(a) 1.67 m/s
For the purpose of the problem, we can consider the two coupled railroad cars has a single car, having a total mass of
[tex]m=2\cdot 3.05\cdot 10^4 kg =6.10\cdot 10^4 kg[/tex]
Initially moving at
v1 = 1.20 m/s
The first car instead has mass
[tex]m = 3.05 \cdot 10^4 kg[/tex]
and velocity
v2 = 2.60 m/s
The law of conservation of momentum states that the total initial momentum of the system must be equal to the total momentum after the collision. So we can write
[tex]m v_1 + (2m) v_2 = (m+2m) v[/tex]
where v is the velocity at which the three cars move together after the collision.
Solving the equation for v, we find:
[tex]v_1 + 2 v_2 = 3 v\\v=\frac{v_1 + 2v_2}{3}=\frac{2.60 m/s + 2(1.20 m/s)}{3}=1.67 m/s[/tex]
(b) [tex]1.9\cdot 10^4 J[/tex]
The kinetic energy is given by
[tex]K=\frac{1}{2}mv^2[/tex]
where m is the mass of the object and v is the speed.
The total kinetic energy before the collision is
[tex]K=\frac{1}{2}mv_1^2 + \frac{1}{2}(2m)v_2^2 = \frac{1}{2}(3.05\cdot 10^4 kg)(2.60 m/s)^2 + \frac{1}{2}(6.10\cdot 10^4 kg)(1.20 m/s)^2=1.47\cdot 10^5 J[/tex]
While the total kinetic energy after the collision is
[tex]K=\frac{1}{2}(3m)v^2 = \frac{1}{2}(3\cdot 3.05\cdot 10^4 kg)(1.67 m/s)^2=1.28\cdot 10^5 J[/tex]
So the kinetic energy lost is
[tex]\Delta E=1.47\cdot 10^5 J -1.28\cdot 10^5 J = 1.9\cdot 10^4 J[/tex]
