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