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(a): At the center of mass will travel .

Part(b): The velocity of the first stone is and the velocity of the second stone is .

Explanation:

Given:

The first stone is dropped at .

The second stone is dropped at,

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 . So the mass of the second stone is .

(a)

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

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

Substituting for , for and or for first stone in equation (1), we have

where is the distance traveled by the first stone.

Substituting for , for and   or for second stone in equation (1), we have

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

(b)

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

Substituting for ,   for and for in equation (2), we have

where is the velocity of the first stone after .

Substituting for ,   for and for in equation (2), we have

where is the velocity of the second stone after .

answer: (431,2)

step-by-step explanation: