15 points! a freight train is being assembled in a switching yard. car 1 has a mass of 65,000 kg and moves with a velocity of +0.80m/s. car 2 with a mass of 92,000 kg and has a velocity of +1.3m/s overtakes car 1 and couples to it. neglecting friction, calculate the final velocity of the two cars after they become coupled.

solve using conservation of momentum.

Pe = 1/2k(x)^2  and .05n/m = .05x10^-9m = x 

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward