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Chapter 9: Problem 31

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)?

Short Answer

Expert verified
Impulses: 2394 Ns, 1764 Ns; Forces: 478800 N, 352800 N.

Step by step solution

01

Determine the final velocity of the elevator

When the elevator free-falls from a height, it gains velocity. Use the kinematic equation \( v^2 = u^2 + 2gh \) to find the final velocity at the moment the elevator hits the bottom. Use \( g = 9.8 \ m/s^2 \), \( u = 0 \ \text{m/s} \), and \( h = 36 \ \text{m}. \)\[v^2 = 0 + 2 \times 9.8 \times 36 \v = \sqrt{2 \times 9.8 \times 36} = \sqrt{705.6} \approx 26.6 \ \text{m/s}.\]
02

Calculate the impulse on the passenger without jumping

Impulse is equal to the change in momentum, \( J = \Delta p = m \Delta v \). Since the passenger is brought to rest, the final velocity is 0, and initial velocity is the same as the cab, \( v_i = 26.6 \ \text{m/s}. \)\[J = m(v_f - v_i) = 90 \times (0 - 26.6) = -2394 \ \text{Ns}. \]The magnitude of impulse is \( 2394 \ \text{Ns}. \)
03

Calculate the average force on the passenger without jumping

The average force can be calculated using the impulse-momentum theorem, \( F_{avg} = \frac{J}{\Delta t} \). The stopping time is \( 5 \ \text{ms} = 0.005 \ \text{s}. \)\[F_{avg} = \frac{2394}{0.005} = 478800 \ \text{N}. \]
04

Calculate the impulse on the passenger when jumping

If the passenger jumps with a speed of \( 7.0 \ \text{m/s} \) upwards relative to the cab just before collision, the total velocity downwards relative to ground is the velocity of the cab minus the jumping speed: \( v = 26.6 - 7 = 19.6 \ \text{m/s}. \)\[J = m(0 - 19.6) = 90 \times (-19.6) = -1764 \ \text{Ns}. \]The magnitude of impulse is \( 1764 \ \text{Ns}. \)
05

Calculate the average force on the passenger when jumping

Apply the impulse-momentum theorem again using the newly calculated impulse and the same stopping time for the calculation.\[F_{avg} = \frac{1764}{0.005} = 352800 \ \text{N}. \]

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Key Concepts

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

Free Fall
The concept of free fall refers to the motion of an object under the influence of gravitational force alone. When an object is in free fall, it moves downward due to gravity, and no other forces act on it. In this scenario, the elevator cab is in free fall after the cable snaps. This means its motion is solely affected by gravity until it hits the ground.

Key points about free fall include:
  • The acceleration of an object in free fall is constant and equal to the gravitational acceleration, which is approximately 9.8 m/s虏 on Earth.
  • In the absence of air resistance, all objects, regardless of mass, experience the same acceleration when in free fall.
  • Velocity increases as the object continues to fall, calculated using kinematic equations to predict its speed before impact.
The elevator cab, starting from rest, gradually gains speed with acceleration of 9.8 m/s虏. We use the kinematic equation to determine its final velocity upon reaching the bottom of the shaft, which is vital for further calculations regarding impulse and force.
Kinematic Equations
Kinematic equations relate the motion of an object with its velocity, acceleration, and time, providing a crucial tool for analyzing motion in physics. These equations help us predict future positions or velocities of objects moving with constant acceleration. In this particular exercise, the elevator's motion can be described using one of these equations.

The most relevant kinematic equation in this context is: \[ v^2 = u^2 + 2gh \]
  • \( v \) is the final velocity of the object.
  • \( u \) is the initial velocity, which is zero for the elevator starting from rest.
  • \( g \) is the acceleration due to gravity.
  • \( h \) is the height from which the object falls.
By substituting the given values, this equation helps calculate the velocity of the elevator right before it reaches the ground as approximately 26.6 m/s.

The kinematic equations not only give us a way to calculate how fast something is moving, but also allow us to understand the inherent physics when a force, such as gravity, acts over time and distance, as with the elevator cab.
Impulse-Momentum Theorem
The Impulse-Momentum Theorem is a fundamental concept in physics that describes the relationship between impulse and the change in an object's momentum. Impulse is defined as the product of a force acting over a specific time interval, while momentum is the product of an object's mass and its velocity.

Mathematically, the theorem is expressed as:\[ J = \Delta p = F_{avg} \times \Delta t \]
  • \( J \) represents impulse, usually measured in Newton-seconds (Ns).
  • \( \Delta p \) is the change in momentum, calculated as the difference between the initial and final momentum of the object.
  • \( F_{avg} \) is the average force exerted on the object during the time interval \( \Delta t \).
In the elevator scenario, the impulse is calculated based on the sudden halt of the passenger's motion as the cab collides with the ground. With an initial velocity matching the elevator and final velocity of zero, we determine the impulse on the system, and use it to calculate the average force during the brief 5 ms halt. This framework shows how forces, even over short periods, dramatically affect an object's state of motion, vital for understanding real-world impacts and safety measures.

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Most popular questions from this chapter

A \(4.0 \mathrm{~kg}\) mess kit sliding on a frictionless surface explodes into two \(2.0 \mathrm{~kg}\) parts: \(3.0 \mathrm{~m} / \mathrm{s}\), due north, and \(5.0 \mathrm{~m} / \mathrm{s}, 30^{\circ}\) north of east. What is the original speed of the mess kit?

A \(6090 \mathrm{~kg}\) space probe moving nose-first toward Jupiter at \(105 \mathrm{~m} / \mathrm{s}\) relative to the Sun fires its rocket engine, ejecting \(80.0 \mathrm{~kg}\) of exhaust at a speed of \(253 \mathrm{~m} / \mathrm{s}\) relative to the space probe. What is the final velocity of the probe?

In February 1955 , a paratrooper fell \(370 \mathrm{~m}\) from an airplane 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?

A spacecraft is separated into two parts by detonating the explosive bolts that hold them together. The masses of the parts are \(1200 \mathrm{~kg}\) and \(1800 \mathrm{~kg} ;\) the magnitude of the impulse on each part from the bolts is \(300 \mathrm{~N} \cdot \mathrm{s}\). With what relative speed do the two parts separate because of the detonation?

An object is tracked by a radar station and determined to have a position vector given by \(\vec{r}=(3500-160 t) \hat{\mathrm{i}}+2700 \hat{\mathrm{j}}+300 \hat{\mathrm{k}}\), with \(\vec{r}\) in meters and \(t\) in seconds. The radar station's \(x\) axis points east. its \(y\) axis north, and its \(z\) axis vertically up. If the object is a \(250 \mathrm{~kg}\) meteorological missile, what are (a) its linear momentum, (b) its direction of motion, and (c) the net force on it?
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