Question

5. 1/4 points Previous Answers SerPSE10 15.A.P.032 My Notes Ask Your Teacher Two gliders are set in motion on an air track. Glider 1 has mass m = 0.370 kg and moves to the right with a speed 0.810 m/s. It will have a rear-end collision with glider 2, of mass m2 = 0.430 kg, which initially moves to the right with a speed 0.240 m/s. A light spring of force constant 53.0 N/m is attached to the back end of glider two as shown in the following figure. When glider 1 touches the spring, superglue instantly and permanently makes it stick to its end of the spring. (a) Find the common speed the two gliders have when the spring is at maximum compression. 0.503625 m/s (b) Find the maximum spring compression distance. 1337625 x Your response differs from the correct answer by more than 10%. Double check your calculations. cm The motion after the gliders become attached consists of a combination of (1) the constant-velocity motion of the center of mass of the two-glider system found in part (a) and (2) simple harmonic motion of the gliders relative to the center of mass. (c) Find the energy of the center-of-mass motion. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m) (d) Find the energy of the oscillation. 0. 1 x Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m) Need Help? Read

Answer 1

(a) Using conservation of momentum, (m + m2 Vmax = m; V: + m2 V2 V = my + m2v2 (m +m) (0.370) (0.81)+(0.43)(0.240) 0.37+0.43 = 0.503625 m/s (b) Using conservation energy (im, + m.) + ( to main) m.v? +km,x 2 (0.37+0.43)(0.503625) L+(0.5)(53) max = (0.5)(0.37)(0.81)* +(0.5)(0.43)(0.24) 0.101455+26.5x24 = 0.1337625 0.1337625 -0.101455 po 26.5 10.1337625-0.101455 Xmax = 26.5 = 0.0355 m =355 cm The energy is, E = mv = (0.5)(0.37 +0.43)(0.503625) = 0.10 J The energy oscillation is, E. = Ha? = (0.5)(53)(0.0355) = 0.0333 J =33.3 mJ