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Chapter 9: Problem 26

In a common but dangerous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land hard on the floor. Suppose the victim falls by \(0.50 \mathrm{~m}\), the mass that moves downward is \(70 \mathrm{~kg}\), and the collision on the floor lasts \(0.082 \mathrm{~s}\). What are the magnitudes of the (a) impulse and (b) average force acting on the victim from the floor during the collision?

Short Answer

Expert verified
Impulse is 219.1 Ns; average force is 2673.2 N.

Step by step solution

01

Determine the Final Velocity

To find the final velocity when the person hits the ground, we'll use the equation of motion without initial velocity. The person falls under the influence of gravity (assuming no air resistance). Use the equation:\[v^2 = u^2 + 2as\]where:- \(v\) is the final velocity,- \(u = 0\) is the initial velocity (since the person starts from rest),- \(a = 9.8 \, \text{m/s}^2\) is the acceleration due to gravity,- \(s = 0.50 \, \text{m}\) is the distance fallen.Plug in the values:\[v^2 = 0 + 2 \times 9.8 \, \text{m/s}^2 \times 0.50 \, \text{m}\]Solving gives:\[v = \sqrt{9.8} \, \text{m/s}\]Thus, the final velocity \(v\) is approximately \(3.13 \, \text{m/s}\).
02

Calculate the Impulse

Impulse is given by the change in momentum, or equivalently, the product of average force and time. The formula for impulse \(J\) is:\[J = \Delta p = m(v - u)\]where:- \(m = 70 \, \text{kg}\) is the mass,- \(v = 3.13 \, \text{m/s}\) is the final velocity (as calculated),- \(u = 0\) is the initial velocity.Substitute the values:\[J = 70 \, \text{kg} \times (3.13 \, \text{m/s} - 0) \]\[J = 219.1 \, \text{Ns}\]So, the impulse is \(219.1 \, \text{Ns}\).
03

Calculate the Average Force

The average force \(F_{avg}\) exerted on the victim can be determined from the impulse-momentum theorem, which relates impulse to the product of force and time:\[F_{avg} \times \Delta t = J\]where:- \(\Delta t = 0.082 \, \text{s}\) is the time duration of the collision,- \(J = 219.1 \, \text{Ns}\) is the impulse from Step 2.Rearranging for \(F_{avg}\):\[F_{avg} = \frac{J}{\Delta t} = \frac{219.1 \, \text{Ns}}{0.082 \, \text{s}} \]\[F_{avg} = 2673.2 \, \text{N}\]Therefore, the average force acting on the victim is approximately \(2673.2 \, \text{N}\).

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Key Concepts

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

The Impulse-Momentum Theorem
When considering how an object changes its motion, the impulse-momentum theorem is an essential concept. It describes how the impulse on an object relates to its change in momentum. Simply put, impulse is the product of the average force applied to an object and the time duration for which the force is applied. This impulse results in a change in momentum, which is the product of the mass and velocity of the object.

For example, in the case of a person falling and colliding with the floor, calculating the impulse involves the following elements:
  • Mass of the person (70 kg in our scenario).
  • Final velocity of the person just before impact (this is found using the equations of motion).
  • Initial velocity, which is zero if the fall started from rest.
  • Time duration of the collision, which is given as 0.082 seconds.
The impulse can be used to determine the resulting average force during the collision by the formula \(J = \Delta p = m(v - u)\).
This shows how the impulse not only helps in quantifying the change in motion but also allows us to calculate the forces involved in such unfortunate accidents.
Equations of Motion
Equations of motion play a pivotal role in solving problems related to moving objects, like our falling person scenario. These equations allow us to calculate various parameters like final velocity, distance, and more, given the initial conditions and constant acceleration.

The main equation used in the solution is:
  • \(v^2 = u^2 + 2as\)
In this case:
  • \(u\) is the initial velocity, which is zero since the person starts from rest.
  • \(a\) is the acceleration due to gravity (9.8 m/s虏).
  • \(s\) is the distance fallen (in the original problem, it is 0.50 meters).
Substituting these values allows us to solve for the final velocity \(v\) just before the collision.
This equation is crucial as it gives insight into how objects behave when subjected to constant forces, such as gravity in free fall. By understanding and applying these equations, you can predict and analyze the motion of objects more effectively.
Acceleration Due to Gravity
Earth鈥檚 gravity is a constant force that pulls objects toward its center, and it's one of the most familiar forces observed in physics. Known as acceleration due to gravity, this constant is denoted by \(g\) and is approximately 9.8 m/s虏 on Earth's surface.

This acceleration affects all free-falling objects, causing them to increase their velocity as they fall. It's a key factor in calculating motion, especially in problems involving motion under gravity. In these calculations:
  • The rate of acceleration \(g\) does not depend on the object's mass.
  • Gravity acts downwards towards the center of the Earth.
In our exercise, we assume that there is no air resistance, meaning gravity is the sole factor altering the person's velocity as they fall. This allows us to use simplistic kinematic equations to find the speed just before impact.

Understanding gravity is fundamental not only in physics problems such as this but also in explaining natural phenomena and even in advanced fields like astrophysics and engineering.

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Most popular questions from this chapter

A force in the negative direction of an \(x\) axis is applied for \(27 \mathrm{~ms}\) to a \(0.40 \mathrm{~kg}\) ball initially moving at \(14 \mathrm{~m} / \mathrm{s}\) in the positive direction of the axis. The force varies in magnitude, and the impulse has magnitude \(32.4 \mathrm{~N} \cdot \mathrm{s}\). What are the ball's (a) speed and (b) direction of travel just after the force is applied? What are (c) the average magnitude of the force and (d) the direction of the impulse on the ball?

A \(6100 \mathrm{~kg}\) rocket is set for vertical firing from the ground. If the exhaust speed is \(1200 \mathrm{~m} / \mathrm{s}\), how much gas must be ejected each second if the thrust (a) is to equal the magnitude of the gravitational force on the rocket and (b) is to give the rocket an initial upward acceleration of \(21 \mathrm{~m} / \mathrm{s}^{2}\) ?

A \(0.30 \mathrm{~kg}\) softball has a velocity of \(15 \mathrm{~m} / \mathrm{s}\) at an angle of \(35^{\circ}\) below the horizontal just before making contact with the bat. What is the magnitude of the change in momentum of the ball while in contact with the bat if the ball leaves with a velocity of (a) \(20 \mathrm{~m} / \mathrm{s}\), vertically downward, and (b) \(20 \mathrm{~m} / \mathrm{s}\), horizontally back toward the pitcher?

A steel ball of mass \(0.500 \mathrm{~kg}\) is fastened to a cord that is \(70.0 \mathrm{~cm}\) long and fixed at the far end. The ball is then released when the cord is horizontal (Fig. \(9-65\) ). At the bottom of its path, the ball strikes a \(2.50 \mathrm{~kg}\) steel block initially at rest on a frictionless surface. The collision is elastic. Find (a) the speed of the ball and (b) the speed of the block, both just after the collision.

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?
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