if the two railroad cars couple and move off together after the collision, what is their velocity?
please show formula
2006-10-28 08:59:40 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Conservation of momentum...
9k kg * 3 m/s + 5k kg * 0 = 14k kg * X m/s
X = 1.93 m/s
2006-10-28 09:04:29 · answer #1 · answered by feanor 7 · 0⤊ 0⤋
this shows conservation of momentum
p initial=m1v1+m2v2=9000*3+5000*0=27000kgm/s
after collision they act as one body therefore
p final=(m1+m2)v=14000v
we know pi=pf
therefore
27000=14000v
v=1.93 m/s (in the forward direction)
2006-10-28 16:26:24 · answer #2 · answered by Anonymous · 0⤊ 0⤋
using the law of momentum,
m1v1 + m2v2 = (m1+m2)V
(9000*3) + 0 = (9000+5000)V
27000= 14000V
V= 27/14
V= 1.928m/s
2006-10-28 16:30:48 · answer #3 · answered by toye7690 1 · 0⤊ 0⤋