Question

A proton (mass 1.67×10⁻²⁷ kg, charge +e=+1.60×10⁻¹⁹ C) follows a path from point B to point A. If its speed at B is 3.00×10⁵ m/s, what is its speed at A?

Answer 1

The speed of the proton at point A is approximately 1.50×10^5 m/s.

Step-by-step explanation:

To solve this problem, we can use the principle of conservation of energy. At point B, the proton has a certain amount of kinetic energy, given by KE = (1/2)mv^2, where m is the mass of the proton and v is its speed. This kinetic energy remains constant as the proton moves from point B to point A because no external forces are doing work on it. Therefore, we can set up the following equation:

(1/2)mvB^2 = (1/2)mVA^2

Where vB is the speed of the proton at point B and VA is the speed of the proton at point A. Rearranging the equation gives:

VA^2 = vB^2 * (m/2m)

VA^2 = vB^2 * (1/2)

VA = √(vB^2 * (1/2))

Plugging in the values for vB and the mass of the proton, we have:

VA = √((3.00×10^5 m/s)^2 * (1/2))

VA = √(9.00×10^10 m^2/s^2 * (1/2))

VA ≈ 1.50×10^5 m/s