9.2 Impulse and collisions (Page 2/10)

University physics volume 1 Page 2 / 10

\vec{J} = m Δ \vec{v} .

Note that the integral form, [link] , applies to constant forces as well; in that case, since the force is independent of time, it comes out of the integral, which can then be trivially evaluated.

The arizona meteor crater

Approximately 50,000 years ago, a large (radius of 25 m) iron-nickel meteorite collided with Earth at an estimated speed of

1.28 \times 10^{4} m/s

in what is now the northern Arizona desert, in the United States. The impact produced a crater that is still visible today ( [link] ); it is approximately 1200 m (three-quarters of a mile) in diameter, 170 m deep, and has a rim that rises 45 m above the surrounding desert plain. Iron-nickel meteorites typically have a density of

ρ = 7970 {kg/m}^{3}

. Use impulse considerations to estimate the average force and the maximum force that the meteor applied to Earth during the impact.

A photo of the Arizona meteor crater. Buildings near the crater are tiny compared to the crater. — The Arizona Meteor Crater in Flagstaff, Arizona (often referred to as the Barringer Crater after the person who first suggested its origin and whose family owns the land). (credit: “Shane.torgerson”/Wikimedia Commons)

Strategy

It is conceptually easier to reverse the question and calculate the force that Earth applied on the meteor in order to stop it. Therefore, we’ll calculate the force on the meteor and then use Newton’s third law to argue that the force from the meteor on Earth was equal in magnitude and opposite in direction.

Using the given data about the meteor, and making reasonable guesses about the shape of the meteor and impact time, we first calculate the impulse using [link] . We then use the relationship between force and impulse [link] to estimate the average force during impact. Next, we choose a reasonable force function for the impact event, calculate the average value of that function [link] , and set the resulting expression equal to the calculated average force. This enables us to solve for the maximum force.

Solution

Define upward to be the + y -direction. For simplicity, assume the meteor is traveling vertically downward prior to impact. In that case, its initial velocity is

{\vec{v}}_{i} = − v_{i} \hat{j}

, and the force Earth exerts on the meteor points upward,

\vec{F} (t) = + F (t) \hat{j}

. The situation at

t = 0

is depicted below.

An x y coordinate system is shown. The region below the x axis is shaded and labeled Earth. A meteor is shown at the origin. An upward arrow at the origin is labeled F vector (t). A downward arrow at the origin is labeled p sub 0 vector equals m times v sub 0 vector.

The average force during the impact is related to the impulse by

{\vec{F}}_{ave} = \frac{\vec{J}}{Δ t} .

From [link] , $\vec{J} = m Δ \vec{v}$ , so we have

{\vec{F}}_{ave} = \frac{m Δ \vec{v}}{Δ t} .

The mass is equal to the product of the meteor’s density and its volume:

m = ρ V .

If we assume (guess) that the meteor was roughly spherical, we have

V = \frac{4}{3} π R^{3} .

Thus we obtain

{\vec{F}}_{ave} = \frac{ρ V Δ \vec{v}}{Δ t} = \frac{ρ (\frac{4}{3} π R^{3}) ({\vec{v}}_{f} - {\vec{v}}_{i})}{Δ t} .

The problem says the velocity at impact was $−1.28 \times 10^{4} m/s \hat{j}$ (the final velocity is zero); also, we guess that the primary impact lasted about $t_{max} = 2 s$ . Substituting these values gives

\begin{matrix} {\vec{F}}_{ave} & = \frac{(7970 \frac{kg}{m^{3}}) [\frac{4}{3} π {(25 m)}^{3}] [0 \frac{m}{s} - (−1.28 \times 10^{4} \frac{m}{s} \hat{j})]}{2 s} \\ = + (3.33 \times 10^{12} N) \hat{j} \end{matrix} .

This is the average force applied during the collision. Notice that this force vector points in the same direction as the change of velocity vector $Δ \vec{v}$ .

Next, we calculate the maximum force. The impulse is related to the force function by

\vec{J} = \int_{t_{i}}^{t_{max}} \vec{F} (t) d t .

We need to make a reasonable choice for the force as a function of time. We define $t = 0$ to be the moment the meteor first touches the ground. Then we assume the force is a maximum at impact, and rapidly drops to zero. A function that does this is

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Source: OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5

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