Answer:
The system would be moving at [tex]17.5\; \rm m \cdot s^{-1}[/tex].
The kinetic energy of this system would be [tex]612500\; \rm J \![/tex] after the collision.
[tex]612500\; \rm J[/tex] (same amount) of kinetic energy would be lost.
Explanation:
The momentum of an object is the product of its mass [tex]m[/tex] and its velocity [tex]v[/tex]. That is: [tex]p = m \cdot v[/tex].
Assume that external forces (e.g., friction) have no effect on this system. The total momentum of this system would stay the same before and after the collision.
Initial momentum of this system:
Hence, the momentum of this system would be [tex]70000\; \rm kg \cdot m \cdot s^{-1}[/tex] before the collision.
Under the assumptions, the collision would not change the momentum of this system. Hence, the momentum of this system would continue to be [tex]70000\; \rm kg \cdot m \cdot s^{-1}[/tex] after the collision.
However, with two identical trains stuck to each other, the mass of this system would be twice that of just one train: [tex]m = 2 \times 2000\; \rm kg[/tex].
Calculate the new velocity of this system:
[tex]\begin{aligned} v &= \frac{p}{m}\\ &= \frac{70000\; \rm kg \cdot m \cdot s^{-1}}{2 \times 2000\; \rm kg} = 17.5\; \rm m\cdot s^{-1}\end{aligned}[/tex].
Calculate the kinetic energy of this system before and after the collision.
Before the collision:
[tex]\begin{aligned}& \text{KE(before)} \\ =\; & \text{KE(moving train)} + \text{KE(stationary train)}\\ =\; & \frac{1}{2} \, m(\text{one train}) \cdot (v(\text{moving train}))^{2} + 0 \\ = \; &\frac{1}{2} \times 2000 \times (35\; \rm m\cdot s^{-1})^{2} \\ = \; & 1225000\; \rm J \end{aligned}[/tex].
After the collision:
[tex]\begin{aligned}& \text{KE(after)} \\ =\; & \frac{1}{2} \, m(\text{two trains}) \cdot v^{2} \\ = \; &\frac{1}{2} \times (2\times 2000\; \rm kg) \times (17.5\; \rm m\cdot s^{-1})^{2} \\ = \; & 612500\; \rm J \end{aligned}[/tex].
Change to the kinetic energy of this system:
[tex]1225000\; \rm J - 612500\; \rm J = 612500\; \rm J[/tex].
