Physics For Scientists And Engineers: A Strategic Approach, 4th Edition
Authors: Randall D. Knight
ISBN-13: 978-0133942651
See our solution for Question 43P from Chapter 25 from Randall Knight's Physics for Scientists and Engineers.
Problem 43P
Chapter:
Problem:
A proton's speed as it passes point A is 50,000 m/s. It follows the trajectory shown in Figure P21.62. What is the proton's speed at point B?
Step-by-Step Solution
Step 1
We will use the law of conservation of energy to solve the problem. According to the law, sum of potential energy ($U$) and kinetic energy ($K$) at any instant is constant. That means, initial total energy is equal to final total energy. In other words, any potential energy that is lost will be converted to kinetic energy.
\[\begin{array}{l}{E_T} = {\rm{constant}}\\\\{U_i} + {K_i} = {U_f} + {K_f} = {\rm{constant}}\end{array}\]
Step 2: Potential difference between point A and B
The potential difference between the points A and B \[\begin{array}{l}\Delta V = {V_B} - {V_A}\\ = - 10 - \left( {30} \right)\\ = - 40 V\end{array}\]Thus while moving from higher potential to lower potential, the proton will gain some speed.
Step 3: Change in potential Energy
When proton moves from A plate to B plate, it loses some potential energy (due to positive charge on proton). Change in potential energy is product of potential difference and charge on proton. \[\begin{array}{l}\Delta U = \left( {\Delta V} \right)\left( e \right)\\ = \left( { - 40\,\,V} \right)\left( {1.6 \times {{10}^{ - 19}}\,\,{\rm{C}}} \right)\\ = 64 \times {10^{ - 19}}\,\,{\rm{C}}{\rm{.}}V\end{array}\]
Step 4: The initial kinetic energy of proton
\[\begin{array}{*{20}{l}}{{K_i} = \dfrac{1}{2}{m_p}v_i^2}\\{ = \dfrac{1}{2}\left( {1.67 \times {{10}^{ - 27}}{\mkern 1mu} {\mkern 1mu} {\rm{kg}}} \right){{\left( {50000} \right)}^2}}\\{ = 20.875 \times {{10}^{ - 19}}{\mkern 1mu} {\mkern 1mu} {\rm{J}}}\end{array}\]
Step 4: The final kinetic energy of electron
The final kinetic energy will be sum of initial kinetic energy and change in potential energy \[\begin{array}{l}{K_f} = \left( {{U_i} - {U_f}} \right) + {K_i}\\ = 64 \times {10^{ - 19}} + 20.875 \times {10^{ - 19}}{\mkern 1mu} {\mkern 1mu} {\rm{J}}\\ = 84.875 \times {10^{ - 19}}{\mkern 1mu} {\mkern 1mu} {\rm{J}}\end{array}\]
Step 5: Final speed of electron
\[\begin{array}{*{20}{l}}{\begin{array}{*{20}{l}}{\dfrac{1}{2}m{v^2} = {K_f}}\\{ = \sqrt {\dfrac{{2{K_f}}}{m}} }\\{ = \sqrt {\dfrac{{2\left( {84.875 \times {{10}^{ - 19}}{\mkern 1mu} {\mkern 1mu} {\rm{J}}} \right)}}{{1.67 \times {{10}^{ - 27}}{\mkern 1mu} {\mkern 1mu} {\rm{kg}}}}} }\\{ = 1.0081999 \times {{10}^5}{\mkern 1mu} {\mkern 1mu} {\rm{m/s}}}\end{array}}\\{ = 1.008 \times {{10}^5}{\mkern 1mu} {\mkern 1mu} {\rm{m/s}}}\\{}\end{array}\]
ANSWER
Final Speed of electron upon reaching point B is
\[v = 1.008\times {10^5}\,\,{\rm{m/s}}\]
We will use the law of conservation of energy to solve the problem. According to the law, sum of potential energy ($U$) and kinetic energy ($K$) at any instant is constant. That means, initial total energy is equal to final total energy. In other words, any potential energy that is lost will be converted to kinetic energy.
\[\begin{array}{l}{E_T} = {\rm{constant}}\\\\{U_i} + {K_i} = {U_f} + {K_f} = {\rm{constant}}\end{array}\]
Step 2: Potential difference between point A and B
The potential difference between the points A and B \[\begin{array}{l}\Delta V = {V_B} - {V_A}\\ = - 10 - \left( {30} \right)\\ = - 40 V\end{array}\]Thus while moving from higher potential to lower potential, the proton will gain some speed.
Step 3: Change in potential Energy
When proton moves from A plate to B plate, it loses some potential energy (due to positive charge on proton). Change in potential energy is product of potential difference and charge on proton. \[\begin{array}{l}\Delta U = \left( {\Delta V} \right)\left( e \right)\\ = \left( { - 40\,\,V} \right)\left( {1.6 \times {{10}^{ - 19}}\,\,{\rm{C}}} \right)\\ = 64 \times {10^{ - 19}}\,\,{\rm{C}}{\rm{.}}V\end{array}\]
Step 4: The initial kinetic energy of proton
\[\begin{array}{*{20}{l}}{{K_i} = \dfrac{1}{2}{m_p}v_i^2}\\{ = \dfrac{1}{2}\left( {1.67 \times {{10}^{ - 27}}{\mkern 1mu} {\mkern 1mu} {\rm{kg}}} \right){{\left( {50000} \right)}^2}}\\{ = 20.875 \times {{10}^{ - 19}}{\mkern 1mu} {\mkern 1mu} {\rm{J}}}\end{array}\]
Step 4: The final kinetic energy of electron
The final kinetic energy will be sum of initial kinetic energy and change in potential energy \[\begin{array}{l}{K_f} = \left( {{U_i} - {U_f}} \right) + {K_i}\\ = 64 \times {10^{ - 19}} + 20.875 \times {10^{ - 19}}{\mkern 1mu} {\mkern 1mu} {\rm{J}}\\ = 84.875 \times {10^{ - 19}}{\mkern 1mu} {\mkern 1mu} {\rm{J}}\end{array}\]
Step 5: Final speed of electron
\[\begin{array}{*{20}{l}}{\begin{array}{*{20}{l}}{\dfrac{1}{2}m{v^2} = {K_f}}\\{ = \sqrt {\dfrac{{2{K_f}}}{m}} }\\{ = \sqrt {\dfrac{{2\left( {84.875 \times {{10}^{ - 19}}{\mkern 1mu} {\mkern 1mu} {\rm{J}}} \right)}}{{1.67 \times {{10}^{ - 27}}{\mkern 1mu} {\mkern 1mu} {\rm{kg}}}}} }\\{ = 1.0081999 \times {{10}^5}{\mkern 1mu} {\mkern 1mu} {\rm{m/s}}}\end{array}}\\{ = 1.008 \times {{10}^5}{\mkern 1mu} {\mkern 1mu} {\rm{m/s}}}\\{}\end{array}\]
ANSWER
Final Speed of electron upon reaching point B is
\[v = 1.008\times {10^5}\,\,{\rm{m/s}}\]