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.

Given: Mass of scooter = 100 kg

Mass of man = 60 kg

Mass of son = 40 kg

Total mass on first day:

$$m_1 = 100 + 60 = 160 \text{ kg}$$

Total mass on second day:

$$m_2 = 100 + 60 + 40 = 200 \text{ kg}$$

The kinetic energy gained by the scooter is:

$$K = \frac{1}{2}mv^2$$

First day:

$$K_1 = \frac{1}{2}(160)v^2$$

Second day:

$$K_2 = \frac{1}{2}(200)v^2$$

Since fuel used is directly proportional to the energy supplied,

Ratio of fuel used = $K_1 : K_2$

$$= 160 : 200$$

$$= 4 : 5$$

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