Tutor profile: Frank D.

A top fuel dragster driver going 200 mph has the brakes and the parachute on their vehicle fail, and she slams into a series of nets and barrels full of sand at the end of the track, crashing to a stop. The driver exits the vehicle uninjured. Later, another driver has a serious concussion from merely tapping into the divider rail between the two lanes (made of concrete). For that driver, they impacted the wall at a mere 30 mph relative to the wall, and bounced back in the opposite direction. Explain why the first driver was not injured while the second one was using the momentum-impulse equation, F • t = m • ∆v. (Concepts, not calculations.) [Note- this is a problem I actually created based on an actual pair of scenarios involving the same top fuel driver, Alexis DeJoria, having two different accidents within a year, with very different results. It is a discrepant event- why did she survive the dramatic crash without injury, but the seemingly small bump into the concrete barrier gave her a serious injury?]

The first driver crashed from a very high speed to a complete stop. She had a large amount of momentum- the right-hands side of the equation- but in looking at the left side F • t , they had a relatively small amount of forces bringing them to a stop over a larger period of time. Simply put- big t, small F. Having lesser forces applied to decelerate a crashing vehicle is highly desirable. Compare the second driver. While they had a smaller amount of momentum to be reversed by the concrete divider, the time (t) over which that Force (F) applied was very small. A smaller time means a larger force, for the same total impulse F • t. Imagine the second driver had instead slammed into a barrier made of very elastic rubber. The time of the impact between the vehicle and barrier would be a lot longer, but if the resulting change in momentum is more or less the same, the force in the equation by necessity MUST be smaller. A smaller force acting over a longer period of time is always desirable to a large force acting over a short period of time, when safety is the main objective. Additionally, the first driver was merely brought to a stop (a certain change in momentum), while the second one had their direction of motion completely reversed....bounced. That's twice as much of a change in momentum (requiring twice the impulse) as merely being brought to a stop. That additional momentum change contributed to the forces being high and injuring the driver.

As you add incandescent light bulbs connected together in a circuit, in which case will they increase in brightness, when connected in series, or in parallel? Explain your answer.

The answer is connected in parallel. The brightness of the bulbs will depend on the current flowing through them. There will be more current for each bulb in a parallel circuit, because there are more pathways for the electrons to flow along in a parallel circuit than a series one. Think of the analogy of water flowing in pipes, or cars moving into and out of a parking structure. If there are more pathways for the water to travel along, in general, the flow rate will be higher. If there are more entrances and exits for cars to get into a parking structure, traffic (current) will flow faster.

Newton's 3rd law of Motion seems simple enough- for every action there's an equal and opposite reaction. No problem. However, in application, even professionals very easily make mistakes in its use. For example- an apple is sitting at rest on a table. Call the force of gravity acting on the apple (its weight) the ACTION force. Question- what is the reaction force to the apple's weight?

Many people will answer that the reaction force must be the table pushing back up on the apple. This is not correct. That force is acting on the apple, but it is not the reaction force to its weight. To use Newton's 3rd Law and get it right every time, 1) isolate the two objects involved in the force described. In this case, "the force of gravity acting on the apple", ask yourself, what object is causing that force on the apple? You hopefully guessed that it is the Earth itself. You have now found your second object in the two-object interaction between apple and Earth. 2) Ignore all other objects in the problem (the table in this case). 3) Remember this rule: If Object A exerts a force on Object B, the reaction force is Object B exerting the same amount of force back on Object A...in the opposite direction. So for our problem- Earth pulls down on apple. Ignore the table, since there was no description of the FORCE that the table puts on the apple (there is a "force pair" between the apple and table as well, but that's not what we're focusing on). Per the rule in step 3 above- If Earth pulls down on apple...apple pulls UP on Earth. That's it! The reaction force is the apple's mass pulling back up on the whole Earth! The above may seem strange, especially when you remember that the apple pulls on the earth just as "hard" (with the same amount of force) as the Earth pulls on the apple. But that roughly 1 newton of force acting on a tiny apple has a big effect (acceleration), compared to the SAME amount of force acting on the huge mass of the Earth (almost no acceleration, one that could never be measured).