Up and away
Two balls of greatly different masses are dropped (almost touching) from a height of 1 metre. They fall through the same path, with the light ball following the heavy ball. After the bounce(s), how high will the lighter ball get to?
If you were able to cascade a series of balls together (with each upper ball being much lighter than the one below it), how many balls would be needed for the uppermost (lightest) ball to achieve escape velocity? Assume that escape velocity = 11 km/sec.
Assume that the balls have neglible diameter compared to 1 metre. Neglect air resistance and only make simplifying assumptions.
If you were able to cascade a series of balls together (with each upper ball being much lighter than the one below it), how many balls would be needed for the uppermost (lightest) ball to achieve escape velocity? Assume that escape velocity = 11 km/sec.
Assume that the balls have neglible diameter compared to 1 metre. Neglect air resistance and only make simplifying assumptions.
Labels: funphysics
posted by Chris at
12:56 PM
3 Comments:
This one's easy!
As nobody seems to be doing this easy peasy problem, here's my solution.
Just before any ball bounces, it will have gained a speed v = Sqrt(2gh) = 4.43 m/s (that's accurate enough). After the first ball bounces, it will have a speed v upwards, and so the relative speed between it and the second ball will be 2v. This relative speed is preserved after the collision (except for the change of sign), so the light ball ends up with speed 2v+v = 3v = 3*4.43 = 13.29 m/s (upwards). It will therefore reach a height h = (3v)^2/(2g) = 9 * 1m = 9 m. If a third ball collided with the second, it would end up with a speed 7 v (= 31.01 m/s). If a fourth ball collided with that, we'd get a speed 15 v, fifth 31 v etc.
Can see the n th would have a final speed, v_n = (2^n - 1) v = 4.43(2^n -1) m/s.
For escape velocity, we want v_n = 11000 m/s. Solving for n =>
n >= log[11004.4/4.4]/log[2] = 11.3 balls. So need (only) 12 balls in cascade.
Just for the heck of it, n balls would give (2^n - 1)^2 metres height. So to achieve 1 km height would need n >= log[1+Sqrt[1000]]/log[2] = 5.03 balls. So a mere 5 balls would cause the last one to reach nearly 1 km height from an initial speed of 494 km/hr.
That amazingness is why I posted the problem.
A minor tidbit. The escape velocity corresponds to how fast you'd have to throw a ball if g stayed constant, i.e. independently of height, and you were trying for a height equal to the Earth's radius.
Did I hear someone ask for more?
I'll oblige. If the 12 balls (escape velocity case) had a mass ratio of 10 between each pair, then that would be a total mass range of 10^11. If the first ball had a mass of 100 kg, then last one would have a mass of 1 microgram. Not only would that be very difficult to achieve, air resistance could not be neglected as the lightest ball would be severely affected by it.
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