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* 10^(-3) s`

`1 ms=10^(-3) s`

t=160 ms=
`160* 10^(-3) s`

`t'=t-t_1=290* 10^(-3)-160* 10^(-3)=130* 10^(-3) s`

Initial velocity=
`u_1`=u'=0

`y_1=(1)/(2) gt^2_1=(1)/(2)(9.8)(290* 10^(-3))^2=0.412 m`

Where
`g=9.8 m/s^2`

`y'=(1)/(2)gt'^2=(1)/(2)(9.8)(130* 10^(-3))^2=0.0828 m`

Center of mass=
`(m_1y_1+m'y')/(m_1+m')`

Center of mass=
`(m(0.412)+6m(0.0828))/(m+6m)=(m(0.412+6(0.0828))/(7m)=0.13 m`

b.
`v=u+gt`

Using the formula

`v_1=9.8(290* 10^(-3))=2.8m/s`

`v'=9.8(130* 10^(-3))=1.27m/s`

Velocity of center of mass=
`(m_1v_1+m'v')/(m_1+m')`

Velocity of center of mass=
`(m(2.8)+6m(1.27))/(m+6m)=(m(2.8+6(1.27))/(7m)=1.49 m/s`