9.3: Impulse and Collisions (Part 1) (2024)

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.

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

\[\vec{F}_{ave} = \frac{\vec{J}}{\Delta t} \ldotp\]

From Equation \ref{9.6}, \(\vec{J}\) = m\(\Delta \vec{v}\), so we have

\[\vec{F}_{ave} = \frac{m \Delta \vec{v}}{\Delta t} \ldotp\]

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

\[m = \rho V \ldotp\]

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

\[V = \frac{4}{3} \pi R^{3} \ldotp\]

Thus we obtain

\[\vec{F}_{ave} = \frac{\rho V \Delta \vec{v}}{\Delta t} = \frac{\rho \left(\dfrac{4}{3} \pi R^{3}\right) (\vec{v}_{f} - \vec{v}_{i})}{\Delta t} \ldotp\]

The problem says the velocity at impact was −1.28 x 10⁴ 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

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

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 \(\Delta \vec{v}\).

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

\[\vec{J} = \int_{t_{i}}^{t_{max}} \vec{F} (t)dt \ldotp\]

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

\[F(t) = F_{max} e^{\frac{-t^{2}}{2 \tau^{2}}} \ldotp\]

The parameter \(\tau\) represents how rapidly the force decreases to zero.) The average force is

\[F_{ave} = \frac{1}{\Delta t} \int_{0}^{t_{max}} F_{max} e^{\frac{-t^{2}}{2 \tau^{2}}} dt\]

where \(\Delta\)t = t_max − 0 s. Since we already have a numeric value for F_ave, we can use the result of the integral to obtain F_max. Choosing \(\tau\) = \(\frac{1}{e}\)t_max (this is a common choice, as you will see in later chapters), and guessing that t_max = 2 s, this integral evaluates to

\[F_{avg} = 0.458\; F_{max} \ldotp\]

Thus, the maximum force has a magnitude of

\[\begin{split} 0.458\; F_{max} & = 3.33 \times 10^{12}\; N \\ F_{max} & = 7.27 \times 10^{12}\; N \ldotp \end{split}\]

The complete force function, including the direction, is

\[\vec{F} (t) = (7.27 \times 10^{12}\; N) e^{\frac{-t^{2}}{8\; s^{2}}} \hat{y} \ldotp\]

This is the force Earth applied to the meteor; by Newton’s third law, the force the meteor applied to Earth is

\[\vec{F} (t) = - (7.27 \times 10^{12}\; N) e^{\frac{-t^{2}}{8\; s^{2}}} \hat{y}\]

which is the answer to the original question.

Significance

The graph of this function contains important information. Let’s graph (the magnitude of) both this function and the average force together (Figure \(\PageIndex{4}\)).

Notice that the area under each plot has been filled in. For the plot of the (constant) force F_ave, the area is a rectangle, corresponding to F_ave \(\Delta\)t = J. As for the plot of F(t), recall from calculus that the area under the plot of a function is numerically equal to the integral of that function, over the specified interval; so here, that is \(\int_{0}^{t_{max}}\)F(t)dt = J. Thus, the areas are equal, and both represent the impulse that the meteor applied to Earth during the two-second impact. The average force on Earth sounds like a huge force, and it is. Nevertheless, Earth barely noticed it. The acceleration Earth obtained was just

\[\vec{a} = \frac{- \vec{F}_{ave}}{M_{Earth}} = \frac{- (3.33 \times 10^{12}\; N) \hat{j}}{5.97 \times 10^{24}\; kg} = - (5.6 \times 10^{-13} m/s^{2}) \hat{j}\]

which is completely immeasurable. That said, the impact created seismic waves that nowadays could be detected by modern monitoring equipment.

Solution

1. Define the +x-direction to be the direction the car is initially moving. We know $$\vec{J} = \vec{F} \Delta t$$and $$\vec{J} = m \Delta \vec{v} \ldotp$$Since J is equal to both those things, they must be equal to each other: $$\vec{F} \Delta t = m \Delta \vec{v} \ldotp$$We need to convert this weight to the equivalent mass, expressed in SI units: $$\frac{860\; N}{9.8\; m/s^{2}} = 87.8\; kg \ldotp$$Remembering that \(\Delta \vec{v} = \vec{v}_{f} − \vec{v}_{i}\), and noting that the final velocity is zero, we solve for the force: $$\vec{F} = m \frac{0 - v_{i}\; \hat{i}}{\Delta t} = (87.8\; kg) \left(\dfrac{-(27\; m/s) \hat{i}}{2.5\; s}\right) = - (948\; N) \hat{i} \ldotp$$The negative sign implies that the force slows him down. For perspective, this is about 1.1 times his own weight.
2. Same calculation, just the different time interval: $$\vec{F} = (87.8\; kg) \left(\dfrac{-(27\; m/s) \hat{i}}{0.20\; s}\right) = - (11,853\; N) \hat{i} \ldotp$$which is about 14 times his own weight. Big difference!

Significance

You see that the value of an airbag is how greatly it reduces the force on the vehicle occupants. For this reason, they have been required on all passenger vehicles in the United States since 1991, and have been commonplace throughout Europe and Asia since the mid-1990s. The change of momentum in a crash is the same, with or without an airbag; the force, however, is vastly different.