Graphical soluton of a two body head on collision
A block `A` of mass `m` moving with a uniform velocity `v_(0)` strikes another identical block `B` kept at rest on a horizontal smooth surface as shown in the figure (i). We can conserve linear momentum.

So `mv_(0)=mv_(A)mv_(B)` (`v_(A)` and `v_(B)` are the velocities of the blocks after collision)
`:. v_(0)=v_(A)+v_(B)`.........(i)
If the collision is perfectly elastic
`1/2 mv_(0)^(2)=1/2 mv_(A)^(2)+1/2 mv_(B)^(2)`
`impliesv_(0)^(2)=v_(A)^(2)+v_(B)^(2)`......(ii)
Both the above equation (i) and (ii) are plotted on `v_(A)-v_(B)` plane as shown in figure (ii). This plot can be used to find the unknowns `v_(A)` and `v_(B)`.
For example the solution of the situation in figure (i) is `v_(A)=0,v_(B)=v_(0)` (point `y` in the plot)
Because `v_(A)=v_(0), v_(B)=0` (point `x` in the plot) is not physically possible.
In a situation block `A` is moving with velocity `2m//s` an strikes another identical block `B` kept at rest. The `v_(A)-v_(B)` plot for the situation is shown. `m` and `l` are the intersection points whose `v_(A), v_(B)` coordinaes are given in the figure. The coefficient of restitution of the collision is

Options

A. `1/2`
B. `1//3`
C. `1`
D. Collision no possible

Answer

B

Explanation

`e=(4/3-2/3)/2=1/3`