Question
A 28000 kg railroad freight car collides with a stationary caboose car. They couple together, and 20 percent of the initial kinetic energy is dissipated as heat, sound, vibrations, and so on. What is the mass of the caboose?
Answers
Damon
initial momentum = 28000 u
final momentum = (m+28000) v
so 28000 u = m v + 28000 v
initial ke = 14000 u^2
final ke = (14000+.5m)v^2
(14000+.5m)v^2 = .8 (14000 u^2)
(14000+.5m)v^2= 11200 u^2
so I have these two equations to work with:
28000 u = m v + 28000 v
and
11200 u^2 = (14000+.5m)v^2
or
.784*10^9 u^2 =(m+28000)^2 v^2
and
11200 u^2 = (14000+.5m)v^2
or
u^2 = 1.28*10^-9 (m+28000)^2 v^2
and
u^2 = 8.93*10^-5 (14000+.5m)v^2
oh, my, v^2 cancels
1.28*10^-9 (m+28000)^2 = 8.93*10^-5 (14000+.5m)
there you go, check arithmetic
final momentum = (m+28000) v
so 28000 u = m v + 28000 v
initial ke = 14000 u^2
final ke = (14000+.5m)v^2
(14000+.5m)v^2 = .8 (14000 u^2)
(14000+.5m)v^2= 11200 u^2
so I have these two equations to work with:
28000 u = m v + 28000 v
and
11200 u^2 = (14000+.5m)v^2
or
.784*10^9 u^2 =(m+28000)^2 v^2
and
11200 u^2 = (14000+.5m)v^2
or
u^2 = 1.28*10^-9 (m+28000)^2 v^2
and
u^2 = 8.93*10^-5 (14000+.5m)v^2
oh, my, v^2 cancels
1.28*10^-9 (m+28000)^2 = 8.93*10^-5 (14000+.5m)
there you go, check arithmetic
Tully
When I solved for m I got 1711, but It was wrong. Was I close?