0.07)
Simplifying the equation, we find:
0 = 0 + 2a(0.07)
0 = 0.14a
Since the final velocity (v) is zero when the proton comes to rest, the term v^2 in the equation becomes zero. Therefore, we only have to solve for time (t), which is given by:
t = (v - u) / a
Plugging in the known values, we get:
t = (0 - 0) / a
t = 0
Hence, the time interval over which the proton comes to rest is 0 seconds.