A railroad car of mass 26500 kg is released from rest in a railway switchyard and rolls to the bottom of a slope Hi =8 m below its original height. At the low point, it collides with and sticks to another car of mass 20000 kg. The two cars roll together up another slope and climb up to a height Hf above the low point, where they come to a stop before rolling back. Ignore the effects of friction and calculate Hf in m.
To calculate the final height, Hf, we need to consider the conservation of mechanical energy. Since no external forces, such as friction, are acting on the system, the total mechanical energy of the system remains constant.
The total mechanical energy of the system is the sum of the potential energy (PE) and the kinetic energy (KE):
Initial mechanical energy (Ei) = PEi + KEi = m1 * g * Hi + 0
Final mechanical energy (Ef) = PEf + KEf = (m1 + m2) * g * Hf + 0
Since the two cars collide and stick together, their masses add up: m1 + m2 = 26500 kg + 20000 kg = 46500 kg
Since the two cars come to a stop at the final height Hf, the kinetic energy is zero:
KEf = 0
Using the principle of conservation of energy, we set the initial mechanical energy Ei equal to the final mechanical energy Ef:
m1 * g * Hi = (m1 + m2) * g * Hf
Simplifying the equation:
m1 * Hi = (m1 + m2) * Hf
Now we can plug in the given values:
26500 kg * 9.8 m/s^2 * 8 m = (26500 kg + 20000 kg) * Hf
Simplifying further:
2063600 N*m = 46500 kg * 9.8 m/s^2 * Hf
Solving for Hf:
Hf = 2063600 N*m / (46500 kg * 9.8 m/s^2)
Hf ≈ 4.34 m
Therefore, the final height Hf is approximately 4.34 meters.