A man of mass 60 kg rides a scooter of mass 100 kg. He accelerates the scooter to a velocity v. The next day, his son with a mass of 40 kg joins him as a passenger. If the scooter reaches the same speed on both days in the same time interval, what is the ratio of the fuel of the tank used on the two days? Assume that the energy transfer to the scooter happens entirely due to fuel, and no other losses occur due to air resistance and friction.
4:5
Explanation
The work done by the fuel is converted entirely into the kinetic energy of the system (scooter + riders) since there are no energy losses. The kinetic energy gained is given by 1/2 . Since the final velocity 'v' is the same on both days, the kinetic energy is directly proportional to the total mass being accelerated. By calculating the total mass for Day 1 (160 kg) and Day 2 (200 kg), we find the ratio of their kinetic energies, which equals the ratio of fuel used.
Solution Steps
Total Mass on Day 1: Mass of man + Mass of scooter = 60 kg + 100 kg = 160 kg.
Total Mass on Day 2: Mass of man + Mass of son + Mass of scooter = 60 kg + 40 kg + 100 kg = 200 kg.
Kinetic Energy Relation: Kinetic Energy (E) = 1/2 × mass × (velocity)^2. Since velocity 'v' is the same on both days, E is directly proportional to mass (E ∝ m).
Ratio of Fuel Used: Fuel used is proportional to Kinetic Energy. Therefore, Ratio of fuel = E₁ / E₂ = m₁ / m₂ = 160 kg / 200 kg = 4/5. The ratio is 4:5.