A body of mass m moving with a constant velocity collides head on with another stationary object at rest of the same mass. If the coefficient of restitution between the bodies is $\dfrac{1}{2}$ then the ratio of their velocities after collision is
A. $\dfrac{1}{3}$
B. $\dfrac{1}{2}$
C. $\dfrac{1}{4}$
D. $1$
Answer
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Hint:In this question based on collision, we need to conserve momentum by taking variables for the different velocities before and after collision. According to the law of conservation of momentum, ${P_i} = {P_f}$ where ${P_i}$ is the initial momentum and ${P_f}$ is the final momentum of the system. Also, we shall apply the formula of coefficient of restitution and take the initial velocity of one of the objects to be 0 since it is at rest. The coefficient of restitution is given by $e = \dfrac{{{v_1} - {v_2}}}{v}$ . We will have two equations from which we will calculate the value of final velocities of both the objects. After calculating the velocities, we shall take their ratio to get the final answer.
Complete step by step answer:
Suppose ${v_1}\,,\,{v_2}$ be the velocities of the bodies after collision.According to the law of conservation of momentum,
${P_i} = {P_f}$
where ${P_i}$ is the initial momentum and ${P_f}$ is the final momentum of the system.
This gives us ${m_1}v + {m_2}.0 = {m_1}{v_1} + {m_2}{v_2}$
Since the masses are given to be equal and after the collision there is no change in the mass of the bodies so,
$mv = m{v_1} + m{v_2}$
Further solving this we get,
$v = {v_1} + {v_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,..............(1)$
Now the coefficient of restitution is given by $e = \dfrac{{{v_1} - {v_2}}}{v}$ (since the second was at rest before the collision).We are given that the coefficient of restitution between the bodies is $\dfrac{1}{2}$. Equating both, we get
$e = \dfrac{{{v_1} - {v_2}}}{v} = \dfrac{1}{2}$
Rearranging the terms, we get,
${v_1} - {v_2} = \dfrac{v}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,............(2)$
Now solving (1) and (2)
Using the method of elimination for solving two equations in two variables we get,
${v_1} = \dfrac{1}{4}v$ and ${v_2} = \dfrac{3}{4}v$
Now the ratio of the velocities after the collision is
$\dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{\dfrac{1}{4}}}{{\dfrac{3}{4}}}$
$ \therefore \dfrac{{{v_1}}}{{{v_2}}} = \dfrac{1}{3}$
Hence, A is the correct answer.
Note:A point worth noting here is that we took only the conservation of momentum and did not use conservation of kinetic energy. This is because the coefficient of restitution was given to be $\dfrac{1}{2}$ which means that the kinetic energy is not conserved as some of it is lost after collision. Also, momentum should be conserved in all the directions.
Complete step by step answer:
Suppose ${v_1}\,,\,{v_2}$ be the velocities of the bodies after collision.According to the law of conservation of momentum,
${P_i} = {P_f}$
where ${P_i}$ is the initial momentum and ${P_f}$ is the final momentum of the system.
This gives us ${m_1}v + {m_2}.0 = {m_1}{v_1} + {m_2}{v_2}$
Since the masses are given to be equal and after the collision there is no change in the mass of the bodies so,
$mv = m{v_1} + m{v_2}$
Further solving this we get,
$v = {v_1} + {v_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,..............(1)$
Now the coefficient of restitution is given by $e = \dfrac{{{v_1} - {v_2}}}{v}$ (since the second was at rest before the collision).We are given that the coefficient of restitution between the bodies is $\dfrac{1}{2}$. Equating both, we get
$e = \dfrac{{{v_1} - {v_2}}}{v} = \dfrac{1}{2}$
Rearranging the terms, we get,
${v_1} - {v_2} = \dfrac{v}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,............(2)$
Now solving (1) and (2)
Using the method of elimination for solving two equations in two variables we get,
${v_1} = \dfrac{1}{4}v$ and ${v_2} = \dfrac{3}{4}v$
Now the ratio of the velocities after the collision is
$\dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{\dfrac{1}{4}}}{{\dfrac{3}{4}}}$
$ \therefore \dfrac{{{v_1}}}{{{v_2}}} = \dfrac{1}{3}$
Hence, A is the correct answer.
Note:A point worth noting here is that we took only the conservation of momentum and did not use conservation of kinetic energy. This is because the coefficient of restitution was given to be $\dfrac{1}{2}$ which means that the kinetic energy is not conserved as some of it is lost after collision. Also, momentum should be conserved in all the directions.
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