As shown in the figure, the coefficient of friction between the horizontal surface and each of the blocks is 0.20 . Given that the collision between the blocks is perfectly elastic, determine the separation between the two blocks when they finally come to rest. Use g = 10 m/s ^2 .

Step-by-step solution

Let the initial velocity of the 2 kg block be u = 1.0 m/s . The retardation due to friction is a = g = 0.20 10 = 2 m/s ^2 . The velocity v of the 2 kg block just before the collision can be found using the kinematic equation: v^2 = u^2 - 2as v^2 = (1.0)^2 - 2(2)(0.16) = 1 - 0.64 = 0.36 v = 0.6 m/s Let v₁' and v₂' be the velocities of the 2 kg and 4 kg blocks respectively, just after the perfectly elastic collision. Using the formulas for one-dimensional elastic collision with the second body at rest: v₁' = m₁ - m₂