Two steel blocks are sliding on a frictionless horizontal surface with the veloc

Question

Two steel blocks are sliding on a frictionless horizontal surface with the velocities shown. Knowing that the coefficient of restitu­tion between the two blocks is 0.75, determine (a) the velocity of each block after impact, (b) the loss of kinetic energy due to the impact.

Explanation / Answer

We first have to decide the cartesian coordinate axes. Let x-axis be towards right and y-axis be vertically upwards. Let Va and Vb be the velocities of thetwo bodies after collision along positive x-axis ( to avoid confusion in this type of problems we assume velocityin any direction then proceed. If our assumption is wrongvelocity will come out to be negative i.e. in the oppositedirection.) Conserving momentum along x-axis(since no external impulsive forceacts along x-axis & ground is smooth) 1.5 x 10 + 0.9 x 6 = 1.5 Va + 0.9Vb or ,20.4 = 1.5 Va + 0.9 Vb    ................... equation (1) and using coefficient of restitution (since the bodies separateafter collision ,Vb >Va) (Vb - Va )/4 = 0.75 or, Vb - Va =3 ..................... equation(2) Solving equation (1) & equation (2) we get, Vb = 10.375 ft/sec Va =7.375 ft/sec Loss of Kinetic Energy after impact =Final Kinetic Energy - InitialKinetic Energy =(.5 x 1.5 x 102 + .5 x .9 x 62 ) - (.5 x 1.5x 7.3752 + .5 x .9 x 10.3752 ) = 1.97 lb ft2 /s2

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