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Chapter 8: Problem 64

A uniform log of length \(2.50 \mathrm{~m}\) has a mass of \(91.0 \mathrm{~kg}\) and is floating in water. Standing on this log is a \(72.0-\mathrm{kg}\) man, located \(22.0 \mathrm{~cm}\) from one end. On the other end is his daughter \((m=20.0 \mathrm{~kg})\), standing \(1.00 \mathrm{~m}\) from the end. a) Find the center of mass of this system. b) If the father jumps off the log backward away from his daughter \((v=3.14 \mathrm{~m} / \mathrm{s}),\) what is the initial speed of \(\log\) and child?

Short Answer

Expert verified
Answer: The center of mass of the system is approximately 0.872 m from the left end of the log. The initial speed of the log and the child is approximately -2.04 m/s, moving in the opposite direction of the father's jump.

Step by step solution

01

Determine the center of mass of the system for part (a)

To find the center of mass of the system, we can consider the log, the man, and the child as individual masses at specific locations. Let the left end of the log be the origin (0). The position of the man is 0.22 m from the origin, the position of the daughter is at 2.50 m - 1.00 m = 1.50 m from the origin, and the center of mass of the log lies in its middle, i.e., at 1.25 m. So, the equation for the center of mass is given by: \(x_{cm} = \frac{m_{man}x_{man} + m_{daughter}x_{daughter} + m_{log}x_{log}}{m_{man} + m_{daughter} + m_{log}}\)
02

Calculate the center of mass

Now, we can plug in the given values into the formula: \(x_{cm} = \frac{(72)(0.22) + (20)(1.5) + (91)(1.25)}{72 + 20 + 91}\) \(x_{cm} = \frac{15.84 + 30 + 113.75}{183}\) \(x_{cm} = \frac{159.59}{183}\) \(x_{cm} \approx 0.872~m\) The center of mass is approximately at 0.872 m from the origin, which is the left end of the log.
03

Set up the equation for conservation of momentum for part (b)

When the father jumps off the log, the total momentum of the system is conserved. Before the jump, the total momentum is 0. After the jump, the father, child, and log will have their own existing momenta. So, we can write the equation for conservation of momentum as: \(m_{man}v_{man} + m_{daughter}v_{daughter} + m_{log}v_{log} = 0\) where \(v_{man}\), \(v_{daughter}\), and \(v_{log}\) are the velocities of man, daughter, and log, respectively, after the jump. Since both the daughter and the log would have the same initial velocity, we can rewrite the equation as: \(-m_{man}v_{man} = (m_{daughter} + m_{log})v\) where v represents the initial speed of the log and the child.
04

Calculate the initial speed of the log and the child

Now we can plug in the given values into the momentum equation: \(-(72)(3.14) = (20+91)v\) Solving for v, we get: \(v = \frac{-226.08}{111}\) \(v\approx -2.04~m/s\) The initial speed of the log and child is approximately -2.04 m/s. The negative sign implies that they will move in the opposite direction of the father's jump.

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Most popular questions from this chapter

An astronaut is performing a space walk outside the International Space Station. The total mass of the astronaut with her space suit and all her gear is \(115 \mathrm{~kg} .\) A small leak develops in her propulsion system and \(7.00 \mathrm{~g}\) of gas are ejected each second into space with a speed of \(800 \mathrm{~m} / \mathrm{s}\). She notices the leak 6.00 s after it starts. How much will the gas leak have caused her to move from her original location in space by that time?

Can the center of mass of an object be located at a point outside the object, that is, at a point in space where no part of the object is located? Explain.

A catapult on a level field tosses a 3 -kg stone a horizontal distance of \(100 \mathrm{~m}\). A second 3 -kg stone tossed in an identical fashion breaks apart in the air into 2 pieces, one with a mass of \(1 \mathrm{~kg}\) and one with a mass of \(2 \mathrm{~kg} .\) Both of the pieces hit the ground at the same time. If the 1 -kg piece lands a distance of \(180 \mathrm{~m}\) away from the catapult, how far away from the catapult does the 2 -kg piece land? Ignore air resistance. a) \(20 \mathrm{~m}\) c) \(100 \mathrm{~m}\) e) \(180 \mathrm{~m}\) b) \(60 \mathrm{~m}\) d) \(120 \mathrm{~m}\)

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The center of mass of an irregular rigid object is always located a) at the geometrical center of c) both of the above the object. d) none of the above b) somewhere within the object.
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