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Chapter 7: Problem 26

Adolf and Ed are wearing harnesses and are hanging at rest from the ceiling by means of ropes attached to them. Face to face, they push off against one another. Adolf has a mass of \(120 \mathrm{kg},\) and \(\mathrm{Ed}\) has a mass of \(78 \mathrm{kg} .\) Following the push, Adolf swings upward to a height of \(0.65 \mathrm{m}\) above his starting point. To what height above his own starting point does Ed rise?

Short Answer

Expert verified
Ed rises 1.54 meters above his starting point.

Step by step solution

01

Identify the Basic Principle

When Adolf and Ed push off each other, they impart equal and opposite momentum to each other, due to the conservation of momentum. This means any increase in kinetic energy resulting from their interaction will be equal and opposite.
02

Apply Conservation of Energy to Adolf

For Adolf, convert initial kinetic energy to gravitational potential energy. Use the formula for gravitational potential energy, \( PE = mgh \), where \( m \) is mass, \( g \) is acceleration due to gravity \((9.81\, \mathrm{m/s^2})\), and \( h \) is height. Adolf's kinetic energy becomes potential energy, \( KE = PE \), meaning:\[ \frac{1}{2}m_{Adolf}v_{Adolf}^2 = m_{Adolf}gh_{Adolf} \]which simplifies to \( v_{Adolf}^2 = 2gh_{Adolf} \). Thus, \( v_{Adolf} = \sqrt{2gh_{Adolf}} \).
03

Calculate Adolf's Initial Velocity

Using \( h_{Adolf} = 0.65 \) m, calculate Adolf's initial velocity:\[ v_{Adolf} = \sqrt{2 \times 9.81 \times 0.65} = \sqrt{12.753} \approx 3.57 \mathrm{m/s} \].
04

Use Conservation of Momentum

The conservation of momentum before and after the push tells us:\[ m_{Adolf}v_{Adolf} = m_{Ed}v_{Ed} \]Using Adolf's velocity from Step 3, substitute to find \( v_{Ed} \):\[ 120 \times 3.57 = 78 \times v_{Ed} \]\[ v_{Ed} = \frac{120 \times 3.57}{78} \approx 5.5 \mathrm{m/s} \].
05

Apply Conservation of Energy to Ed

Now, apply energy conservation to Ed to find his maximum height:\[ \frac{1}{2}m_{Ed}v_{Ed}^2 = m_{Ed}gh_{Ed} \]This simplifies to \( v_{Ed}^2 = 2gh_{Ed} \), so:\[ h_{Ed} = \frac{v_{Ed}^2}{2g} \]Substitute \( v_{Ed} = 5.5 \mathrm{m/s} \) to calculate \( h_{Ed} \):\[ h_{Ed} = \frac{5.5^2}{2 \times 9.81} = \frac{30.25}{19.62} \approx 1.54 \mathrm{m} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conservation of Energy
The principle of conservation of energy states that energy cannot be created or destroyed; it can only be transferred or transformed from one form to another. In this scenario, when Adolf and Ed push off each other, they convert their stored energy into another form. Initially, they have potential energy since they are at rest. After the push, this energy is converted into kinetic energy as they move, demonstrating conservation of energy.

Later, as they reach the height of their swing, this kinetic energy is transformed back into gravitational potential energy. It is essential to note that the total mechanical energy of the system (combination of kinetic and potential energies) remains constant, if we ignore other forces like air resistance. This conversion is tightly linked to equations used in calculations, such as moving from kinetic to potential energy in the formulas applied here. Understanding these processes highlights how conservation of energy governs motion.
Gravitational Potential Energy
Gravitational potential energy (\[ PE = mgh \]reflects the energy stored in an object due to its position relative to the earth. It depends on three factors:
  • The object's mass \( m \)
  • The height \( h \) of the object above a reference point
  • The gravitational acceleration \( g \), typically \(9.81 \text{ m/s}^2 \) on Earth's surface


In our exercise, initially Adolf and Ed are at rest, lacking kinetic energy, with all energy present as gravitational potential energy due to their elevated position. When Adolf swings up to a height of 0.65 m, his gravitational potential energy increases, given the increased height. This increase highlights the transformation of his initial kinetic energy, derived from pushing off Ed, into gravitational potential energy. Studying such energy changes clarifies the relationship between Adolf's increase in height and gravitational potential energy.
Kinetic Energy
Kinetic energy (\[ KE = \frac{1}{2} mv^2 \]represents the energy possessed by an object due to its motion. It relies on two main aspects:
  • The mass of the object \( m \)
  • The speed of the object \( v \)


