Question 8.8: Collision analysis, continued Let’s revisit Example 8.1, in ...
Collision analysis, continued
Let’s revisit Example 8.1, in which we considered the impending collision of a small car with a larger truck. It’s now 2 seconds later, and the collision has occurred. Fortunately, all occupants were wearing seat belts and there were no injuries, but the two vehicles became thoroughly tangled and moved away from the point of impact as one mass. The insurance adjuster has asked you to help find the velocity of the wreckage just after impact.
SET UP Figure 8.14 shows our sketches. The “before” diagram is like the one in Example 8.1. We worked out the initial momentum components in that example. Review that calculation before proceeding. To find out what happens after the collision, we’ll assume that we can treat the vehicles as an isolated system during the collision. This approach may seem implausible; the pavement certainly exerts substantial friction forces on the tires. But the interaction forces between the vehicles during the collision are much larger—so much larger, in fact, that they crumple the vehicles. So it’s reasonable to ignore the friction forces and consider the two vehicles together as an isolated system during the impact.
SOLVE The total momentum \overrightarrow{\pmb{P}} of the vehicles just after the collision
is the same as we calculated in Example 8.1: 2.5 \times 10^4 kg \cdot m/s in a direction 36.9° north from straight east. Assuming that no parts fall off the wreckage, the total mass of the wreckage is M = 3000 kg. Calling the final velocity \overrightarrow{\pmb{V}}, we have \overrightarrow{\pmb{P}}= M \overrightarrow{\pmb{V}}. The direction of the velocity \overrightarrow{\pmb{V}}, is the same as that of the momentum \overrightarrow{\pmb{P}} , and its magnitude is
V=\frac{P}{M}=\frac{2.5 \times 10^4 kg \cdot m/s}{8.3 m/s}= 8.3 m/s (final velocity).
This is an inelastic collision, so we expect the total kinetic energy to be less after the collision than before. When you make the calculations, you’ll find that the initial kinetic energy is 2.1 \times 10^5 J and the final kinetic energy is 1.0 \times 10^5 J. More than half of the initial kinetic energy is converted to other forms of energy.
REFLECT Were you tempted to find the final velocity by taking the vector sum of the initial velocities? If so, ask yourself why you would expect that to work. Is there a law of conservation of velocities? Absolutely not; the conserved quantity is the total momentum of the system.
Let’s look again at the matter of ignoring the friction of tires against pavement and treating the two vehicles as an isolated system.
The mass of the truck is 2000 kg, so its weight is about 20,000 N. If the coefficient of friction is 0.5, the friction force is about 10,000 N, which sounds like a large force. But suppose the truck runs into a brick wall while going 10 m>s. Its kinetic energy just before impact is \frac{1}{2} (2000 kg)(10 m/s)^2 = 1.0 \times 10^5 J, so the force applied by the wall must do -1.0 \times 10^5 J of work on the truck to stop (and crumple) it. The truck may crumple 0.2 m or so; for the stopping force to do -1.0 \times 10^5 J of work in 0.2 m, the force must have a magnitude of 5.0 \times 10^5 N, or 50 times as great as the friction force. So, it is reasonable to ignore the friction forces on both vehicles and assumethat during the collision they form an isolated system.
Practice Problem: Suppose that just before the collision, the driver of the car swerves, so that the car is still traveling at a speed of 15 m/s, but in the direction of 15° east of north when the car collides with the truck. If the two vehicles stick together after the collision, find the velocity (magnitude and direction) of the wreckage. Answers: 9.3 m/s, 31° N of E.
