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
604.2k+ 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.
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