Question

A4.5-kg block moving at 2.0 m/s west on a frictionless surface collides totally inelastically with a second 1.0-kg block traveling east at 3.0m/s. Part A Determine the final velocity of the blocks Express your answer with the appropriate units. op Value Units Submit Request Answer Part B Determine the kinetic energy of the first block before th Express your answer with the appropriate units.

Answer 1

The system is the two blocks and the surface. The initial state of the system is the instant before they collide and the final state is the instant after they collide totally inelastically. Let the eastward direction be the positive direction. Apply the principle of momentum conservation from Equation (1) to the system of the two blocks and the surface between the described initial and final states: mi(-v1) + m202 = (mi + m2) uf m202 – m1v1 = (mi + m2) uf where us is the final velocity of the two blocks. Solve for uf: m202 – m1v1 mi + m2 Substitute numerical values from given data: (1 kg)(3 m/s) – (4.5kg) (2 m/s) 1 kg +4.5kg = - 1.09 m/s The negative sign indicates that the final speed of the two blocks is directed westward.

The kinetic energy of the first block before and after the collision is found from Equation (2): Ki; = 5 mxv1 = }(4.5kg)(2 m/s)2 = 9 J Kus = zmu} = f(4.5 kg) 1.1m/s)2 =2.73 The kinetic energy of the second block before and after the collision is found from Equation (2): K2; = 5 muž = 3(1 kg)(3m/s)2 = 4.5 J Kas = 15m2v; = (1 kg)(1.Im/s)?= 0.61