Question

For perfectly elastic collision, $e=$ ________ and for perfectly inelastic collision, $e=$ _____
(A) $1,0$
(B) $2,0$
(C) $4,2$
(D) $5,3$

Answer 1

Hint To find the right option out of the given four options, you first need to understand what the given quantity, $e$ is. $e$ given in the question is nothing but the coefficient of restitution. It is defined as the ratio of velocity of separation after the collision of two colliding particles to the velocity of approach before the collision of the same particles. Then you need to understand what perfectly elastic and perfectly inelastic collision mean. In a perfectly elastic collision, no loss of energy or mass happens, while in a perfectly inelastic collision, the two particles combine and act as a new third particle with combined mass.

Complete step by step answer
As explained in the hint section of the solution, perfectly elastic collision is the collision in which no loss of energy or mass happens while in perfectly inelastic collision, the two colliding particles combine and act as a new third particle with mass as the sum of the masses of the two particles.
From this, we can deduce that since there is no loss of energy or mass or momentum in a perfectly elastic collision, the velocity of separation after the collision must be exactly the same as the velocity of approach before the collision. Let us first given the expression of coefficient of restitution $\left(e\right)$ :
$e={\displaystyle \frac{v_{separation}}{v_{approach}}}$
As mentioned above, in a perfectly elastic collision:
$v_{separation}=v_{approach}$
Substituting this in the equation of coefficient of restitution, we get:
$e=1$
Hence, for a perfectly elastic collision, the value of coefficient of restitution is: $e=1$
Now, if we have a look at a perfectly inelastic collision, the two colliding particles combine and act as a new third mass with mass as the sum of both of the masses. As we can deduce, the two colliding particles get stuck to each other and do not separate, this means:
$v_{separation}=0$
Now no matter what the value of velocity of approach of the two colliding particles is, if we substitute the value of velocity of separation after the collision as zero, we are always bound to get the value of coefficient of restitution as zero:
$$\begin{gathered} e={\displaystyle \frac{0}{v_{approach}}} \\ \Rightarrow e=0 \end{gathered}$$
As we can see, the value of coefficient of restitution for a perfectly inelastic collision is $e=0$

If we check the options, only option (A) matches with the values of coefficient of restitution as we found out above, hence, the correct answer to the question is option (A).

Note A common mistake is assuming that even though no mass, energy or momentum is lost in a perfectly elastic collision, the value of velocity of separation is not bound to be the same as that of the velocity of approach. If you solve the equation of the conditions of a perfectly elastic collision, you will find that it actually is so. So, you must always remember this result to gain quick and easy marks in any exam.