91Ó°ÊÓ

Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height of \(36 \mathrm{~m}\). During the collision at the bottom of the elevator shaft, a \(90 \mathrm{~kg}\) passenger is stopped in \(5.0 \mathrm{~ms}\). (Assume that neither the passenger nor the cab rebounds.) What are the magnitudes of the (a) impulse and (b) average force on the passenger during the collision? If the passenger were to jump upward with a speed of \(7.0\) \(\mathrm{m} / \mathrm{s}\) relative to the cab floor just before the cab hits the bottom of the shaft, what are the magnitudes of the (c) impulse and (d) average force (assuming the same stopping time)?

Step by step solution

01

Calculate the final velocity of the cab

First, let's calculate the final velocity of the elevator cab just before it hits the ground. Since the cab is in free-fall, we can use the equation \(v^2 = u^2 + 2gh\), where \(u = 0\) (initial velocity), \(g = 9.8\, \mathrm{m/s^2}\) (acceleration due to gravity), and \(h = 36\, \mathrm{m}\).\[ v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 36} \approx 26.57\, \mathrm{m/s} \]

02

Calculate impulse for part (a)

Impulse is the change in momentum. The initial momentum of the passenger is \(mv\), and the final momentum is \(0\), since the passenger stops.\[ \text{Impulse} = m(v_f - v_i) = 90 \times (0 - (-26.57)) = 2391.3\, \mathrm{Ns} \]

03

Calculate average force for part (b)

The average force exerted on the passenger can be found using the impulse-momentum theorem \( \text{Impulse} = F_{\text{avg}} \times \Delta t \). Rearrange for \(F_{\text{avg}}\):\[ F_{\text{avg}} = \frac{\text{Impulse}}{\Delta t} = \frac{2391.3}{5 \times 10^{-3}} = 478260\, \mathrm{N} \]

04

Calculate impulse for part (c) with jump

The passenger jumps upwards with a speed of \(v_j = 7\, \mathrm{m/s}\) relative to the cab. Hence, his velocity relative to the ground just before impact is \(v_i = 26.57 - 7 = 19.57\, \mathrm{m/s}\).\[ \text{Impulse with jump} = 90 \times (0 - (-19.57)) = 1761.3\, \mathrm{Ns} \]

05

Calculate average force for part (d) with jump

Again using the impulse-momentum theorem for the scenario where the passenger jumps:\[ F_{\text{avg\_jump}} = \frac{1761.3}{5 \times 10^{-3}} = 352260\, \mathrm{N} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Impulse

Impulse is a key concept in physics that refers to the change in momentum of an object when a force acts on it for a specific time period. In simple terms, it's the product of the average force and the time duration the force is applied. This concept is crucial for understanding how forces cause changes in motion in practical scenarios.
When we calculate impulse, we are looking for the change in velocity of an object, which could be caused by any event that applies a force over time, like a collision.

• Formula: \( ext{Impulse} = ext{Force} imes ext{Time} \)
• It's also equivalent to the change in momentum: \( ext{Impulse} = m(v_f - v_i) \), where \(m\) is mass, \(v_f\) is the final velocity, and \(v_i\) is the initial velocity.
• In our elevator scenario, impulse helps us understand the effect of the collision on the passenger.

This understanding can also tell us more about the dynamics of the free-falling elevator and the passenger's interaction with it when it comes to an abrupt stop.

Average Force

When a force is applied over a specific time period causing a change in momentum, it is often analyzed in terms of 'average force'. In physics, especially in collision scenarios like our exercise, average force provides a helpful way to evaluate the impact over the action duration.
To find the average force exerted, we can divide the impulse by the time interval during which the force was applied.

• Formula: \( F_{\text{avg}} = \frac{\text{Impulse}}{\Delta t} \)
• This helps us determine the intensity of the force over time, and is crucial for designing safety measures.
• In practical applications, understanding average force can guide safety and construction measures, such as those found in vehicle crash safety tests.

In the example, calculating the average force on the passenger when the elevator impacts the ground shows us just how intense this event is, emphasizing the extreme forces involved in a collision scenario.

Free-fall

The term free-fall describes the motion of a body where it is only under the influence of gravity, with no other forces acting on it. This concept is central when calculating how objects behave when they drop from height, including the moment just before they hit the ground.
Free-fall calculations help us determine how fast an object is going just before hitting the surface, using the height it fell from and gravitational acceleration.

• In an ideal free-fall scenario, air resistance is negligible, and the only force acting on the object is gravity.
• We use the equation \( v^2 = u^2 + 2gh \) (where \( u \) is initial velocity, \( v \) is final velocity, \( g \) is gravitational acceleration, and \( h \) is height) to find the final speed.
• Our elevator example shows the cab accelerating until just before impact, giving us insights into the final conditions before the collision occurs.

These details are vital to analyzing how forces would later act when the cab hits the ground, affecting anyone inside it, such as the passenger in our scenario.

Momentum Change

Momentum, a fundamental concept in physics, is the product of mass and velocity of an object. When an object's velocity changes during an event, such as a collision, there's a resultant change in momentum. This change illuminates the effects of forces acting on an object.
Understanding momentum change is pivotal when studying impacts because it helps quantify the effect of forces. It's directly tied to impulse, as previously mentioned, since impulse is the measure of momentum change during an event.

• Formula for momentum: \( p = m imes v \) (where \( p \) is momentum, \( m \) is mass, \( v \) is velocity)
• Change in momentum: \( \Delta p = m(v_f - v_i) \)
• By examining momentum change, we can derive vital information regarding the forces involved and the expected outcomes of a collision.

In our case, the passenger's velocity before and after the crash leads us to the momentum change, an essential consideration for understanding the dynamics of the impact they experience.