A proton has an initial speed of 3.9×105 m/s.

What potential difference is required to bring the proton to rest?

What potential difference is required to reduce the initial speed of the proton by a factor of 2?

What potential difference is required to reduce the initial kinetic energy of the proton by a factor of 2?

To answer these questions, you need to understand the concepts of electric potential energy and kinetic energy.

First, let's address the concept of potential difference. Potential difference, also known as voltage, is the amount of work needed to move a unit charge between two points in an electric field.

Now, we can calculate the potential difference required to bring the proton to rest.

1. What potential difference is required to bring the proton to rest?

When the proton comes to rest, all of its kinetic energy is converted to potential energy. Therefore, the potential difference required to bring the proton to rest is equal to its initial kinetic energy.

The kinetic energy (KE) of the proton can be calculated using the equation:

KE = (1/2)mv^2

Where:
KE is the kinetic energy,
m is the mass of the proton (approximately 1.67 × 10^-27 kg), and
v is the initial velocity of the proton (given as 3.9 × 10^5 m/s).

Plugging in the values, we can calculate the initial kinetic energy of the proton.

KE = (1/2)(1.67 × 10^-27 kg)(3.9 × 10^5 m/s)^2

Once we have the value of the kinetic energy, we know that it is equal to the potential difference (V):

V = KE

Calculating the potential difference will give you the answer.

Now, let's move on to the next question.

2. What potential difference is required to reduce the initial speed of the proton by a factor of 2?

Reducing the speed of the proton by a factor of 2 means that the final velocity (vf) will be half of the initial velocity (vi).

vf = vi/2 = (3.9 × 10^5 m/s)/2 = 1.95 × 10^5 m/s

To determine the potential difference, we need to calculate the change in kinetic energy (ΔKE). The initial kinetic energy (KEi) is given by the equation:

KEi = (1/2)mv^2

And the final kinetic energy (KEf) is given by:

KEf = (1/2)mvf^2

The change in kinetic energy (ΔKE) is obtained by subtracting the final kinetic energy from the initial:

ΔKE = KEf - KEi

Next, we can calculate the potential difference, which is equal to the change in kinetic energy:

V = ΔKE

Finally, substitute the calculated values into the equation to find the potential difference.

For the last question:

3. What potential difference is required to reduce the initial kinetic energy of the proton by a factor of 2?

Reducing the initial kinetic energy by a factor of 2 means that the final kinetic energy (KEf) will be half of the initial kinetic energy (KEi).

KEf = KEi/2

We can calculate the final kinetic energy using the equation mentioned earlier:

KEf = (1/2)mv^2

Since we need to solve for the potential difference, we can use the equation:

V = KEf

Plug in the values and calculate the potential difference.

By following these steps and using the provided values, you can find the potential difference required to bring the proton to rest, reduce its initial speed by a factor of 2, and decrease its initial kinetic energy by a factor of 2.

To find the potential difference required to bring the proton to rest, we can equate the initial kinetic energy to the final potential energy.

1. Potential difference required to bring the proton to rest:
Initially, the proton has kinetic energy given by:

KE_i = (1/2) * m * v_i^2

where m is the mass of the proton and v_i is its initial speed. Since the proton comes to rest, its final kinetic energy is zero, and the final potential energy is given by:

PE_f = q * V

where q is the charge of the proton and V is the potential difference across which it is being accelerated.

Equating the initial kinetic energy to the final potential energy, we have:

KE_i = PE_f
(1/2) * m * v_i^2 = q * V

Solving for V, we get:

V = (1/2) * (m/q) * v_i^2

Substituting with the given values:
m = mass of proton = 1.67 × 10^-27 kg,
q = charge of proton = 1.6 × 10^-19 C,
v_i = initial speed = 3.9 × 10^5 m/s,

V = (1/2) * (1.67 × 10^-27 kg / 1.6 × 10^-19 C) * (3.9 × 10^5 m/s)^2

Calculating this, we find that the potential difference required to bring the proton to rest is approximately 3.03 × 10^4 V.

To find the potential difference required to reduce the initial speed of the proton by a factor of 2:

2. Potential difference required to reduce the initial speed of the proton by a factor of 2:
Since the initial speed is being reduced by a factor of 2, the final speed of the proton is (1/2) * v_i.

Using the same equation as before:
KE_i = PE_f
(1/2) * m * v_i^2 = q * V

Now we can substitute v_f = (1/2) * v_i:

(1/2) * m * [(1/2) * v_i]^2 = q * V

Simplifying this equation, we have:

(1/2) * m * (1/4) * v_i^2 = q * V
(1/8) * m * v_i^2 = q * V

Multiplying both sides by 8:

m * v_i^2 = 8 * q * V

Dividing both sides by v_i^2:

m = 8 * q * V / v_i^2

Solving for V:

V = m * v_i^2 / (8 * q)

Substituting with the given values, we get:

V = (1.67 × 10^-27 kg) * (3.9 × 10^5 m/s)^2 / (8 * 1.6 × 10^-19 C)

Calculating this, we find that the potential difference required to reduce the initial speed of the proton by a factor of 2 is approximately 6.96 × 10^4 V.

To find the potential difference required to reduce the initial kinetic energy of the proton by a factor of 2:

3. Potential difference required to reduce the initial kinetic energy of the proton by a factor of 2:
Since the kinetic energy is proportional to the square of the velocity, if we want to reduce the initial kinetic energy by a factor of 2, we need to reduce the velocity by sqrt(2) (i.e., reduce the speed by sqrt(2)).

Using the same equation as before:
KE_i = PE_f
(1/2) * m * v_i^2 = q * V

Now we substitute v_f = sqrt(2) * v_i:

(1/2) * m * (sqrt(2) * v_i)^2 = q * V

Simplifying this equation:

(1/2) * m * 2 * v_i^2 = q * V
m * v_i^2 = q * V

Since the kinetic energy is reduced by a factor of 2, the final potential energy will also be reduced by a factor of 2. Therefore, the potential difference required will be the same as in the first question.

V = (1/2) * (1.67 × 10^-27 kg / 1.6 × 10^-19 C) * (3.9 × 10^5 m/s)^2

Calculating this, we find that the potential difference required to reduce the initial kinetic energy of the proton by a factor of 2 is approximately 3.03 × 10^4 V, the same as the answer for the first question.

Figure initial kinetic energy (1/2) m v^2

(ok because speed well below that of light)

volts * charge = energy = (1/2) m v^2

I guess you can divide v by two and v^2 by 2