When Adolf and Ed pushed apart, they transformed potential energy into kinetic energy as each began to move. The energy they gain from their respective motion can be calculated using the formula above, which shows how velocity affects kinetic energy. For instance, we derive Adolf's velocity by knowing his rise in height and potential energy, leading to transformation equations showing \( v_{Adolf} = \sqrt{2gh_{Adolf}} \).

Then, employing conservation of momentum allows calculating Ed's velocity, which further translates to kinetic energy, later converting back into potential energy as Ed ascends to a peak. This cyclical transformation between kinetic and potential energy under conservation principles serves as a roadmap to predict and comprehend motion aspects.

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Most popular questions from this chapter

Two particles are moving along the \(x\) axis. Particle 1 has a mass \(m_{1}\) and a velocity \(v_{1}=+4.6 \mathrm{m} / \mathrm{s} .\) Particle 2 has a mass \(m_{2}\) and a velocity \(v_{2}=\) \(-6.1 \mathrm{m} / \mathrm{s} .\) The velocity of the center of mass of these two particles is zero. In other words, the center of mass of the particles remains stationary, even though each particle is moving. Find the ratio \(m_{1} / m_{2}\) of the masses of the particles.

In a science fiction novel two enemies, Bonzo and Ender, are fighting in outer space. From stationary positions they push against each other. Bonzo flies off with a velocity of \(+1.5 \mathrm{m} / \mathrm{s},\) while Ender recoils with a velocity of \(-2.5 \mathrm{m} / \mathrm{s} .\) (a) Without doing any calculations, decide which person has the greater mass. Give your reasoning. (b) Determine the ratio \(m_{\text {Bonno }}\) \(m_{\text {Ender }}\) of the masses of these two enemies.

A 40.0 -kg boy, riding a 2.50 -kg skateboard at a velocity of \(+5.30 \mathrm{m} / \mathrm{s}\) across a level sidewalk, jumps forward to leap over a wall. Just after leaving contact with the board, the boy's velocity relative to the sidewalk is \(6.00 \mathrm{m} / \mathrm{s}, 9.50^{\circ}\) above the horizontal. Ignore any friction between the skateboard and the sidewalk. What is the skateboard"s velocity relative to the sidewalk at this instant? Be sure to include the correct algebraic sign with your answer.

You and your crew must dock your \(25000 \mathrm{kg}\) spaceship at Spaceport Alpha, which is orbiting Mars. In the process, Alpha's control tower has requested that you ram another vessel, a freight ship of mass \(16500 \mathrm{kg},\) latch onto it, and use your combined momentum to bring it into dock. The freight ship is not moving with respect to the colossal Spaceport Alpha, which has a mass of \(1.85 \times 10^{7} \mathrm{kg} .\) Alpha's automated system that guides incoming spacecraft into dock requires that the incoming speed is less than \(2.0 \mathrm{m} / \mathrm{s}\). (a) Assuming a perfectly linear alignment of your ship's velocity vector with the freight ship (which is stationary with respect to Alpha) and Alpha's docking port, what must be your ship's speed (before colliding with the freight ship) so that the combination of the freight ship and your ship arrives at Alpha's docking port with a speed of \(1.50 \mathrm{m} / \mathrm{s} ?\) (b) How does the velocity of Spaceport Alpha change when the combination of your vessel and the freight ship successfully docks with it? (c) Suppose you made a mistake while maneuvering your vessel in an attempt to ram the freight ship and, rather than latching on to it and making a perfectly inelastic collision, you strike it and knock it in the direction of the spaceport with a perfectly elastic collision. What is the speed of the freight ship in that case (assuming your ship had the same initial velocity as that calculated in part (a))?

A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The incident water stream has a velocity of \(+16.0 \mathrm{m} / \mathrm{s},\) while the exiting water stream has a velocity of \(-16.0 \mathrm{m} / \mathrm{s} .\) The mass of water per second that strikes the blade is \(30.0 \mathrm{kg} / \mathrm{s}\). Find the magnitude of the average force exerted on the water by the blade.
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