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Getting the most of out standard kinematics questions

January 26, 2012

A question I’ve gotten a lot of leverage out the past two semesters is the following one:

You toss your keys straight up to a friend, who is 30m above you leaning out over a balcony. They keys leave your hand with a speed of 25 m/s. Will it get to your friend?

Sure this is a standard boring question. What makes it work is how the show is run. We start off by listing our best guesses about whether it makes it up and the top height they think it gets to: Their answers this semester ranged between 19m and 40m.

In my class, I actually work out this first answer for them (because I’m supposed to model a sample problem), but I ask for their help along the way.

First, I draw a motion map showing how the speed changes at 1s intervals, and we talk about the speed going from 25m/s to 15m/s to 5 m/s, etc, and how the time to the top is when v = 0 m/s. We talk about how much time it takes to get to 0 m/s if you are losing 10 m/s each second: it takes 2.5s to lose 25 m/s. We also talk about the average speed during the trip (12.5 m/s, half way in between 0 m/s and 25 m/s). This, of course, all builds on ideas we built up last week when talking about 1D acceleration problems.

The answer is immediately given as 12.5 m/s * 2.5s = 31.25m

The best guess this time was 32m, and kudos were given to that group.

Lot’s of students then want to talk about why it’s not 40 m (25m + 15m + 10m), and we get to talk about what constantly changing velocity means.

Because of class constraints, I typically re-derive the 31.25m in a way that is more typical of how they are expected to do it: Write down your knowns and unknowns and pick an equation or two to plug away with.

I then send them off to work on the next question. How fast are the keys moving by the time they reach your friend’s hand? Our guesses range between 1.25 m/s and 2.5 m/s.

The right answer is 5 m/s. And students are pretty surprised to find out that we all underestimated the speed. Every group got the right answer. Most students solved the problem by plugging away into equations. One group did so, but didn’t believe that 5 m/s was right, and so they took another approach, using two equations instead of one.

One group took this approach:

In the first second, the ball slows from 25 to 15, with an average velocity of 20 m/s. Thus in the first second, the ball covers 20 m. In the second second, the balls slows from 15 to 5, with an average velocity of 10 m/s. Thus in the second second, the ball covers 10m. That’s 30m covered, with a final speed of 5 m/s. That same group realized that for the first 2 seconds, the average speed was 15 m/s for 2 seconds, also giving 30m of travel.

Last semester, I had a group solve the problem by finding the speed of a ball dropped 1.25 m/s, arguing on the ground of symmetry that it had to be the same.

We ended the problem this semester by talking about the last 1/2 second, where the ball has an average speed of 2.5 m/s for 1/2 second, thus covering the final 1.25m, and why our guesses for the speed were so off.

Simple problem, but lots of places for intuition, lots of places for multiple approaches, and lots of opportunities to talk about velocity, distance, average velocity, and acceleration.

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