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Chapter 7: Problem 81

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

Many nuclear collisions are truly elastic. If a proton with kinetic energy \(E_{0}\) collides elastically with another proton at rest and travels at an angle of \(25^{\circ}\) with respect to its initial path, what is its energy after the collision with respect to its original energy? What is the final energy of the proton that was originally at rest?

An uncovered hopper car from a freight train rolls without friction or air resistance along a level track at a constant speed of \(6.70 \mathrm{~m} / \mathrm{s}\) in the positive \(x\) -direction. The mass of the car is \(1.18 \cdot 10^{5} \mathrm{~kg}\). a) As the car rolls, a monsoon rainstorm begins, and the car begins to collect water in its hopper (see the figure). What is the speed of the car after \(1.62 \cdot 10^{4} \mathrm{~kg}\) of water collects in the car's hopper? Assume that the rain is falling vertically in the negative \(y\) -direction. b) The rain stops, and a valve at the bottom of the hopper is opened to release the water. The speed of the car when the valve is opened is again \(6.70 \mathrm{~m} / \mathrm{s}\) in the positive \(x\) -direction (see the figure). The water drains out vertically in the negative \(y\) -direction. What is the speed of the car after all the water has drained out?

Consider these three situations: (i) A ball moving to the right at speed \(v\) is brought to rest. (ii) The same ball at rest is projected at speed \(v\) toward the left. (iii) The same ball moving to the left at speed \(v\) speeds up to \(2 v\). In which situation(s) does the ball undergo the largest change in momentum? a) situation (i) d) situations (i) and (ii) b) situation (ii) e) all three situations c) situation (iii)

A baseball pitcher delivers a fastball that crosses the plate at an angle of \(7.25^{\circ}\) relative to the horizontal and a speed of \(88.5 \mathrm{mph}\). The ball (of mass \(0.149 \mathrm{~kg}\) ) is hit back over the head of the pitcher at an angle of \(35.53^{\circ}\) with respect to the horizontal and a speed of \(102.7 \mathrm{mph}\). What is the magnitude of the impulse received by the ball?

A 3.0 -kg ball of clay with a speed of \(21 \mathrm{~m} / \mathrm{s}\) is thrown against a wall and sticks to the wall. What is the magnitude of the impulse exerted on the ball?
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