Q23P - Fundamentals Of Physics

Question

An electron with an initial velocity${\mathbf{\text{V}}}_{\mathbf{0}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{ }\mathbf{m}\mathbf{/}\mathbf{s}$enters a region of length L=1.00cm where it is electrically accelerated. It emerges with$\mathbf{v}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{70}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{m}\mathbf{/}\mathbf{s}$. What is its acceleration, assumed constant?

Step-by-Step Solution

Answer

The acceleration of electron to be assumed constant is $1.62\times {10}^{15}\mathrm{m}/{\mathrm{s}}^2$ .

Step 1: Given Data

Initial velocity, ${\mathrm{v}}_0=1.50\times {10}^5\mathrm{m}/\mathrm{s}$ 

Final velocity, $\mathrm{v}=5.70\times {10}^5\mathrm{m}/\mathrm{s}$ 

Distance travelled, x=1.00 cm

Step 2: Understanding the concept of acceleration

Acceleration can be found by using the given data in the kinematic equation which involves initial velocity, final velocity, acceleration, and displacement terms. 

The formula to find acceleration can be obtained with the help of final as well as initial velocity.

$$\begin{gathered} v^2=v_0^2+2ax \\ \mathrm{Here}, \mathrm{v} \mathrm{is} \mathrm{the} \mathrm{final} \mathrm{velocity},{\mathrm{v}}_0 \mathrm{is} \mathrm{the} \mathrm{initial} \mathrm{velocity},\mathrm{a} \mathrm{is} \mathrm{the} \mathrm{acceleration} \mathrm{and} \mathrm{x} \mathrm{is} \mathrm{the} \mathrm{distance} \mathrm{travelled}. \end{gathered}$$

Step 3: Determination of the acceleration.

Convert 1.00 cm into i.e. in S.I. units.

$$\begin{gathered} 1\mathrm{cm}=\frac{1\mathrm{cm}}{100\mathrm{cm}}\times 1\mathrm{m} \\ =0.01m \\ \mathrm{Rearrange} \mathrm{the} \mathrm{equation} \left(\mathrm{i}\right) \mathrm{to} \mathrm{calculate} \mathrm{the} \mathrm{acceleration} \mathrm{as}, \\ \mathrm{a}=\frac{{\mathrm{v}}^2-{\mathrm{v}}_0^2}{2\mathrm{x}} \\ \mathrm{Substitute} \mathrm{the} \mathrm{values} \mathrm{in} \mathrm{the} \mathrm{above} \mathrm{expression}. \\ \mathrm{a}=\frac{{\left(5.7\times {10}^6\mathrm{m}/\mathrm{s}\right)}^2-{\left(1.5\times {10}^5\mathrm{m}/\mathrm{s}\right)}^2}{2\times 0.01\mathrm{m}} \\ =1.62\times {10}^{15} \mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{Therefore}, \mathrm{the} \mathrm{acceleration} \mathrm{of} \mathrm{the} \mathrm{electron} \mathrm{to} \mathrm{be} \mathrm{assumed} \mathrm{constant} \mathrm{is} 1.62\times {10}^{15} \mathrm{m}/{\mathrm{s}}^2 . \end{gathered}$$