A block of mass 1 kg is pushed towards another block of mass 2 kg from a 6 m distance as shown in the figure…

A block of mass $1 \mathrm{kg}$ is pushed towards another block of mass $2 \mathrm{kg}$ from a $6 \mathrm{m}$ distance as shown in the figure. If just after the collision, the velocity of $2 \mathrm{kg}$ block is $4 \mathrm{m} {\mathrm{s}}^{-1}$, then

1. Coefficient of restitution between the blocks is $1$
2. Coefficient of restitution between two blocks is $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$
3. Coefficient of restitution between two blocks is $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$
4. Coefficient of restitution between two blocks is $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

Solution

From conservation of linear momentum we can see that velocity of $1 \mathrm{kg}$ block just after collision is $2 \mathrm{m}/\mathrm{s}$ leftwards.





Now relative velocity of approach = $6 \mathrm{m}/\mathrm{s}$

and relative velocity of separation = $6 \mathrm{m}/\mathrm{s}$

$\text{e}=\frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}}=1$

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