While stuck at a railroad crossing, you notice a full train car (m1 = 9000kg) is

Question

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*

Explanation / Answer

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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