One of the events in the Scottish Highland Games is the sheaf toss, in which a \(9.09-\mathrm{kg}\) bag of hay is tossed straight up into the air using a pitchfork. During one throw, the sheaf is launched straight up with an initial speed of \(2.7 \mathrm{~m} / \mathrm{s}\). a) What is the impulse exerted on the sheaf by gravity during the upward motion of the sheaf (from launch to maximum height)? b) Neglecting air resistance, what is the impulse exerted by gravity on the sheaf during its downward motion (from maximum height until it hits the ground)? c) Using the total impulse produced by gravity, determine how long the sheaf is airborne.

Short Answer

Expert verified
Answer: The total impulse exerted on the sheaf by gravity during its upward and downward motion is -24.5 Ns (upward) and 24.5 Ns (downward). The sheaf is airborne for a total of 0.550 seconds.

Step by step solution

01

Find the time it takes for the sheaf to reach maximum height

To find the time it takes for the sheaf to reach maximum height, we can use the following kinematic equation: v = u + at, where v is the final velocity (0 \(\mathrm{m/s}\), since the sheaf stops momentarily at its highest point), u is the initial velocity (2.7 \(\mathrm{m/s}\)), a is the acceleration due to gravity (-9.81 \(\mathrm{m/s^2}\)), and t is the time we want to find. Rearrange the equation for t: t = (v - u) / a Plug in the given values: t = (0 - 2.7) / (-9.81) = 0.275 s The time it takes for the sheaf to reach its maximum height is 0.275 seconds.
02

Calculate the impulse exerted on the sheaf during its upward motion

Impulse is given by the equation: Impulse = Force × Time In this case, the force acting on the sheaf is gravity, which is the mass of the sheaf multiplied by the acceleration due to gravity: Force = mass × acceleration = 9.09 kg × 9.81 \(\mathrm{m/s^2} = 89.2 N\) (gravity acts downward, so the force is negative) Time = 0.275 s (determined in Step 1) Now, we can calculate the impulse exerted on the sheaf during its upward motion: Impulse = -89.2 N × 0.275 s = -24.5 Ns The impulse exerted on the sheaf by gravity during its upward motion is -24.5 Ns.
03

Calculate the impulse exerted on the sheaf during its downward motion

During the downward motion, the sheaf falls from the maximum height until it hits the ground. In this case, the time of the downward motion is equal to the time of the upward motion, which we already found in Step 1 (0.275 s). As gravity acts in the same direction as the sheaf's motion, the impulse during the downward motion has the same magnitude as that during the upward motion but opposite direction: Impulse = Force × Time = 89.2 N × 0.275 s = 24.5 Ns The impulse exerted on the sheaf by gravity during its downward motion is 24.5 Ns.
04

Calculate the total time the sheaf is airborne

To find the total time the sheaf is airborne, we need to add the time it takes for the sheaf to travel upwards and downwards. Since these times are equal, we can simply double the time we found in Step 1: Total time = 2 × 0.275 s = 0.550 s The sheaf is airborne for a total of 0.550 seconds.

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 Vaia!

Key Concepts

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

Kinematic Equations
Kinematic equations are the cornerstone of classical mechanics. They allow us to predict the future position, velocity, and acceleration of an object moving under the influence of a constant acceleration, such as gravity. Typically, these equations are used to solve problems involving projectiles, like the sheaf toss in the Scottish Highland Games.

One of the fundamental kinematic equations is given by:
\[ v = u + at \]
This equation links the final velocity (\(v\)) of an object to its initial velocity (\(u\)), the acceleration (\(a\)), and the time (\(t\)) it has been accelerating. When discussing the sheaf toss in the problem at hand, the equation allows us to calculate the time taken to reach the maximum height by rearranging it to solve for \(t\). Knowing this time is crucial for calculating other parameters like impulse.
Gravity Force
Gravity force, often represented as \( F = mg \), is an essential concept in physics—it's the force that pulls objects towards the center of the Earth. In this formula, \( m \) stands for the mass of the object and \( g \) stands for the acceleration due to gravity, which is approximately \( 9.81 m/s^2 \) on Earth's surface.

In the sheaf toss scenario, gravity acts continuously on the bag of hay, thus exerting a force that can be calculated by multiplying the sheaf's mass by the gravitational acceleration. This force is what creates the impulses during the upward and downward phases of the sheaf's flight. Even though gravity's pull is constant, the impulses differ because the direction of motion changes. Understanding this force is crucial for analyzing the sheaf's motion and calculating the total impulse exerted on it during its flight.
Airborne Time Calculation
Calculating the time an object remains airborne is a practical application of kinematic equations in physics, especially when analyzing projectile motion. The time an object spends in the air is directly related to its initial launch conditions and the forces acting on it—in most cases, gravity.

For objects projected vertically, like the sheaf, airborne time calculation involves finding the time it takes to reach the maximum height and then doubling it, since the ascent and descent durations are symmetrical in the absence of air resistance. By using the rearranged kinematic equation to find the time to the peak, we determined the sheaf's ascent time. Multiplying it by two gives us the total time the sheaf spends airborne. This duration is fundamental to understanding the dynamics of the toss, such as the peak height and the sheaf's impact velocity when it returns to the ground.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Moessbauer spectroscopy is a technique for studying molecules by looking at a particular atom within them. For example, Moessbauer measurements of iron (Fe) inside hemoglobin, the molecule responsible for transporting oxygen in the blood, can be used to determine the hemoglobin's flexibility. The technique starts with X-rays emitted from the nuclei of \({ }^{57}\) Co atoms. These X-rays are then used to study the Fe in the hemoglobin. The energy and momentum of each X-ray are \(14 \mathrm{keV}\) and \(14 \mathrm{keV} / \mathrm{c}\) (see Example 7.5 for an explanation of the units). \(\mathrm{A}^{57}\) Co nucleus recoils as an \(\mathrm{X}\) -ray is emitted. A single \({ }^{57}\) Co nucleus has a mass of \(9.52 \cdot 10^{-26} \mathrm{~kg} .\) What are the final momentum and kinetic energy of the \({ }^{57}\) Co nucleus? How do these compare to the values for the X-ray?

A particle \(\left(M_{1}=1.00 \mathrm{~kg}\right)\) moving at \(30.0^{\circ}\) downward from the horizontal with \(v_{1}=2.50 \mathrm{~m} / \mathrm{s}\) hits a second particle \(\left(M_{2}=2.00 \mathrm{~kg}\right),\) which was at rest momentarily. After the collision, the speed of \(M_{1}\) was reduced to \(.500 \mathrm{~m} / \mathrm{s}\), and it was moving at an angle of \(32^{\circ}\) downward with respect to the horizontal. Assume the collision is elastic. a) What is the speed of \(M_{2}\) after the collision? b) What is the angle between the velocity vectors of \(M_{1}\) and \(M_{2}\) after the collision?

7.57 A 1439 -kg railroad car traveling at a speed of \(12 \mathrm{~m} / \mathrm{s}\) strikes an identical car at rest. If the cars lock together as a result of the collision, what is their common speed (in \(\mathrm{m} / \mathrm{s})\) afterward?

A golf ball is released from rest from a height of \(0.811 \mathrm{~m}\) above the ground and has a collision with the ground, for which the coefficient of restitution is \(0.601 .\) What is the maximum height reached by this ball as it bounces back up after this collision?

An open train car moves with speed \(v_{0}\) on a flat frictionless railroad track, with no engine pulling it. It begins to rain. The rain falls straight down and begins to fill the train car. Does the speed of the car decrease, increase, or stay the same? Explain.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free