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`

Step-by-step explanation:

Given

first stone mass is m

second stone mass is 6 m

distance traveled by first stone in 430 ms

`y_1=ut+(at^2)/(2)`

`y_1=0+(g(0.43)^2)/(2)`

`y_1=0.9069 m`

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

`y_2=ut+(at^2)/(2)`

`y_2=0+(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* 0.43=4.214 m/s`

velocity of second stone after t=0.371 s

`v_2=u+at`

`v_2=0+9.8* 0.371=3.63 m/s`

Position of Center of mass of system

`y=(y_1m_1+y_2m_2)/(m_1+m_2)`

`y=(0.9069* m+0.674* 6m)/(m+6m)`

`y=(4.95m)/(7m)=0.707 m`

Velocity of COM

`v_(com)=(v_1m_1+v_2m_2)/(m_1+m_2)`

`v_(com)=(4.214* m+3.63* 6m)/(m+6m)`

`v_(com)=3.713 m/s`