Question

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.

Answer 1

All of the fuel's energy is converted into the scooter's (and its riders') kinetic energy.

kinetic energy = `1/2 "Mv"^2`

Because the amount of fuel required is proportionate to the energy (and, thus, to the total mass for the same speed),

Day 1:

Total mass, M₁ = mass of man + mass of scooter = 60 + 100 = 160 kg

Energy, E₁ = `1/2 xx 160 xx v^2`

Day 2:

Total mass, M₂ = mass of man + mass of scooter + mass of son = 60 + 100 + 40 = 200 kg

Energy, E₂ = `1/2 xx 200 xx v^2`

Ratio of fuel used = `(E_1)/(E_2) = 160/200 = 4/5`

Hence, the ratio of the fuel used on the two days is 4 : 5.