Physics, 22.04.2020 04:52 whitakers87

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?

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Answer from: episodegirl903

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 $\bf{4.067~m.s^{-1}}$ .

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 . So the mass of the second stone is .

(a)

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

$y = \dfrac{1}{2}gt^{2}~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

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

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

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

where $y_{1}$ is the distance traveled by the first stone.

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

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

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

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

(b)

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

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

Substituting $v_{1}$ for , $9.8~m.s^{-2}$ for and 0.47~s for 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 , $9.8~m.s^{-2}$ for and 0.415~s for 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 .

Answer from: Quest

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Answer from: Quest

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j

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A stone is dropped at t = 0. A second stone, with 3 times the mass of the first, is dropped from the...