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 = 55 ms. (a) How far below the release point is the center of mass of the two stones at t = 470 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

Part(a): At
`\bf{t = 470~ms}` the center of mass will travel
`\bf{0.903~m}`.

Part(b): The velocity of the first stone is
`\bf{4.606~m.s^(-1)}` and the velocity of the second stone is
`\bf{4.067~m.s^(-1)}`.

Step-by-step explanation:

Given:

The first stone is dropped at
`t_(1)=0~s`.

The second stone is dropped at,
`t_(2)=55~ms=0.055~s`

The mass of the second stone is 3 times the mass of the first.

Both the stones are dropped from the same point.

Consider the mass of the first stone be
`m`. So the mass of the second stone is
`3m`.

(a)

The formula to calculate the distance traveled by each stone is given by

`y = (1)/(2)gt^(2)~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)`

where
`g` is the acceleration due to gravity and
`t` is the time taken by each stone.

Substituting
`9.8~m.s^(-2)` for
`g`,
`y_(1)` for
`y` and
`470~ms` or
`0.47~s` for first stone in equation (1), we have

`y_(1)&=& (1)/(2)(9.8~m.s^(-2))(0.47~s)^(2)\\~~~~&=& 1.08~m`

where
`y_(1)` is the distance traveled by the first stone.

Substituting
`9.8~m.s^(-2)` for
`g`,
`y_(2)` for
`y` and
`(470-55)~ms = 415~ms` or
`0.415~s` for second stone in equation (1), we have

`y_(2)&=& (1)/(2)(9.8~m.s^(-2))(0.415~s)^(2)\\~~~~&=& 0.844~m`

The formula to calculate the distance traveled by the center of mass is given by

`y_(c) &=& (my_(1)+3my_(2))/(m+3m) \\~~~~&=& (1.08m + 0.844(3m))/(4m)\\~~~~&=& 0.903~m`

(b)

The formula to calculate the velocity of each stone is given by

`v=gt~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)`

Substituting
`v_(1)` for
`v`,
`9.8~m.s^(-2)` for
`g` and
`0.47~s` for
`t` in equation (2), we have

`v_(1) &=& (9.8~m.s^(-2))(0.47~s)\\~~~~&=& 4.606~m.s^(-1)`

where
`v_(1)` is the velocity of the first stone after
`470~ms`.

Substituting
`v_(2)` for
`v`,
`9.8~m.s^(-2)` for
`g` and
`0.415~s` for
`t` in equation (2), we have

`v_(2) &=& (9.8~m.s^(-2))(0.415~s)\\~~~~&=& 4.067~m.s^(-1)`

where
`v_(2)` is the velocity of the second stone after
`470~ms`.