A railroad car of mass 3.50 x10^ 4 kg moving at 2.00 m/s collides and couples with two coupled railroad cars, each of the same mass as the single car and moving in the same direction at 1.20 m/s.
(a) What is the speed of the three coupled cars after the collision in m/s
(b) How much kinetic energy is lost in the collision in J
2007-06-20 15:07:11 · 2 answers · asked by Idaly M 2 in Science & Mathematics ➔ Physics
a) m1V1 + 2m2V2=3m1V3
V3=( m1V1 + 2m2V2)/(3m1)
V3= ( V1 + 2V2)/(3)
V3=(2.00 + 2 (1.20))/3
V3=1.47m/s
b) K1+ K2 =0.5 m(V1^2 + 2V2^2)= Ke before
Ke before = 0.5 x 3.50 E+ 4 x (2^2 + 2(1.2)^2)=
Ke before= 120,400 Joules
K3 = (3/2)m (V3)^2=Ke after
Ke after=(3/2)3.50 E+ 4 (1.47)^2=113447 Joules
Ke lost = 120,400 - 113447= 6953 joules
2007-06-20 16:27:09 · answer #1 · answered by Edward 7 · 1⤊ 0⤋
2 cents worth right here. whilst an "X" is a component of the reporting marks, it shows a rail corporation as defined above, even with the undeniable fact that it extra identifies that corporation as possessing no locomotives.
2016-12-13 08:49:57 · answer #2 · answered by Anonymous · 0⤊ 0⤋