Answer:
The mass of the locomotive is 288,000 kg
Step-by-step explanation:
The question relates to the law of conservation of energy
The given parameters are;
The form of engine on the locomotive = Rocket engine
The mass of the locomotive = m
The initial velocity of the locomotive, v = 0 m/s
The duration at which the engine is fired = 20 s
The mass of the kerosene expelled, m₁ = 500 kg
The average speed with which the kerosene is expelled, v₁ = 1,200 m/s
The final speed of the locomotive engine after 20-s, v₂ = 50 m/s
Kinetic energy, K.E. = 1/2·m·v²
The total kinetic energy of the kerosene particles, K.E.
= 1/2 × 500 × 1,200² = 360,000,000
∴ K.E.
= 360,000,000 J
By the conservation of energy, the kinetic energy of the expelled kerosene particles is equal to the kinetic energy gained by the moving locomotive
Therefore, we have;
K.E.
= K.E.
K.E.
= 1/2 × m × v₂² = 1/2 × m × 50²
∴ K.E.
= 1/2 × m × 50² = K.E.
= 360,000,000 J
1/2 × m × 50² = 360,000,000 J
m = 360,000,000 J/(1/2 × (50 m/s)²) = 288,000 kg
m = 288,000 kg
The mass of the locomotive, m = 288,000 kg