Question

A stone is dropped at t = 0. A second stone, with 6 times the mass of the first, is dropped from the same point at t = 59 ms. (a) How far below the release point is the center of mass of the two stones at t = 430 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

Answer:$y_{com}=0.707 m$

$v_{com}=3.713 m/s$

Explanation:

Given

first stone mass is m

second stone mass is 6 m

distance traveled by  first stone in 430 ms

$y_1=ut+\frac{at^2}{2}$

$y_1=0+\frac{g(0.43)^2}{2}$

$y_1=0.9069 m$

Distance traveled by stone 2 in t=430-59=371 ms

$y_2=ut+\frac{at^2}{2}$

$y_2=0+\frac{g(0.0.371)^2}{2}$

$y_2=0.674 m$

velocity of first stone after t=0.43 s

$v_1=u+at$

$v_1=0+9.8\times 0.43=4.214 m/s$

velocity of second stone after t=0.371 s

$v_2=u+at$

$v_2=0+9.8\times 0.371=3.63 m/s$

Position of Center of mass of system

$y=\frac{y_1m_1+y_2m_2}{m_1+m_2}$

$y=\frac{0.9069\times m+0.674\times 6m}{m+6m}$

$y=\frac{4.95m}{7m}=0.707 m$

Velocity of COM

$v_{com}=\frac{v_1m_1+v_2m_2}{m_1+m_2}$

$v_{com}=\frac{4.214\times m+3.63\times 6m}{m+6m}$

$v_{com}=3.713 m/s$