To determine the initial speed of the proton, we can use the equation:
vf^2 = vi^2 + 2ad,
where:
- vf represents the final velocity, which is zero since the proton comes to rest,
- vi represents the initial velocity (what we want to find),
- a represents the acceleration of the proton,
- d represents the distance traveled by the proton.
First, let's find the acceleration of the proton using the electric field.
The force experienced by the proton due to the electric field is given by:
F = qE,
where:
- F represents the force experienced by the proton,
- q represents the charge of the proton,
- E represents the electric field strength.
The motion of the proton can be considered 1D, so the force can be related to the acceleration using Newton's second law:
F = ma,
where:
- m represents the mass of the proton,
- a represents the acceleration.
Since the problem doesn't provide the mass of the proton, we'll use the known value: m = 1.67 × 10^(-27) kg
Combining the equations for force, acceleration, and electric field:
qE = ma.
Solving for 'a':
a = qE / m.
The charge of a proton is given by q = 1.6 × 10^(-19) C.
Substituting the values:
a = (1.6 × 10^(-19) C)(-6 × 10^5 N/C) / (1.67 × 10^(-27) kg).
Now, let's calculate 'a':
a ≈ -5.76 × 10^13 m/s².
Now that we have the acceleration, we can use the motion equation to find the initial velocity.
vf^2 = vi^2 + 2ad.
Since vf is zero:
0 = vi^2 + 2ad.
Rearranging the equation to solve for 'vi':
vi^2 = -2ad.
Taking the square root of both sides:
vi = √(-2ad).
Substituting the values:
vi = √(-2)(-5.76 × 10^13 m/s²)(0.07 m).
Solving for 'vi':
vi ≈ √(8.064 × 10^12 m²/s²).
Simplifying:
vi ≈ 2.84 × 10^6 m/s.
Therefore, the initial speed of the proton is approximately 2.84 × 10^6 m/s.
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 Its initial speed,
in simple steps
1 answer