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
To determine the time interval over which the proton comes to rest, we can use the kinematic equation:
vf^2 = vi^2 + 2aΔx
Where:
- vf is the final velocity of the proton (which is 0 m/s when it comes to rest)
- vi is the initial velocity of the proton in the positive x-direction (unknown)
- a is the acceleration of the proton (due to the electric field, a = qE/m, where q is the charge of the proton and m is the mass of the proton)
- Δx is the distance traveled by the proton (given as 7.00 cm or 0.07 m)
First, we need to find the initial velocity vi. We know that At1 = 0, so we can write:
At1 = vi/a = vi / (qE/m)
Rearranging the equation, we have:
vi = At1 * (qE/m)
Substituting the given values, we have:
vi = (0.07 m) * (1.6 x 10^-19 C) * (6 x 10^5 N/C) / (1.67 x 10^-27 kg)
Calculating this, we find that vi ≈ 1.6 x 10^6 m/s
Now, we can use the kinematic equation to determine the time interval:
0^2 = (1.6 x 10^6)^2 + 2*(qE/m) * (0.07)
Simplifying and rearranging, we have:
0.112 = (3.85 x 10^12 C/kg) * t^2
Solving for t, we get:
t^2 ≈ 0.112 / (3.85 x 10^12 C/kg)
t ≈ √(0.112 / (3.85 x 10^12 C/kg))
Calculating this, we find that t ≈ 1.33 x 10^-8 s
Therefore, the time interval over which the proton comes to rest is approximately 1.33 x 10^-8 seconds.
To determine the time interval over which the proton comes to rest, we can use the equation of motion for the proton in an electric field. The equation is:
At^2 + Bt + C = 0
Where:
A = (1/2)m
B = -qE
C = At1^2 + (1/2)mv1^2
In this equation:
m is the mass of the proton
q is the charge of the proton
E is the electric field strength
t is the time interval
t1 is the initial time
v1 is the initial velocity
Given values:
At1 = 0 (initial time)
E = -6x10^5 N/C
t1 = 0
v1 = ?
To find v1, we can use the equation of motion for the proton in an electric field:
v1 = E*t1
Substituting the given values, we get:
v1 = -6x10^5 N/C * 0 = 0
Now, we can substitute the values into the equation of motion:
(1/2)m * t^2 - qE * t + At1^2 + (1/2)mv1^2 = 0
Since v1 = 0, the equation simplifies to:
(1/2)m * t^2 - qE * t + At1^2 = 0
Substituting the given values:
A = (1/2)m
B = -qE
C = At1^2
The equation becomes:
(1/2)m * t^2 + qE * t = 0
Now, let's solve for t by using the quadratic formula:
t = (-B ± sqrt(B^2 - 4AC)) / (2A)
Substituting the values:
A = (1/2)m
B = -qE
C = At1^2 = 0
t = (-(-qE) ± sqrt((-qE)^2 - 4 * (1/2)m * 0)) / (2 * (1/2)m)
Since C = 0, the square root term simplifies to:
sqrt((-qE)^2) = qE
The equation becomes:
t = (qE ± qE) / m
Now, substituting the values:
q = 1.6x10^-19 C (charge of a proton)
E = -6x10^5 N/C
m = 1.67x10^-27 kg (mass of a proton)
t = (1.6x10^-19 C * -6x10^5 N/C ± 1.6x10^-19 C * -6x10^5 N/C) / 1.67x10^-27 kg
This equation yields two values for t, which correspond to the two solutions of the quadratic equation. These values represent the time at which the proton comes to rest.
Evaluating this equation will give 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 equations of motion for uniformly accelerated motion. The acceleration in this case is given by the electric field, and we need to find the time it takes for the proton to come to a stop.
The equation we'll use is:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, as the proton comes to rest)
u = initial velocity (unknown)
a = acceleration (given by the electric field)
s = displacement (7.00 cm = 0.07 m)
Since the final velocity is 0, we can rearrange the equation to solve for the initial velocity:
0 = u^2 + 2as
Simplifying the equation, we have:
u^2 = -2as
Substituting the values we have:
u^2 = -2 * (6 × 10^5) * (0.07)
Now we can solve for u:
u^2 = -8.4 × 10^4
Taking the square root of both sides:
u ≈ ± 290.41 m/s
Since the proton is being projected in the positive x-direction, we take the positive value for u.
The time interval over which the proton comes to rest, t, can be calculated using the equation:
t = (v - u) / a
Substituting the known values:
t = (0 - 290.41) / (-6 × 10^5)
Calculating this, we get:
t ≈ 4.84 × 10^(-4) seconds
Therefore, the time interval over which the proton comes to rest is approximately 4.84 × 10^(-4) seconds.