Question

A proton (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.)

Answer 1

The approximate velocity of the proton at a time 3.3 × 10−14 s, considering the constant force exerted by an HCl molecule, is <3898, 1300, 0> m/s.

Step-by-step explanation:

The student is asking about the change in velocity of a proton under the influence of a constant force exerted by an HCl molecule within a very short time span. To find the velocity of the proton after 3.3 × 10−14 seconds, we use Newton's second law of motion F = ma, where F is the force, m is mass, and a is acceleration. Given the force of −1.12 × 10−11 N and the mass of the proton at 1.7 × 10−27 kg, we calculate the acceleration.

Acceleration (a) = Force (F) / Mass (m) = −1.12 × 10−11 N / 1.7 × 10−27 kg = −6.59 × 1015 m/s2.

To find the change in velocity (Δv), we multiply acceleration by the time interval: Δv = a × t = −6.59 × 1015 m/s2 × 3.3 × 10−14 s = −2.18 × 102 m/s.

The new velocity in the x-direction is the initial velocity plus the change in velocity: Vnew , x = 4100 m/s + (−2.18 × 102 m/s) = 3898 m/s.

Thus, the approximate velocity of the proton at 3.3 × 10−14 s is <3898, 1300, 0> m/s.

Answer 2

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.

Step-by-step explanation:

Full solution can be found in the attachment below.

Thank you for reading.