While stuck at a railroad crossing, you notice a full train car (m1 = 9000kg) is moving to your left with a speed of 9m/s. Two seperated empty train cars (m2 & m3 = 3000kg each) are on the same track moving to the right, each with a speed of 3m/s. There are two succesive collisions. First m1 and m2 hit and stick together. That pair (m1 & m2) then runs into m3. In this second collision, the air bag safety system on m3 blows up an air bag in a poof of carbon dioxide gas. The car m3 then bounces off and goes to your left with a final speed of 5m/s.
a. Determine the velocity of m1 & m2, stuck together, after the first collision.
b. Show that the speed of the coupled cars (m1 & m2) after the second collision is 4m/s to the left.
c. Find the initial energy (Eo) before any collision, and the final energy (Ef) after the second collison. *need given info and data from part B*
let velocity to the right be positive and velocity to the left be negative.
initial velocities are:
m1: -9 m/s
m2: 3 m/s
m3: 3 m/s
part a:
as m1 and m2 stuck each other, it is an inelastic collision.
let speed of combined masses is v m/s.
using conservation of momentum principle:
momentum before collision=momentum after collision
==>m1*(-9)+m2*3=(m1+m2)*v
==>v=(m1*(-9)+m2*3)/(m1+m2)=-6 m/s
i.e. 6 m/s to the left.
part b:
conserving momentum for 2nd collision:
let speed of coupled cars be v1.
(m1+m2)*v+m3*3=(m1+m2)*v1+m3*(-5)
==>v1=((m1+m2)*(-6)+m3*3+m3*5)/(m1+m2)=-4 m/s
which is 4 m/s to the left. (proved.)
part c:
kinetic energy=0.5*mass*speed^2
hence,
initial energy before collision=0.5*m1*9^2+0.5*m2*3^2+0.5*m3*3^2
=391500 J
final energy after the second collision=0.5*(m1+m2)*v1^2+0.5*m3*5^2
=133500 J
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