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?

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

Explanation:

Given:

The first stone is dropped at [tex]t_{1}=0~s[/tex].

The second stone is dropped at, [tex]t_{2}=55~ms=0.055~s[/tex]

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 [tex]m[/tex]. So the mass of the second stone is [tex]3m[/tex].

(a)

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

[tex]y = \dfrac{1}{2}gt^{2}~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]

where [tex]g[/tex] is the acceleration due to gravity and [tex]t[/tex] is the time taken by each stone.

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

[tex]y_{1}&=& \dfrac{1}{2}(9.8~m.s^{-2})(0.47~s)^{2}\\~~~~&=& 1.08~m[/tex]

where [tex]y_{1}[/tex] is the distance traveled by the first stone.

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

[tex]y_{2}&=& \dfrac{1}{2}(9.8~m.s^{-2})(0.415~s)^{2}\\~~~~&=& 0.844~m[/tex]

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

[tex]y_{c} &=& \dfrac{my_{1}+3my_{2}}{m+3m} \\~~~~&=& \dfrac{1.08m + 0.844(3m)}{4m}\\~~~~&=& 0.903~m[/tex]

(b)

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

[tex]v=gt~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)[/tex]

Substituting [tex]v_{1}[/tex] for [tex]v[/tex], [tex]9.8~m.s^{-2}[/tex] for [tex]g[/tex] and [tex]0.47~s[/tex] for [tex]t[/tex] in equation (2), we have

[tex]v_{1} &=& (9.8~m.s^{-2})(0.47~s)\\~~~~&=& 4.606~m.s^{-1}[/tex]

where [tex]v_{1}[/tex] is the velocity of the first stone after [tex]470~ms[/tex].

Substituting [tex]v_{2}[/tex] for [tex]v[/tex], [tex]9.8~m.s^{-2}[/tex] for [tex]g[/tex] and [tex]0.415~s[/tex] for [tex]t[/tex] in equation (2), we have

[tex]v_{2} &=& (9.8~m.s^{-2})(0.415~s)\\~~~~&=& 4.067~m.s^{-1}[/tex]

where [tex]v_{2}[/tex] is the velocity of the second stone after [tex]470~ms[/tex].