Aproton (mass 1.7 × 10−27 kg) interacts electrically with a neutral hcl molecule located at the origin. at a certain time , the proton's position is < 1.6 × 10−9, 0, 0> m and the proton's velocity is < 4100, 1300, 0> m/s. the force exerted on the proton by the hcl molecule is < −1.12 × 10−11, 0, 0> n. at a time 3.3 × 10−14 s what is the approximate velocity of the proton? (you may assume that the force was approximately constant during this interval.)

The approximatel velocity of the proton after the time interval t = 3.3 × 10^-14s is Vp = (3883, 1300, 0)m/s. This result was gotten on the assumption that the force of the HCl molecule on the proton was constant during the given time interval. This allows us to apply Newton's second law of motion easily which is stated mathematically as

F = ma = m(v - u)/t

Using that formula and rearranging gives the formula for v which the required variable.

Explanation:

Full solution can be found in the attachment below.

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step-by-step explanation: