When an object falls through a liquid, three forces act on it: its weight, the buoyancy and the resistance of the liquid. Two spheres, of mass 0.5 kg and 1.5 kg respectively, have the same radius, so that they have the same buoyancy of 3.2 newtons, and the same resistance formula, 5v newtons when falling at speed v ms^-1. Both spheres enter the liquid falling vertically at 1 ms^-1. Calculate the terminal speeds of the two spheres, and the acceleration or deceleration when they enter the liquid.
(Here I found: for mass 0.5 kg, terminal speed 0.36 ms^1 and acceleration -6.4 ms^2; for mass 1.5 kg, terminal speed 2.36, acceleration 4.53 (to 3 s.f.))
If the depth of the liquid is 10m, show that the heavier sphere reaches the bottom after a time between 4 and 10 seconds. Find bounds for the time that the lighter sphere takes to reach the bottom.
Ok to starting with the heavier sphere I thought it would be logical to proceed like this:
To find higher bound: distance = 10, u=1, v<2.36, a= (15 – 3.2 – 5v)/1.5 = (11.8 -5v)/1.5, equation s=ut + 0.5at^2, find v in terms of t and find interval for which v<2.36 is true
No problem here, I found t<10
I used a similar method to find the lower bound using s=vt-0.5at^2 and v=1, u>1, but this time the expression had no real roots, so I’m wondering if I’m not using the wrong method altogether, and if my first result was just lucky.
Can someone show me how to work through this?