Answer to Question #135427 in Mechanics | Relativity for Vanessam

Question

An electron of mass 9.11 x 10-31 kg has an initial speed of 3.00 x 105 m/s. It travels in a straight line, and its speed increases to 105 m/s in a distance of 5.00 cm. Assuming its acceleration is constant, determine the magnitude of the force exerted on the electron.

Answer 1

Maybe it's a typo in the task, because the initial velocity is greater than the final, but it is stated that the velocity increased. We may consider several possible cases.

1) Let the final velocity be exactly $10^5$ m/s, so the velocity decreased.

The acceleration $a = \dfrac{\Delta v}{\Delta t}, \; \Delta t = \dfrac{\Delta v}{a}$ , the distance travelled is $s = v_0\Delta t + \dfrac12\cdot a\cdot (\Delta t)^2 = v_0\cdot\dfrac{\Delta v}{a} + \dfrac12\cdot a\cdot \dfrac{(\Delta v)^2}{a^2} = \dfrac{v_1^2-v_0^2}{2a}.$ Therefore, the acceleration is $a = \dfrac{(\Delta v)^2}{a^2} = \dfrac{v_1^2-v_0^2}{2s}$ .

The force is $|F| = |ma| =| m\cdot \dfrac{v_1^2-v_0^2}{2s}| =\left |9.11 \cdot 10^{-31}\,\text{kg} \cdot \dfrac{(10^5\,\mathrm{m/s})^2 - (3\cdot10^5\,\mathrm{m/s})^2}{2\cdot0.05\,\mathrm{m}}\right| = 7.3\cdot10^{-19}\,\mathrm{N}.$

2) If the final velocity is $a\times 10^5$ m/s, then

$|F| = |ma| =| m\cdot \dfrac{v_1^2-v_0^2}{2s}| =\left |9.11 \cdot 10^{-31}\,\text{kg} \cdot \dfrac{(a\cdot10^5\,\mathrm{m/s})^2 - (3\cdot10^5\,\mathrm{m/s})^2}{2\cdot0.05\,\mathrm{m}}\right| =9.11\cdot10^{-20}\cdot(a^2-1)\,\mathrm{N}.$

For every value of a we may obtain the necessary value of force