Question

-17 points 13. Tipler6 8.P.058. Ask Your Teacher My Notes A 3.2 kg block moving with a velocity of +4.7 m/s makes an elastic collision with a stationary block of mass 2.3 kg. (a) Use conservation of momentum and the fact that the relative speed of recession equals the relative speed of approach to find the velocity of each block after the collision. m/s (for the 3.2 kg block) m/s (for the 2.3 kg block) (b) Check your answer by calculating the initial and final kinetic energies of each block. J (initially for the 3.2 kg block) J (initially for the 2.3 kg block) J (finally for the 3.2 kg block) J (finally for the 2.3 kg block) Are the two total kinetic energies the same? Yes No eBook

Answer 1

I using conservation of momentum 3.2x 4.7 +0= 3.264+ 2-3 V2 3.2V, +2.342= 3.2x4.7 - and var = 4.7m/se o (2.3+3.2)/2 = 243-2X4-7_m/sec 1%= 5.47m/sec y = 5047 - 4.7 )misec = 0.77 misec 10.77 mlsee for 3.2 kg block | 5-47 misec for 2.3 kg block b) kez= *(3-2)^(4.1)? J = 35-3445 KEze=0 kiegf = *3-2)(0.71) J = 0.9495 K.54= 4x(2-3) (5.42)J = 34-4095 * Yes. Kinetic energy is same before and after collision since 35.344 a 0.949 +34.909