Built on Facts

I think it’s time for a good old practice problem. This is a pretty basic one, which you might encounter as a freshman physics major. The general concept of dealing with inelastic collisions is one that you never escape, and from special relativity to quantum mechanics this type of thing keeps appearing in new contexts. This particular problem is from an old GRE practice exam, but I think solving it is kosher since it’s not original to them and you can’t copyright facts.

In a nonrelativistic, one-dimensional collision, a particle of mass 2m collides with a particle of mass m at rest. If the particles stick together after the collision, what fraction of the initial kinetic energy is lost in the collision?

Let’s look at the concept first. Why should any energy be lost in the collision? Imagine sticking a magnet to your refrigerator. As you bring the magnet close, you’ll feel a force from the attraction of the magnet and the refrigerator surface. That force pulls the magnet in, and a force through a distance is work. The energy involved in this process is the binding energy, and you’ll have to do that much work to pull the magnet away from the refrigerator. On the level of individual particles this happens too. Particles that stick together will have a binding energy associated with their being bound. That’s not the only way colliding things can lose kinetic energy, of course especially for macroscopic objects, heat, sound, and elastic deformation are all possibilities as well.

On to the problem. The initial energy is (1/2)(2m)v^2The initial momentum by the same token is mv. After the collision, the momentum must be conserved. Let’s denote the initial velocity with a 0 and the final velocity of the two particle system with the letter f and write down the conservation of momentum.

Which after collecting terms becomes

Plug that into the kinetic energy equation and we get

Which is 2/3 of the original kinetic energy, meaning 1/3 of the original kinetic energy was lost.

Comments

#1 CRM-114
August 14, 2008

Conservation laws are about the coolest thing in physics. It’s always fun to say, “Momentum is conserved.” That is a pretty big hammer.
#2 Tom Jackson
August 14, 2008

That reminds me of the first time I figured out how a ballistic pendulum works, back in high school physics class.

It was one of those brick-to-the-head moments of total amazement when I realized you could measure something as unmeasurable as bullet velocity with nothing more than a big block of wood and some string, as long as you knew the basic rules and a little math.
#3 IBY
August 14, 2008

What a coincidence, I was just watching a video lecture of mit about momentum and its conservation.It is cool how these equations add up into certain conclusions of how the world works, many of them being somewhat unintuitive.
#4 Mark Harrison
August 14, 2008

When I taught high school physics last year, I did a lecture on a two-dimensional inelastic collision between equal masses with one starting at rest. While I find the fact that the masses’ velocities are perpendicular afterwards to be interesting in its own right, my students weren’t as enthusiastic. So, I took them on a field trip to a billiards hall to demonstrate an important application.
#5 Paul Murray
August 14, 2008

Here’s a practice problem that thinking about whenever I turn a shap corner on the motorbike.

If you had a tall stack of small boxes, and were turning a corner (radius r metres, velocity w rad/sec), what shape would the stack have to be so as not to fall over? Obviously, the stack would tilt in towards the centre of rotation and asymotopically aproach that axis. But that’s as far as I have gotten.

Does the height of the stack of boxes matter? Does their weight?
#6 Paul Murray
August 14, 2008

Further to my little practice problem of my tall stack of small boxes, the stack will be in unstable equilibrium (pretty obviously). Is there a neat way to prove this?
#7 Matt Springer
August 14, 2008

You could leave them as just a straight tower and tilt it inward as you suggest. The angle would need to be such that the tower was pointed along the same direction as the vector sum of gravity and centripetal force. So… the angle would be Arctan((mv^2/r)/(mg)). The mass will cancel, so that gives Arctan((v^2)/(gr)).

As for other possible tower shape configurations, I think the Leaning Tower of Lire is a good starting point. But generally speaking if it’s stable in a turn it’s not necessarily unstable when not turning, and vice versa. The center of gravity actually has to be over the edge of one of the boxes. The unstable equilibrium question actually depends on how you define unstable equilibrium. Formally it’s usually defined as a local maximum of potential energy, which the stack of boxes actually isn’t quite or it would topple from any perturbation.
#8 James Brennan
August 18, 2008

Matt, I have to disagree with you. I think what Paul was suggesting was that the height of the stacked boxes is appreciable with respect to the radius of the curve. Hence, as the stack tilts inward the boxes will no longer all have the same r in the centripetal force law. The r for any given box will depend on the angle of the tilt as well as its height in the stack.
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