Two-body problem quantitative demonstration

Consider the classic two-body problem, in which a cart on a horizontal low-friction track is attached via rope-and-pulley to a hanging mass.  The cart has mass 340 g, and the hanginging mass is 100 g.  I release the cart, which speeds up.

Is the tension in the rope...
(A) greater than 1 N
(B) less than 1 N
(C) equal to 1 N?

"Equal," say a majority of the students -- because the rope is attached to the 100 g hanging mass, which has a weight of 1 N.

"Less than 1 N," the cleverer ones respond.  The hanging mass has a downward acceleration, because it is moving down and speeding up.  So the down forces must be GREATER THAN the up forces, meaning the tension is less than the 1 N weight.

"Let's see," I say.  The picture shows my cart with a Vernier force probe taped inelegantly on top.  The string is attached to the force probe, is run over a low-friction Pasco pulley, and connected to a 100 g hanging mass.  I tell Vernier's Logger Pro software to collect force probe data.  Before I let go of the cart, the probe reads 1.0 N.  As soon as I let go, the force probe's reading very obviously dips, to something like 0.7 - 0.8 N.  Looks like the cleverer ones were right.

Next, I use free body diagrams and Newton's second law to predict the acceleration of the cart and the tension in the rope -- I get 2.3 m/s² and 0.77 N.  Sure enough, that's just about what the force probe and a motion detector read.  (I get the acceration from the slope of a velocity-time graph, which I make with the vernier motion detector.)

Next question:  Instead of letting the cart go from rest, I give the cart a shove to the right, away from the rope. 

After I let go but as the cart is still moving to the right, is the tension...
(A) greater than 1 N
(B) less than 1 N
(C) equal to 1 N

"Greater than 1 N," say the majority.  "The hanging mass is now moving upward, so the up forces must be greater than the down forces."

"Nonsense," I say.  The mass is slowing down while it moves upward.  Slowing down means that acceleration and velocity are in opposite directions.  Thus, the acceleration must still be in the downward direction, and the tension must still be less than the weight.

In fact, the entire set of free bodies is unchanged from the previous problem -- the force that I pushed with doesn't act once I let go!  So, all calculations are the same, and the acceleration and tension should be unchanged.

Of course, I finish the class by doing the experiment -- sure enough, the tension and acceleration readings are the same as before.

GCJ