Conservation of Momentum and Energy

Transcription

1 Department of Physics & Astronomy David Rittenhouse Laboratory Friday March 1, 2013 Conservation of Momentum and Energy In this lab we will study how the momentum and kinetic energy of a moving object are changed by the application of a braking force. All the major elements of the apparatus are seen in Figure 1 (below). The moving object is a low-friction glider (A) moving on a grooved track that ensures straight-line motion. To document the glider s position and velocity, an ultrasonic motion detector (B) is connected to a computer using the Vernier Logger Pro software. The repeated high-pitched chirps from the motion detector echo off the black metal flag (C), and return to the detector. Using the time needed for the echo to return, and the known speed of sound, the computer calculates the position and velocity of the glider 50 times every second and graphs the data. To apply and measure a stopping force on the glider we will use a force sensor (D) connected to a wooden rod or dowel (E). The dowel runs easily through a hole in the metal plate on the glider. A wooden clothespin (F) is placed at about the 2/3rds point on the dowel.

2 When the metal plate with the hole hits the clothespin, the clothespin is pushed along the dowel until friction brings the glider to a stop, as shown in Figure 2. The applied friction force on the rod pushes on the force sensor. The computer records all the data (position, velocity, Figure 2. force, and time) in a single table. Clicking on the either axis label of any graph produces a drop-down menu of all variables. A graph of any two data variables can be produced, including calculated columns inserted after the data is collected. Momentum and Impulse: If the mass of the glider is measured, the momentum (p) of the glider just before it hits the clothespin can be calculated (p = mv). That permits a check on the impulse (I) formula for change in momentum: (impulse) = (change in momentum) = (mass) x (change in velocity) or I = p = m v (eq. 1.) Since the glider comes to a stop, the change in velocity is equal to the velocity value when it strikes the clothespin. Equation 1 permits us to calculate the change but it does not indicate why it happens. To understand why motion changes we must take forces into account, and we also have that data. There is a second way to calculate impulse (or the change in momentum) that includes force: I = p = F t (eq. 2.) Since we have collected both force and time data we can graph them by changing the plotted variables as described to the right of figure 2. Keep time in its traditional place on the horizontal axis, and insert force on the vertical axis. If force were constant, the graph would be a horizontal line producing a rectangle with a height of F and a width of t. That would make the product F t equal the area under the force vs. time graph. More importantly, that would make the change in momentum equal to the area under the force vs. time graph.

3 And most importantly, that would be true even if the graph were not a neat rectangle. The area under any force vs. time graph indicates how much the momentum changed during that time. As shown in figure 3, our calculated loss of momentum can be checked against the force vs. time data collected by the force sensor. The Figure 3. Logger Pro software will calculate the area under any region of the graph you highlight by clicking and dragging that mouse from lower left to upper right of the box you want to include (the shaded box around the graph in fig 3.) Once that is highlighted, go to the top toolbar and click the icon that shows an area under a curve (it s labeled integral in fig 3). Within realistic limits you should find that momentum change based on result (equation 1) equals momentum change based on cause (equation 2). That would mean we could combine equations 1 and 2 to produce the impulse equation: m v = F t (eq. 3.) In drawing your conclusions, discuss how close your data came to verifying equation 3. a. What measurement challenges might have kept the product on the left from being exactly equal to the product on the right? Be specific and if a value is too big, discuss things that may have pushed it in that direction. b. How big would the difference in the two sides of equation 3 have to be to make the outcome questionable? To disprove the hypothesis?

4 Energy and Work: In the same way that physicists have realized the change or transfer of momentum is important enough to have the separate name impulse, they have realized the change in energy is also so important that it needs its own term: work. These are both conserved quantities, so making sure all the changes total to zero is a very good check on how well we understand and measure the thing we are studying. (The same idea is central to finance. The change in money is called pay. Both are measured in the same unit, dollars. But what Apple store employees get paid is very different from the total value of the Apple corporation.) Although the formulas are not quite the same, the way we think about energy is similar to our approach to momentum. As with momentum, the kinetic energy (KE) of the glider just before it hits the clothespin can be calculated (KE = ½ mass vel 2 ). That permits a check on the work (w) formula for change in energy: (work) = (change in KE) = ( ½ mass x (final vel) 2 )- ( ½ mass x (initial vel) 2 ) or w = KE = ½ mv 2 2 f - ½ mv i (eq. 4.) Note: we can t just insert v into the formula because in the case of energy the velocity term gets squared, and squaring changes subtraction: (vf vi) 2 (vf 2 -vi 2 ) Like equation 1 in the momentum section, equation 4 above lets us calculate an outcome but does not deal with the forces that caused it. So as with impulse, there is also a way to calculate work using force: w = KE = F x (eq. 5.) Very similar to the momentum case, except that we multiply force times change in position rather than change in time. The similarities extend to the fact that work will equal the area under a force vs. distance graph. So use the same data you just used for your momentum-impulse work in the first part, and do a similar analysis of energy and work. You will need to change the horizontal axis of your force graph from time to position before finding the area under it. As you did with momentum, see if the combination of equations 4 and 5 is correct, and discuss how close your results come to expectations: F x = ½ mv f 2 - ½ mv i 2 (eq. 6.)

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