Question

A proton (mass\(1.7 \times {10^{ - 27}}\;{\rm{kg}}\)) interacts electrically with a neutral\({\rm{HCl}}\)molecule located at the origin. At a certain time\(t\)the proton's position is\(\left\langle {1.6 \times {{10}^{ - 9}},0,0} \right\rangle \;{\rm{m}}\)and the proton's velocity is\(\left\langle {3600,600,0} \right\rangle \;{\rm{m}}/{\rm{s}}\). The force exerted on the proton by the HCl molecule is\(\left\langle { - 1.12 \times {{10}^{ - 11}},0,0} \right\rangle \;{\rm{N}}\). At a time\(t = 3.4 \times {10^{ - 14}}\;{\rm{s}}\), what is the approximate velocity of the proton? (You may assume that the force was approximately constant during this interval.)

Short answer

The velocity is\(\left\langle {3.8 \times {{10}^3},6 \times {{10}^2},0} \right\rangle \;{\rm{m/s}}\).

Step-by-step solution

Definition and formula for Newton’s second law

Newton's second law states that the force imposed on a body equals the time rate of change of its momentum in both magnitude and direction.

\(F = ma\).

The formula for calculating the proton's ultimate velocity, which involves

Multiplying the beginning velocity by the area and time interval.

\({v_f} = {v_i} + a\Delta t\).

Finding the value of acceleration

Substitute the values \(F = \left\langle { - 1.12 \times {{10}^{ - 11}},0,0} \right\rangle \;{\rm{N}}\) and \(m = 1.7 \times {10^{ - 27}}\;{\rm{kg}}\,\) into \(F = ma\).

\(\begin{aligned}{c}a &= \frac{F}{m}\\ &= \frac{{\left\langle { - 1.12 \times {{10}^{ - 11}},0,0} \right\rangle \;{\rm{N}}}}{{1.7 \times {{10}^{ - 27}}\;{\rm{kg}}}}\\ &= \left\langle { - 6.58 \times {{10}^{15}},0,0} \right\rangle \;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\,\end{aligned}\)

Find the nearest value of proton velocity

Substitute\(a = \left\langle { - 6.58 \times {{10}^{15}},0,0} \right\rangle \;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\), \(t = 3.4 \times {10^{ - 14}}{\rm{ sec}}\) and \({v_i} = \left\langle {3,600,0} \right\rangle \;{\rm{m/s}}\) into Newton’s first equation of motion,

\(\begin{aligned}{c}{v_f} &= {v_i} + a\Delta t\\ &= \left\langle {3,600,0} \right\rangle \;{\rm{m/s}} + \left( {\left\langle { - 6.58 \times {{10}^{15}},0,0} \right\rangle \;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {3.4 \times {{10}^{ - 14}}\;{\rm{s}}} \right)\\ &= \left\langle {3.8 \times {{10}^3},6 \times {{10}^2},0} \right\rangle \;{\rm{m/s}}\,\,\end{aligned}\)