Question

A stone is dropped at t = 0. A second stone, with 3 times the mass of the first, is dropped from the same point at t = 100 ms. (a) How far below the release point is the center of mass of the two stones at t = 390 ms? (Neither stone has yet reached the ground.) (b) How fast is the center of mass of the two-stone system moving at that time?

Answer 1

from the given equation

stone 1 falls for 390 ms

stone 2 falls for 390 - 100 = 290 ms

Let m1 be the mass of first stone and m2 be the mass of second stone

m2 = 3 * m1

distance covered by first stone

h = 1/2 * 9.8 * 0.39² = 0.74529 m

h = 1/2 * 9.8 * 0.29² = 0.41209 m

so,

h_c.o.m = m1 * 0.74529 + 3m1 * 0.41209 / m1 + 3m1

h_c.o.m = 1.98156 m1 / 4 m1

h_c.o.m = 0.49539 m

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to find velocity of center of mass

final velocities must be found for both stones

v1 = 9.8 * 0.39 = 3.822 m/s

v2 = 9.8 * 0.29 = 2.842 m/s

so,

v_c.o.m = 3.822 + 3* 2.842 / 4

v_c.o.m = 3.087 m/s