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# 9.3: Impulse and Collisions (Part 1)

Strategy

We are given the driver’s weight, his initial and final velocities, and the time of collision; we are asked to calculate a force. Impulse seems the right way to tackle this; we can combine Equation \ref{9.5} and Equation \ref{9.6}.

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.

# Contributors

• Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).