http://youtu.be/wLpR-RvPZfY

http://youtu.be/8IdRhQM_Lkw

Careful observations have shown that when a 'doubled' coffee filter is dropped from a height of 200 cm, the single filter dropped simultaneously from a height of 128 cm will strike the floor at the same time.

• The possibility that one filter presents a greater area to air friction was controlled by switching filters and confirming that this has no effect on the relative heights from which the filters must be dropped.
• If we assume that the filters always reach terminal velocity in negligible time, we conclude that the average velocity of a falling filter is equal to its terminal velocity.

• To the extent that our assumption is incorrect, the terminal velocity will be somewhat greater than the average velocity.
• We also assume that every filter has mass m, which is the same for all filters.

http://youtu.be/SO0nKsYkTSg

Noting that the force of air resistance for a freely falling object is equal and opposite to the weight of the object, and that the weight of a single filter is the force m * g exerted on it by gravity, we see that the velocities v1, 1.6 v1, 2.2 v1 and 2.6 v1 are associated with respective forces m * g, 2 m * g, 3 m * g and 4 m * g.

• Graphing these results we obtain the data points and approximate curve sketched at the right of the figure below.
• This result is in poor agreement with the result expected in light of carefully done experiments conducted over the last two or three centuries, obviously reflecting some sort of perverse behavior by the coffee filters used for the experiment.

http://youtu.be/jssj_--mpjI

Today's quiz problem concerned a 10 kg object moving at 20 m/second at 230 degrees with respect to the positive x axis, colliding with a 20 kg object moving at 12 m/second at 130 degrees with respect to the positive x axis.

We begin by finding the components of the momentum of each object.

• The momentum of the first object is clearly 200 kg m/second at 230 degrees.
• The x and y components of this momentum are easily found using the cosine and sine functions.

Adding the x and y components, we obtain a total x momentum component of -282 kg m/s and a total y component of 31 kg m/s.

• These momenta are sketched in the 'green' graph below.
• The magnitude and angle of this total momentum are calculated using the Pythagorean Theorem and the inverse tangent, as in the lower right-hand corner of the figure below.
• Note that since the x component of the total momentum is negative we add 180 degrees to the inverse tangent, obtaining angle `theta = 174 degrees. This angle is reasonably consistent with our sketch.

http://youtu.be/s75qX9S17so

If the collision is inelastic we end up with a 30 kg object with a total momentum of 284 kg m/second at 174 degrees.

• From this we easily calculate the velocity, equal to the momentum divided by the mass.
• The velocity we obtain is 9.47 m/second at 174 degrees.

To determine whether energy is conserved in this collision we first calculate the kinetic energy before collision, which is the sum of the kinetic energies of the 10 kg object moving at 20 m/second and the 20 kg object moving at 10 m/second.

Depending on the angle at which the objects collide, if the collision is not inelastic the second object might end up with a wide variety of speeds and directions.

• In fact, the only limit on possible collisions (except for those involving a release of additional energy, as in explosions), is that momentum must be conserved and the total kinetic energy after collision cannot be greater than that before collisions.

Suppose for example that after collision the second object was moving at 4 m/second at 310 degrees from a positive x axis.

• We first calculate the components of the momentum of the second object, obtaining something like p2x = 25 kg m/second and p2y = -27 kg m/second (verify these calculations, which have not been validated).

http://youtu.be/XcLSA8AIc04

We therefore have the situation depicted in the somewhat poorly drawn figure below.

• The total momentum has components -282 kg m/second and 31 kg m/second in the x and y directions, respectively.
• This total is made up of the momentum of the second object, with its x and y components of 25 kg m/second and -27 kg m/second, and the unknown momentum of the first object.

• It follows that the x momentum of the first object, when added to the existing 25 kg m/second momentum of the second, must give a result of -282 kg m/second.
• We easily conclude that the x momentum of the first object is therefore -307 kg m/second.

• Similarly, in order to obtain the 31 kg m/second total momentum we must add 58 kg m/second to the existing -27 kg m/second, so that the y momentum of the first object must be 58 kg m/second.

http://youtu.be/hn2RSav4yS8

We could have calculated the momentum p1 more formally by noting that since p1 + p2 = pTotal, p1 = pTotal - p2.

• This indicates that in order to get the components of p1, we must therefore subtract the components of p2 from those of pTotal.
• We would therefore obtain p1x = -282 kg m/second - 25 kg m/second = -307 kg m/second, and p1y = 31 kg m/second - (-27 kg m/second) = 58 kg m/second, in agreement with the previous results.