Two smooth objects with a coefficient of restitution u, collide directly and bounce as shown. Newton’s law of restitution gives
Just before impact
Just after impact
$
{\text{A}}{\text{. }}e \times 4u = {v_2} + {v_1} \\
{\text{B}}{\text{. }}e \times 2u = {v_1} - {v_2} \\
{\text{C}}{\text{. }}e \times 2u = {v_2} - {v_1} \\
{\text{D}}{\text{. It cannot be applied as the mass are not known}} \\
$
Answer
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Hint: The coefficient of restitution is defined as the ratio of the relative velocities of the objects after collision and the relative velocities of the objects before the collision. Then depending on given values and directions of the objects, we can obtain the required condition for the coefficient of restitution.
Complete answer:
We are given two smooth objects which collide with each other and bounce off. Before collision takes place between them, the initial velocity of first object is given as
${u_1} = 3u$
The initial velocity of the second object is given as
${u_2} = u$
After the collision has taken place, the final velocity of the first object is ${v_1}$ while the final velocity of the second object is ${v_2}$.
We are also given that the coefficient of restitution for this collision is u. The definition of coefficient of restitution is that it is the ratio of relative velocities after collision and the relative velocities before the collision.
The relative velocity before collision is ${u_1} - {u_2} = 3u - u = 2u$.
The relative velocity in the final state is ${v_1} - {v_2}$.
Now we can write the coefficient of restitution in the following way.
$
e = \dfrac{{{v_1} - {v_2}}}{{2u}} \\
\Rightarrow e \times 2u = {v_1} - {v_2} \\
$
This is the required relation.
Hence the correct answer is option B.
Note:
It should be noted that the two objects move in the same direction both before and after the collision. When two objects are moving in the same direction then their relative velocities are equal to the difference between their velocities. In this case, the smaller velocity gets subtracted from the larger velocity. Before collision the first object moves with larger velocity while the second object moves with smaller velocity.
Complete answer:
We are given two smooth objects which collide with each other and bounce off. Before collision takes place between them, the initial velocity of first object is given as
${u_1} = 3u$
The initial velocity of the second object is given as
${u_2} = u$
After the collision has taken place, the final velocity of the first object is ${v_1}$ while the final velocity of the second object is ${v_2}$.
We are also given that the coefficient of restitution for this collision is u. The definition of coefficient of restitution is that it is the ratio of relative velocities after collision and the relative velocities before the collision.
The relative velocity before collision is ${u_1} - {u_2} = 3u - u = 2u$.
The relative velocity in the final state is ${v_1} - {v_2}$.
Now we can write the coefficient of restitution in the following way.
$
e = \dfrac{{{v_1} - {v_2}}}{{2u}} \\
\Rightarrow e \times 2u = {v_1} - {v_2} \\
$
This is the required relation.
Hence the correct answer is option B.
Note:
It should be noted that the two objects move in the same direction both before and after the collision. When two objects are moving in the same direction then their relative velocities are equal to the difference between their velocities. In this case, the smaller velocity gets subtracted from the larger velocity. Before collision the first object moves with larger velocity while the second object moves with smaller velocity.
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