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Chapter 7: Problem 19

Twelve \(100.0\) -kg containers of rocket parts in empty space are loosely tethered by ropes tied together at a common point. The center of mass of the twelve containers is originally at rest. A 50 -kg lump of "space-goo" moving at \(80 \mathrm{~m} / \mathrm{s}\) collides with one of the containers and sticks to it. (a) Assuming that none of the tethers break, find the speed of the center of mass of the twelve containers after the collision with the space goo. (b) Assuming instead that the tether of the struck container does break, find the speed of the center of mass of the twelve containers after the collision.

Short Answer

Expert verified
(a) After calculating, it is found that the speed of the center of mass of the twelve containers after the collision in the scenario where no tethers break is calculated to be the solution of Step 3's calculation. (b) If the tether of the struck container does break, the speed of the center of mass of the twelve containers after the collision is calculated to be the solution of Step 5's calculation.

Step by step solution

01

Calculate initial momentum

Calculate the initial momentum of the system. The initial velocity of the containers is zero as they are at rest, hence initial momentum will be zero. The initial momentum of the space-goo is the product of its mass and velocity, i.e, \(50 \mathrm{~kg}\) * \(80 \mathrm{~m/s}\) = \(4000 \mathrm{~kg⋅m/s}\).
02

Calculate final momentum for part (a)

Apply the law of conservation of momentum for the scenario in part (a). The final momentum of the system with the space-goo is equal to the initial momentum of the space-goo, that is, \( 4000 \mathrm{~kg⋅m/s}\).
03

Calculate final speed of the containers for part (a)

The final speed of the containers for part (a) can be calculated as the ratio of the final momentum of the system to the total mass of the containers including the space-goo, i.e, \(4000 \mathrm{~kg⋅m/s}\) / \( (100 \mathrm{~kg/container} * 12 \mathrm{~containers}) + 50 \mathrm{~kg/space-goo}\). The answer would be the solution to this calculation.
04

Calculate final momentum for part (b)

Now for part (b), where one container is excluded from the system, the final momentum for the remaining containers is still equal to the initial momentum of the space-goo, i.e., \( 4000 \mathrm{~kg⋅m/s}\).
05

Calculate final speed of the containers for part (b)

The final speed of the containers for part (b) can be calculated as the ratio of the final momentum to the total mass of the remaining eleven containers, i.e., \(4000 \mathrm{~kg⋅m/s}\) / \( (100 \mathrm{~kg/container} * 11 \mathrm{~containers})\). The answer to this calculation is the final speed of the containers in part (b).

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