Question

5) A 200 gram box slides along a frictionless table at 2m/s. It hits a stationary 300 gram box. The two boxes stick together and continue to travel. What is the velocity of the combined boxes and how much energy was lost in the collision ?

Answer 1

m - 200 g - U = initial velocity 0.2 kg of n = 2mls m2 = 300g 42 = initial - 0.3 kg velocity of my - 0 m/s. v e velocity of system (mitm) final together often sticking often sticking together. → => 1984 Alan PRU121211212LLLLLLLLLLLLLL Conserving momentum before8 m, u + m2U2 = (m, + m2) v (0.2 x 2) + 0 = (0.2 +0.3) v V = 0.2 x 2___ mis 0.2 +0.3) 0.2 x 2 m/s - 0.5 Velocity of_combined boxes = 0.8 más - K.E. sticking KE; of - m, 8 m2 before Muiz+ m2 43? --( 0.2222) + (x 0.3x03) - 0.4 I

R.E. m, sticking KE. of - 8 m₂ after 1 (m. + m2) v2 _ = 1 (0.2 + 0.3) x 0.82_J - 0.16 J Energy__ Jost__in_ Fone collision =_ K.E. i - K.E.F = J 0.4 -0.16 0.24. Il
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