/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 In February 1955 , a paratrooper... [FREE SOLUTION] | 91Ó°ÊÓ

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In February 1955 , a paratrooper fell \(370 \mathrm{~m}\) from an air- plane without being able to open his chute but happened to land in snow, suffering only minor injuries. Assume that his speed at impact was \(56 \mathrm{~m} / \mathrm{s}\) (terminal speed), that his mass (including gear) was \(85 \mathrm{~kg}\). and that the magnitude of the force on him from the snow was at the survivable limit of \(1.2 \times 10^{5} \mathrm{~N}\). What are (a) the minimum depth of snow that would have stopped him safely and (b) the magnitude of the impulse on him from the snow?

Short Answer

Expert verified
(a) Minimum snow depth: 1.11 m, (b) Impulse: 4760 Ns.

Step by step solution

01

Understand the Problem

A paratrooper falls 370 m without opening his parachute. His speed upon impact is 56 m/s, mass (including gear) is 85 kg, and the maximum force the snow can exert on him is \(1.2 \times 10^5\, \mathrm N\). We need to calculate the minimum depth of the snow that can completely stop him and the impulse exerted by the snow.
02

Calculate Kinetic Energy at Impact

The kinetic energy (KE) of the paratrooper at impact is given by the formula: \( KE = \frac{1}{2} m v^2 \), where \(m = 85\, \mathrm{kg}\) and \(v = 56\, \mathrm{m/s}\). So, \( KE = \frac{1}{2}\times 85 \times 56^2 = 133280 \, \mathrm{J} \).
03

Calculate the Minimum Depth of Snow

The work done by the snow to stop the paratrooper is equal to his initial kinetic energy. This work is given by the formula: \( W = F \cdot d \), where \( F = 1.2 \times 10^5 \, \mathrm{N} \). Solving for \(d\), we have: \( d = \frac{KE}{F} = \frac{133280}{1.2 \times 10^5} \approx 1.11 \, \mathrm{m} \).
04

Calculate the Impulse

Impulse \(J\) is given by the change in momentum, which can also be calculated by the formula \( J = F \cdot \Delta t \), but here it's more straightforward to use \( J = \Delta p = m \cdot v \), where \(v\) is the impact speed. Thus, \( J = 85 \times 56 = 4760 \, \mathrm{Ns} \).

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Key Concepts

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

Kinetic Energy
Kinetic energy is a fundamental concept in physics, describing the energy an object possesses due to its motion. It's expressed with the formula: \[ KE = \frac{1}{2} mv^2 \]where \( m \) is the mass of the object and \( v \) is its velocity. In the context of our paratrooper, when he hits the snow, he has a substantial amount of kinetic energy due to his high speed of 56 m/s.
This energy results from the motion acquired during his fall from 370 meters without a parachute.
Understanding kinetic energy helps us make sense of why the snow had to do so much work, i.e., exert such a large force, to stop him. The energy had to be absorbed fully to bring him to a safe stop. When the snow pushes back with force \( F \), this force performs work on the paratrooper to gradually reduce his kinetic energy to zero.
Consequently, the depth of snow required to stop him can be calculated by equating the kinetic energy to the work done by this force over the distance, or depth, \( d \). This relationship is given by:\[ W = F \cdot d = 133280 \, \mathrm{J} \] resulting in \( d \approx 1.11 \, \mathrm{m} \).
Thus, kinetic energy gives insight into the magnitude of the stopping mechanism required, highlighting the snow's crucial role in preventing injury.
Impulse and Momentum
Impulse and momentum are closely interconnected concepts in physics, helping us understand the effect of forces over time. Momentum is the product of an object's mass and velocity, calculated by the formula:\[ p = m \cdot v \]where \( m \) denotes mass and \( v \) the velocity. For our parachute-less paratrooper, his momentum right before hitting the snow is calculated as: \[ p = 85 \, \mathrm{kg} \times 56 \, \mathrm{m/s} = 4760 \, \mathrm{kg\cdot m/s} \].
This indicates the amount of motion the paratrooper has just before impact.
Impulse, meanwhile, is the change in momentum, and is produced when a force acts over a time interval. It can be expressed as:\[ J = \Delta p = F \cdot \Delta t \] or, more directly relevant to this scenario, \( J = m \cdot v \). Here, the impulse exerted by the snow is equivalent to the change in the paratrooper's momentum as he comes to a stop. The total impulse imparted by the snow equals the initial momentum, equalling to 4760 Ns.
This means the snow had to enact an impulse equal to the entire momentum of the moving paratrooper in order to bring him to a stop. Understanding impulse allows us to grasp how forces applied over time can safely decelerate objects, crucial for analyzing impacts and ensuring safety in emergencies.
Force and Motion
Force and motion are key concepts needed to explain various physical interactions, such as the stopping of a moving object. A force can be thought of as a push or pull acting on an object, which can cause it to accelerate, decelerate, come to a stop, or change direction. The relationship between force, mass, and acceleration is elegantly described by Newton's second law:\[ F = m \cdot a \]where \( F \) is the force applied, \( m \) is the mass, and \( a \) is the acceleration produced.
In this scenario, the paratrooper experienced a large deceleration imposed by the force of the snow, which had a specific survivable limit. This force was \( 1.2 \times 10^5 \mathrm{~N} \), allowing the paratrooper to come to a stop safely. The use of such strong force in snow demonstrates how it effectively reduces speed to prevent severe injury.
A crucial insight from force and motion is recognizing that the same force applied over a different time period changes the outcome of a collision – a principle used to design features like car airbags and crumple zones. In essence, while the paratrooper's fall was dangerous, the force exerted by the snow and the depth over which it acted played critical roles in saving his life. Understanding the dynamics of force and motion helps in engineering solutions for minimizing impact forces and improving safety standards.

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Most popular questions from this chapter

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