A ball of mass m moving at a speed ν makes a head-on collision with an identical ball at rest. The energy of the balls after collision is \[\dfrac{1}{4}\]th of the original value. Coefficient of restitution is
A. 1
B. \[\dfrac{1}{2}\]
C. \[\dfrac{1}{3}\]
D. Zero
Answer
599.1k+ views
Hint: In the solution, it is given that the balls are identical; it means the two balls have the same mass. So assume \[{m_1} = {m_2}\]. The energy mentioned in the question is Kinetic energy because the balls are having certain velocity, so the kinetic energy is the energy required by a body to move according to its mass.
Complete step by step answer:
Given-
The mass of the ball is m
The speed of the ball moving is v
Given that the balls are identical, then masses will be\[{m_1} = {m_2}\], as the\[{m_1},\,\,{m_2} = m\].
The energy of the ball after the collision is \[e = \dfrac{1}{4}\]of original value.
Let the velocity of the first ball is \[{v_1}\].
The velocity of the second ball is \[{v_2}\].
The equation to find the coefficient of restitution is
The final kinetic energy of the two balls is equal to the kinetic energy of the ball, then
\[\dfrac{1}{2}m{v_1}^2 + \dfrac{1}{2}m{v_2}^2 = \dfrac{1}{4} \times \dfrac{1}{2}m{v^2}\]
Substituting the values in the above equation we will get
\[\begin{array}{l}
\dfrac{1}{2}m{v_1}^2 + \dfrac{1}{2}m{v_2}^2 = \dfrac{1}{4} \times \dfrac{1}{2}m{v^2}\\
\Rightarrow {v_1}^2 + {v_2}^2 = \dfrac{1}{4}{v^2}\\
\Rightarrow \dfrac{{{{\left( {{v_1} + {v_2}} \right)}^2} + {{\left( {{v_1} - {v_2}} \right)}^2}}}{2} = \dfrac{{{v^2}}}{4}
\end{array}\]
Taking e as the coefficient of restitution, then
\[\begin{array}{l}
\dfrac{{\left( {1 + {e^2}} \right){v^2}}}{2} = \dfrac{{{v^2}}}{4}\\
1 + {e^2} = \dfrac{1}{2}\\
e = - \dfrac{1}{{\sqrt 2 }}
\end{array}\]
Therefore, the coefficient of restitution is \[e = - \dfrac{1}{{\sqrt 2 }}\], the answer is not given in the options. So no option is correct.
Note:In the solution, we may be struck at the masses of the two balls, so if we consciously focus on the question, we can observe that the balls are identical. And in the question, we may be confused to assume the type of energy, as the balls collide due to the motion so the balls will have a certain velocity, which means the energy is kinetic energy. We must be aware of the kinetic energy formula to solve this question.
Complete step by step answer:
Given-
The mass of the ball is m
The speed of the ball moving is v
Given that the balls are identical, then masses will be\[{m_1} = {m_2}\], as the\[{m_1},\,\,{m_2} = m\].
The energy of the ball after the collision is \[e = \dfrac{1}{4}\]of original value.
Let the velocity of the first ball is \[{v_1}\].
The velocity of the second ball is \[{v_2}\].
The equation to find the coefficient of restitution is
The final kinetic energy of the two balls is equal to the kinetic energy of the ball, then
\[\dfrac{1}{2}m{v_1}^2 + \dfrac{1}{2}m{v_2}^2 = \dfrac{1}{4} \times \dfrac{1}{2}m{v^2}\]
Substituting the values in the above equation we will get
\[\begin{array}{l}
\dfrac{1}{2}m{v_1}^2 + \dfrac{1}{2}m{v_2}^2 = \dfrac{1}{4} \times \dfrac{1}{2}m{v^2}\\
\Rightarrow {v_1}^2 + {v_2}^2 = \dfrac{1}{4}{v^2}\\
\Rightarrow \dfrac{{{{\left( {{v_1} + {v_2}} \right)}^2} + {{\left( {{v_1} - {v_2}} \right)}^2}}}{2} = \dfrac{{{v^2}}}{4}
\end{array}\]
Taking e as the coefficient of restitution, then
\[\begin{array}{l}
\dfrac{{\left( {1 + {e^2}} \right){v^2}}}{2} = \dfrac{{{v^2}}}{4}\\
1 + {e^2} = \dfrac{1}{2}\\
e = - \dfrac{1}{{\sqrt 2 }}
\end{array}\]
Therefore, the coefficient of restitution is \[e = - \dfrac{1}{{\sqrt 2 }}\], the answer is not given in the options. So no option is correct.
Note:In the solution, we may be struck at the masses of the two balls, so if we consciously focus on the question, we can observe that the balls are identical. And in the question, we may be confused to assume the type of energy, as the balls collide due to the motion so the balls will have a certain velocity, which means the energy is kinetic energy. We must be aware of the kinetic energy formula to solve this question.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

