Question

A proton is moving at 105 m/s at a point where the potential is 10 V. Later, it is at a place where the potential is 5 V. What is its speed there, assuming energy is conserved?

Answer 1

Answer:

The speed is $7.07\times10^{4}\ m/s$

Explanation:

Given that,

Speed of proton $v= 10^{5}\ m/s$

Final potential = 10 v

Initial potential = 5 V

We need to calculate the speed

Using formula of energy

$\dfrac{1}{2}mv^2=eV$

$v^2=\dfrac{2eV}{m}$

The speed of the particle is directly proportional to the potential.

$v^2\propto V$

Put the value into the formula

$(10^{5})\propto 10$....(I)

For 5 V,

$v^2\propto 5$.....(II)

From equation (I) and (II)

$\dfrac{(10^{5})^2}{v^2}=\dfrac{10}{5}$

$v=70710.67\ m/s$

$v=7.07\times10^{4}\ m/s$

Hence, The speed is $7.07\times10^{4}\ m/s$

Answer 2

The speed of the proton in the second place is 74.3 m/s.

To calculate the speed of the proton in the second place, first, we need to find the mass of the proton.

Using,

• P.E = mv²/2............ Equation 1

Where:

• P.E = potential energy of the proton
• m = mass of the proton
• v = speed of the proton.

Make m the subject of the equation

• m = 2P.E/v²............. Equation 2

Given:

• P.E = 10 V
• v = 105 m/s

Substitute these values into equation 2

• m = 2×10/(105²)
• m = 1.81×10⁻³ kg.

Finally, to calculate the speed in the second place, we make v the subject of equation 1

• v = √(2P.E/m)................. Equation 3

Given:

• P.E = 5 V
• m = 1.81×10⁻³ kg

Substitute these values into equation 3

• v = √[(2×5)/(1.81×10⁻³)]
• v = 74.3 m/s

Hence, The speed of the proton in second place is 74.3 m/s.

Learn more about speed here: https://brainly.com/question/6504879