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 = 160 ms. (a) How far below the release point is the center of mass of the two stones at t = 290 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 with Explanation:

Let mass of one stone=$m_1=m$

Mass of second stone,m'=6 m

a.$t_1=290 ms=290\times 10^{-3} s$

$1 ms=10^{-3} s$

t=160 ms=$160\times 10^{-3} s$

$t'=t-t_1=290\times 10^{-3}-160\times 10^{-3}=130\times 10^{-3} s$

Initial velocity=$u_1$=u'=0

$y_1=\frac{1}{2} gt^2_1=\frac{1}{2}(9.8)(290\times 10^{-3})^2=0.412 m$

Where $g=9.8 m/s^2$

$y'=\frac{1}{2}gt'^2=\frac{1}{2}(9.8)(130\times 10^{-3})^2=0.0828 m$

Center of mass=$\frac{m_1y_1+m'y'}{m_1+m'}$

Center of mass=$\frac{m(0.412)+6m(0.0828)}{m+6m}=\frac{m(0.412+6(0.0828)}{7m}=0.13 m$

b.$v=u+gt$

Using the formula

$v_1=9.8(290\times 10^{-3})=2.8m/s$

$v'=9.8(130\times 10^{-3})=1.27m/s$

Velocity of center of mass=$\frac{m_1v_1+m'v'}{m_1+m'}$

Velocity of center of mass=$\frac{m(2.8)+6m(1.27)}{m+6m}=\frac{m(2.8+6(1.27)}{7m}=1.49 m/s$