Question
Two air carts of mass m1= 0.84 kg and m2=0.42 kg are placed on africtionless track. Cart 1 is at rest initially and hasa spring bumper with a force constant of 610 N/m.Cart 2 has a flat metal surface for a bumper and moves toward thebumper of the stationary cart with an initial speed of v=0.68 m/s.A) what is the speed of the two carts at the moment when theirspeeds are equal? B) how much energy is stored in the springbumper when the carts have the same speed? C) what is the finalsspeed of the carts after the collision?
Explanation / Answer
(a) momentum must be conserved, so... . initial momentum (cart 2 only) = mv = 0.42 * 0.68 = 0.2856 . final momentum (carts moving at same speed) = mv = (0.84 + 0.42) * v = 0.2856 . v = 0.226667 m/s . (b) energy is conserved, so... . initial KE of cart 2 = (1/2) mv2 = (1/2) * 0.42 * 0.682 = 0.097104 . final KE of both carts = (1/2) * (0.42 + 0.84) *0.2266672 = 0.032368 . energy stored in spring = 0.097104 -0.032368 = 0.064736 Joules . (c) energy is conserved and momentum isconserved... . momentum: 0.2856 = 0.42 *u + 0.84 * w . KE: 0.097104 = (1/2) * 0.42 * u2 + (1/2) *0.84 * w2 . 0.68 = u + 2w . 0.4624 = u2 + 2w2 . u = 0.68 - 2w u2 = 0.4624 - 2.72 w + 4w2 . 0.4624 = 0.4624 - 2.72 w + 4 w2 + 2w2 . 2.72 = 6 w . w = 0.45333 m/s final speed of block#1 . u = 0.68 - 2 * 0.45333 = -0.226667 m/s final speed of block #2 (minus indicates it bounces back opposite direction) . 2.72 = 6 w . w = 0.45333 m/s final speed of block#1 . u = 0.68 - 2 * 0.45333 = -0.226667 m/s final speed of block #2 (minus indicates it bounces back opposite direction)