
The coefficient of restitution of a perfectly elastic collision is
$A)\text{ }1$
$B)\text{ 0}$
$C)\text{ }\infty $
$D)\text{ -}1$
Answer
510.3k+ views
Hint: This problem can be solved by using the definition of the coefficient of restitution of a collision in terms of the relative velocity of recede and the relative velocity of approach and put in the conditions for an elastic collision into the equation.
Formula used:
$e=\dfrac{v}{u}$
Complete step by step answer:
We will use the definition for the coefficient of restitution of a collision and put in the criteria for an elastic collision into the equation.
The coefficient of restitution $e$ of a collision between two bodies is the ratio of the relative velocity $v$ of receding of the two bodies just after the collision to the relative velocity $u$ of approach of the two bodies just before the collision.
$\therefore e=\dfrac{v}{u}$ --(1)
Now, for a perfectly elastic collision, the kinetic energy is fully conserved and there is no dissipation of the energy of the two bodies during the collision in the form of heat, sound or any permanent deformation. Therefore, the relative speed of separation of the two bodies just after the collision is equal to the relative speed of approach of the two bodies just before the collision.
Therefore, for an elastic collision,
$\therefore v=u$ --(2)
Putting (2) in (1), we get
$e=\dfrac{u}{u}=1$
Therefore, the coefficient of restitution for a perfectly elastic collision is $1$.
Hence, the correct option is $A)\text{ }1$.
Note:
Students should not think that the coefficient of restitution should be negative since the relative velocity of approach and the relative velocity of separation can be in opposite directions. This is because the direction is already taken care of in the definition for the coefficient of restitution and hence, only the magnitudes of relative velocity, that is, the relative speeds should be considered. Therefore, the coefficient of restitution can only be between $0$ (for a perfectly inelastic collision) and $1$ (for a perfectly elastic collision).
Formula used:
$e=\dfrac{v}{u}$
Complete step by step answer:
We will use the definition for the coefficient of restitution of a collision and put in the criteria for an elastic collision into the equation.
The coefficient of restitution $e$ of a collision between two bodies is the ratio of the relative velocity $v$ of receding of the two bodies just after the collision to the relative velocity $u$ of approach of the two bodies just before the collision.
$\therefore e=\dfrac{v}{u}$ --(1)
Now, for a perfectly elastic collision, the kinetic energy is fully conserved and there is no dissipation of the energy of the two bodies during the collision in the form of heat, sound or any permanent deformation. Therefore, the relative speed of separation of the two bodies just after the collision is equal to the relative speed of approach of the two bodies just before the collision.
Therefore, for an elastic collision,
$\therefore v=u$ --(2)
Putting (2) in (1), we get
$e=\dfrac{u}{u}=1$
Therefore, the coefficient of restitution for a perfectly elastic collision is $1$.
Hence, the correct option is $A)\text{ }1$.
Note:
Students should not think that the coefficient of restitution should be negative since the relative velocity of approach and the relative velocity of separation can be in opposite directions. This is because the direction is already taken care of in the definition for the coefficient of restitution and hence, only the magnitudes of relative velocity, that is, the relative speeds should be considered. Therefore, the coefficient of restitution can only be between $0$ (for a perfectly inelastic collision) and $1$ (for a perfectly elastic collision).
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