Recently our physics class tried an experiment in class where we measured impulse by having a cart slide off a basically frictionless ramp that's inclined. The cart hits a spring which is attached to a force sensor. We let the cart collide, and then stop the cart and watch the force sensor record the force vs. time graph.
We get the usual impulse peak, but immediately after that there are some really small bumps on the graph, and after that the graph levels out and has a slope of 0 again.
What do these bumps indicate, because this is happening after the cart has already gotten off the spring? How does this affect the momentum of the system?
2007-11-11 04:38:09 · 3 answers · asked by J Z 4 in Science & Mathematics ➔ Physics
I'd agree with the first answer except it sounds like a real, not computer-modelled system. Normally a spring is modelled as massless, and here you saw actual forces due to the spring vibrating alone, because it actually does have some mass. These vibrations don't affect the system if you define it to include the cart, spring and world. Obviously the cart alone does not see momentum conserved as it bounces off the spring with opposite momentum. But when it loses contact with the spring the vibration doesn't affect its momentum. The total measured impulse shouldn't be affected either, since the vibrating-spring reaction forces are oscillatory with a zero mean.
2007-11-11 05:19:18 · answer #1 · answered by kirchwey 7 · 0⤊ 0⤋
I as assuming that this is a computer driven model as a completely frictionless surface is not possible. Based on your explanation of the events I would have to conclued that the 'bumps' in the gragh would be caused from the expansion and contraction of the spring after the impact of the cart. Have you ever pushed on a door then someone pulled on it from the other side? You get thrown forward under your own momentum. The same happens with the spring. The force from the impact of the cart compressed the spring. Because of the frictionless system, when the cart was pushed away, the spring over extends itself. It is then snapped back, slightly compressing, then re-extending. This is repeated until all the energy from the impact is released.
As for how does this affect the momentum, All of the cart's forward energy is transfered into the spring on the impact. The spring then absorbs and re-transfers part of the energy back to the cart, causing it be pushed away. The left over energy causes the small impulses that I explained above, and thus reducing the total momentum and engery in the system.
2007-11-11 04:59:01 · answer #2 · answered by Brad W 1 · 0⤊ 0⤋
this subject is fairly difficult me as properly.what's the question? i can in user-friendly terms wager that the question is: at what attitude is the rope proper to the field. is that this the question? if so, answer as follows: by means of fact the field strikes at a continuing velocity, the ensuing or information superhighway rigidity is 0. as a result the utilized rigidity in a horizontal course must be equivalent to the friction rigidity: utilized rigidity horizontal course = 25N 25N is the horizontal factor of a 43N rigidity at an attitude ? with the floor. scientific guard ?. cos ? = 25/40 3 cos ? = 0.5814 ? = fifty 4.40 5° above the horizontal The rope grew to become at an attitude of 50 4.40 5° above the horizontal
2016-12-16 05:18:57 · answer #3 · answered by carra 4 · 0⤊ 0⤋