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A bungee jumper with mass \(55.0 \mathrm{~kg}\) reaches a speed of \(13.3 \mathrm{~m} / \mathrm{s}\) moving straight down when the elastic cord tied to her feet starts pulling her back up. After \(0.0250 \mathrm{~s},\) the jumper is heading back up at a speed of \(10.5 \mathrm{~m} / \mathrm{s}\). What is the average force that the bungee cord exerts on the jumper? What is the average number of \(g\) 's that the jumper experiences during this direction change?

Short Answer

Expert verified
Answer: ____________ N, ____________ g's

Step by step solution

01

Determine initial and final momentum

Calculate the momentum before and after the elastic cord starts pulling. Momentum (p) is given by the product of mass (m) and velocity (v). Initial momentum (\(p_1\)): \(p_1 = m * v_1\), with \(m = 55.0 kg\) and \(v_1 = -13.3 m/s\). Final momentum (\(p_2\)): \(p_2 = m * v_2\), with \(m = 55.0 kg\) and \(v_2 = 10.5 m/s\).
02

Calculate the impulse

The impulse (J) on the jumper can be found by the change in momentum \(\Delta p\): \(J = \Delta p = p_2 - p_1\). Calculate the impulse on the jumper by substituting the values.
03

Calculate the average force

The impulse-momentum theorem states that the impulse is equal to the average force (F) multiplied by the time interval \(\Delta t\). Therefore, \(J = F * \Delta t\) We can now solve for the average force acting on the jumper: \(F = \frac{J}{\Delta t}\), where \(\Delta t = 0.0250 s\).
04

Determine the average number of g's

To find the average number of g's experienced by the jumper, we need to divide the average force by the jumper's weight: Average number of g's = \(\frac{F}{mg}\), where \(g \approx 9.81 m/s^2\). Calculate the value using the average force found in Step 3.

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Key Concepts

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

Momentum
Momentum is a physical quantity that represents the motion contained within an object. It is directly related to the object's mass and velocity, described with the equation,
\( p = m \times v \),
where \( p \) is the momentum, \( m \) is the mass, and \( v \) is the velocity. In the context of bungee jumping, as the jumper hurtles downwards, the mass remains constant, but the velocity changes notably when the cord begins to stretch and decelerate the jumper. This change causes a shift in momentum, which is pivotal to understanding the forces at play during the jump.
Impulse
Impulse is defined as the change in momentum of an object when a force is applied over a period of time. It is given by,
\( J = m \times \Delta v \),
where \( J \) is the impulse, \( m \) is the mass, and \( \Delta v \) is the change in velocity. For the bungee jumper, the impulse would be the product of their mass and the difference between their final and initial velocity. This concept explains the effectiveness of the bungee cord in altering the jumper's speed in such a short interval.
Impulse-Momentum Theorem
The impulse-momentum theorem is integral to the physics of bungee jumping. It states that the impulse on an object is equal to the change in its momentum. This can be expressed with the equation,
\( J = \Delta p \),
where \( J \) is the impulse and \( \Delta p \) is the change in momentum. By using the initial and final velocity of our bungee jumper, and knowing their mass, we can calculate the impulse required to change their direction. This impulse is provided by the tension in the bungee cord, which can then be used to find out the force exerted on the jumper.
Average Force
Average force is the total force executed on an object divided by the time it was applied. It's especially useful to find the average force exerted by the bungee cord as it brings the jumper to a stop and then propels them upwards. According to the impulse-momentum theorem, you can calculate the average force using the following relationship,
\( F = \frac{J}{\Delta t} \),
where \( F \) is the average force, \( J \) is the impulse, and \( \Delta t \) is the time over which the impulse is applied. By deciphering the average force, we estimate the immense stresses a bungee jumper's body undergoes in the brief moment of direction change.
Acceleration Due to Gravity
The acceleration due to gravity, denoted as \( g \), is approximately \( 9.81 \frac{m}{s^2} \) on Earth's surface. It's a constant acceleration that acts on objects in freefall, governing their increase in speed as they descend. For a bungee jumper, this acceleration plays a crucial role until the cord begins to stretch. When analyzing the force felt by the jumper, we consider it in relation to their weight, a product of mass and gravity, to express it in g-forces. These g-forces are a measure of the acceleration and provide insight into the physical experience of the jumper.

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Most popular questions from this chapter

A sled initially at rest has a mass of \(52.0 \mathrm{~kg}\), including all of its contents. A block with a mass of \(13.5 \mathrm{~kg}\) is ejected to the left at a speed of \(13.6 \mathrm{~m} / \mathrm{s} .\) What is the speed of the sled and the remaining contents?

Two bumper cars moving on a frictionless surface collide elastically. The first bumper car is moving to the right with a speed of \(20.4 \mathrm{~m} / \mathrm{s}\) and rear-ends the second bumper car, which is also moving to the right but with a speed of \(9.00 \mathrm{~m} / \mathrm{s} .\) What is the speed of the first bumper car after the collision? The mass of the first bumper car is \(188 \mathrm{~kg}\), and the mass of the second bumper car is \(143 \mathrm{~kg}\). Assume that the collision takes place in one dimension.

An astronaut becomes stranded during a space walk after her jet pack malfunctions. Fortunately, there are two objects close to her that she can push to propel herself back to the International Space Station (ISS). Object A has the same mass as the astronaut, and Object \(\mathrm{B}\) is 10 times more massive. To achieve a given momentum toward the ISS by pushing one of the objects away from the ISS, which object should she push? That is, which one requires less work to produce the same impulse? Initially, the astronaut and the two objects are at rest with respect to the ISS.

Current measurements and cosmological theories suggest that only about \(4 \%\) of the total mass of the universe is composed of ordinary matter. About \(22 \%\) of the mass is composed of dark matter, which does not emit or reflect light and can only be observed through its gravitational interaction with its surroundings (see Chapter 12). Suppose a galaxy with mass \(M_{\mathrm{G}}\) is moving in a straight line in the \(x\) -direction. After it interacts with an invisible clump of dark matter with mass \(M_{\mathrm{DM}}\), the galaxy moves with \(50 \%\) of its initial speed in a straight line in a direction that is rotated by an angle \(\theta\) from its initial velocity. Assume that initial and final velocities are given for positions where the galaxy is very far from the clump of dark matter, that the gravitational attraction can be neglected at those positions, and that the dark matter is initially at rest. Determine \(M_{\mathrm{DM}}\) in terms of \(M_{\mathrm{G}}, v_{0},\) and \(\theta\).

The electron-volt, \(\mathrm{eV},\) is a unit of energy \((1 \mathrm{eV}=\) \(\left.1.602 \cdot 10^{-19} \mathrm{~J}, 1 \mathrm{MeV}=1.602 \cdot 10^{-13} \mathrm{~J}\right) .\) Since the unit of \(\mathrm{mo}\) mentum is an energy unit divided by a velocity unit, nuclear physicists usually specify momenta of nuclei in units of \(\mathrm{MeV} / c,\) where \(c\) is the speed of light \(\left(c=2.998 \cdot 10^{9} \mathrm{~m} / \mathrm{s}\right) .\) In the same units, the mass of a proton \(\left(1.673 \cdot 10^{-27} \mathrm{~kg}\right)\) is given as \(938.3 \mathrm{MeV} / \mathrm{c}^{2} .\) If a proton moves with a speed of \(17,400 \mathrm{~km} / \mathrm{s}\) what is its momentum in units of \(\mathrm{MeV} / \mathrm{c}\) ?

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