Problem 62 - University Physics with Modern Physics

Question

People (especially the elderly) who are prone to falling can wear hip pads to cushion the impact on their hip from a fall. Experiments have shown that if the speed at impact can be reduced to 1.3 \(\mathrm{m} / \mathrm{s}\) or less, the hip will usually not fracture. Let us investigate the worst-case scenario in which a 55 -kg person completely loses her footing (such as on icy pavement) and falls a distance of \(1.0 \mathrm{m},\) the distance from her hip to the ground. We shall assume that the person's entire body has the same acceleration, which, in reality, would not quite be true. (a) With what speed does her hip reach the ground? (b) A typical hip pad can reduce the person's speed to 1.3 \(\mathrm{m} / \mathrm{s}\) over a distance of 2.0 \(\mathrm{cm} .\) Find the acceleration (assumed to be constant) of this person's hip while she is slowing down and the force the pad exerts on it. (c) The force in part (b) is very large. To see whether it is likely to cause injury, calculate how long it lasts.

Step-by-Step Solution

Answer

The hip reaches the ground at 4.43 m/s. Deceleration is -447.75 m/s², force is -24626.25 N, lasting 0.007 s.

Step 1: Understanding the Problem

We need to calculate the speed at which a 55-kg person hits the ground after falling from a height of 1.0 m. Then, find the acceleration when her speed is reduced by a hip pad. Lastly, determine the duration the force from the pad is applied.

Step 1: Calculating Speed Upon Impact

To find the speed just before impact, we use the kinematic equation: \\[ v^2 = u^2 + 2as \] \where \( u = 0 \ \mathrm{m/s} \) (initial speed), \( a = 9.8 \ \mathrm{m/s^2} \) (acceleration due to gravity), and \( s = 1.0 \ \mathrm{m} \) (fall distance). \ Plugging in the values, we get: \\[ v^2 = 0 + 2 \times 9.8 \times 1.0 \] \\[ v = \sqrt{19.6} \approx 4.43 \ \mathrm{m/s} \] \Thus, the person's hip will hit the ground at approximately \( 4.43 \ \mathrm{m/s} \).

Step 2: Finding Deceleration Due to Hip Pad

Using the kinematic equation: \\[ v_f^2 = v_i^2 + 2as \] \where \( v_f = 1.3 \ \mathrm{m/s} \), \( v_i = 4.43 \ \mathrm{m/s} \), and \( s = 0.02 \ \mathrm{m} \), solving for acceleration \( a \): \\[ 1.3^2 = 4.43^2 + 2a \times 0.02 \] \\[ 1.69 = 19.6 + 0.04a \] \\[ 0.04a = 1.69 - 19.6 \] \\[ a = \frac{-17.91}{0.04} \approx -447.75 \ \mathrm{m/s^2} \] \Thus, the acceleration is approximately \(-447.75 \ \mathrm{m/s^2}\).

Step 3: Calculating Force Exerted by Hip Pad

Using Newton's second law, \( F = ma \), where \( m = 55 \ \mathrm{kg} \) and \( a = -447.75 \ \mathrm{m/s^2} \): \\[ F = 55 \times (-447.75) \approx -24626.25 \ \mathrm{N} \] \The force exerted by the hip pad is approximately \(-24626.25 \ \mathrm{N}\).

Step 4: Determining Duration of Force Application

Using the kinematic equation: \\[ v_f = v_i + at \] \where \( v_f = 1.3 \ \mathrm{m/s} \), \( v_i = 4.43 \ \mathrm{m/s} \), and \( a = -447.75 \ \mathrm{m/s^2} \): \\[ 1.3 = 4.43 + (-447.75)t \] \\[ -3.13 = -447.75t \] \\[ t = \frac{-3.13}{-447.75} \approx 0.007 \ \mathrm{s} \] \The force from the pad lasts approximately \(0.007 \ \mathrm{s}\).

Key Concepts

Kinematics

Kinematics is the branch of physics that deals with the motion of objects without considering the causes of this motion. It involves parameters such as displacement, velocity, and acceleration. In the context of our problem, understanding kinematics is crucial in determining the speed at which a person hits the ground after falling from a certain height.

To calculate this speed, we use one of the fundamental kinematic equations: \[v^2 = u^2 + 2as\]where

• \( v \) is the final velocity, which we want to find,
• \( u \) is the initial velocity,
• \( a \) is the acceleration due to gravity (approximately \( 9.8 \ \mathrm{m/s^2} \)),
• \( s \) is the displacement (or height from which the person falls).

In this scenario, the person falls from rest, so the initial velocity \( u = 0 \ \mathrm{m/s} \). Given the downward displacement \( s = 1.0 \ \mathrm{m} \), we can compute the final velocity \( v \) just before impact. By substituting these values into the equation, we find \( v \approx 4.43 \ \mathrm{m/s} \), indicating the speed with which the person reaches the ground.

Newton's Laws of Motion

Newton's Laws of Motion form the cornerstone of classical mechanics, and they play a vital role in understanding the interactions between forces and the motion of objects. In this problem, Newton's second law is particularly important, as it helps us connect forces with acceleration.

According to Newton's second law, the net force \( F \) acting on an object is equal to the mass \( m \) of the object multiplied by its acceleration \( a \):\[F = ma\]In the case of the hip pad reducing the person's speed, this formula allows us to calculate the amount of force exerted by the hip pad on the person during the deceleration process. Here, the person has a mass of \( 55 \ \mathrm{kg} \), and we have previously calculated the deceleration as \( a \approx -447.75 \ \mathrm{m/s^2} \).

Using these values, the force \( F \) exerted by the hip pad can be calculated. This results in a force of approximately \(-24626.25 \ \mathrm{N} \). Newton's second law reveals why impact forces, such as those experienced in falls, can be very large and potentially injurious.

Impact Forces

Impact forces are intense forces that occur during a collision or sudden stop, such as when a person falls and hits the ground or when a hip pad slows them down. The magnitude of these forces and their duration are critical factors in determining the potential for injury.

When an object strikes a surface, it often does so at high speed, and impact forces can be quite significant if the deceleration happens over a very short distance or time. In our problem, although the hip pad reduces the speed to a safer 1.3 \( \mathrm{m/s} \) over a 2.0 \( \mathrm{cm} \) distance, the deceleration required is very high, calculated at about \(-447.75 \ \mathrm{m/s^2} \). This significant deceleration leads to a large force of \(-24626.25 \ \mathrm{N} \) exerted by the hip pad.

The duration of force application is crucial as well, as prolonged exposure to high forces increases the risk of injury. In the given exercise, the duration of this force is approximately \( 0.007 \ \mathrm{s} \), which is relatively short. Short durations generally help reduce the likelihood of severe injury, as forces are less likely to have a lasting damaging effect.

• Impact force magnitude: Dependent on deceleration distance/time.
• Key to injury prevention: Shorter duration and effective cushioning.

Understanding these parameters allows us to design better protective gear, such as hip pads, to enhance safety.