Conservation of Momentum in Two Dimensions

Transcription

1 Conservation of Momentum in Two Dimensions Consider the two-dimensional (glancing) collision shown below. Here, mass m 1 travels to the right along the x-axis with velocity v 1o and strikes mass m 2 initially at rest. After the collision, m 1 continues to the right with velocity v 1f at an angle θ 1 from the x-axis; m 2 acquires velocity v 2f and travels at an angle θ 2 from the x-axis. Figure 1: Experimental Collision Momentum is a vector quantity; therefore the analysis of momentum involves vectors. p o = p f m 1 v 1o + m 2 v 2o = m 1 v 1f + m 2 v 2f However, this analysis can be simplified if we look at the momentum in each direction separately. m 1 v 1ox + m 2 v 2ox = m 1 v 1fx + m 2 v 2fx m 1 v 1oy + m 2 v 2oy = m 1 v 1fy + m 2 v 2fy With the collision defined as such, there are two things worth noting. First, m 2 is initially at rest, so it has no initial x or y momentum. Second, m 1 is initially traveling along the x-axis, so the initial momentum in the y direction is zero. Thus, the final total momentum in this direction will be zero (even though the individual momenta are not zero). If the masses are equivalent (m 1 = m 2 ), then this collision reduces to 1

2 v 1ox = v 1fx + v 2fx (1) in the x-direction, and in the y-direction. 0 = v 1fy + v 2fy (2) The collision in this experiment is between 2 small steel spheres. One is projected horizontally (parallel to the floor) from a launcher and collides with a second, stationary sphere placed in its path. Both spheres are projected horizontally from here and fall onto the floor below. Since there is no horizontal force on the spheres, they move in this direction with constant velocity. If a sphere travels distance d across the newsprint, then its velocity is v = d t (3) where t its travel time. But, each sphere is in free fall once projected, so that t = 2h g (4) where h is the free fall distance. Apparatus Mini projectile launcher, Clamp, Steel spheres, Plastic ramrod, Target support, Paper, Carbon paper, Masking tape, Meter stick, Protractor, Plumb bob. Procedure In collecting data, it is imperative that: The launcher be securely fastened to the table, so that it does not move during the experiment as it is being fired. The height of the stationary target sphere is the same as that of the projected sphere. Less than half of the spheres overlap (when viewed from above) when they do collide. Use the short range setting (one click on the launcher) for all shots. 1. Measure the free fall distance h from the bottom of the launcher barrel to the floor. Calculate the free fall time. 2. Adjust the target plate so that it extends straight out from the barrel of the launcher (the target screw should be between 1cm and 2cm from the end of the barrel). 2

3 3. Tape a sheet of paper to the floor directly below the target screw. Use the plumb bob to locate the position of the target screw and mark it on the paper. [This takes 15s; suspend the plumb bob and mark the point - you are then done with it. Do not attach the plumb bob line to the apparatus!] This is the origin of your collision coordinate system. 4. Lower the height of the target screw so that as m 1 is projected from the launcher, it will pass over without hitting it. Fire m 1 from the launcher and note where it strikes the floor. Tape another sheet of paper to the floor at this location and place a piece of carbon paper over it. 5. Fire m 1 five times and make sure the sphere lands in roughly the same place each time. When done, remove the carbon paper and draw a circle around your five shots; not a circle around each but the smallest circle containing all five. Using the meter stick, draw a line from the origin to the center of this circle (on the two sheets of paper containing the meter stick at least). This is distance d 1o and is, by definition, the x-axis of your collision coordinate stytem. Calculate v 1o. 6. Readjust the height of the target screw so that m 2 will be the same height as m 1 as it leaves the launcher. Tighten the bolt on the screw to lock this height in place. Loosen the bolt holding the target plate to the launcher and rotate it so that less than half the spheres will overlap when they collide (you want a glancing collision). Tighten the bolt holding the plate to the launcher to lock this location in place. 7. Fire m 1 at m 2 and note where each strikes the floor. Tape sheets of paper to the floor at these locations and place pieces of carbon paper over them. 8. Fire m 1 five times at m 2 and again make sure that each of the spheres lands in roughly the same place each time. 9. Using the same technique as in obtaining d 1o, get d 1f and d 2f. You will also need the angle that each makes with your x-axis. Calculate v 1f and v 2f. 3

4 Data Sheet Free fall distance h (m) Free fall time t (s) Distance d 1o (m) Velocity v 1o of m 1 before collision (m/s) Distance d 1f (m) Velocity v 1f of m 1 after collision (m/s) Angle θ 1 between v 1f and x-axis () Distance d 2f (m) Velocity v 2f of m 2 after collision (m/s) Angle θ 2 between v 2f and x-axis () 4

5 Analysis 1. Calculate total momentum in the x-direction before and after the collision (show all work). a. Before: b. After: c. Compare (% difference) the momentum from a and b. Is momentum conserved in this direction? 5

6 2. Calculate total momentum in the y-direction before and after the collision (show all work). a. Before: b. After: c. What can you compare (% difference) to show that momentum was conserved in this direction? Do so below. 6

7 Pre-Lab: Conservation of Momentum in Two Dimensions Name Section Answer the questions at the bottom of this sheet, below the line (only) - continue on the back if you need more room. Any calculations should be shown in full. 1. An object falls from rest (v oy = 0) 1.250m to the floor. How much time does it take for this to happen? 2. Rather than falling straight down, the object from Question 1 is projected horizontally with some velocity v o. If it travels 3.250m horizontally before striking the floor, what is v o? 3. Consider the collision shown in Figure 1 of the handout (m 1 = m 2 ; v 1o = 5.00m/s, v 2o = 0). After the collision, m 1 has a velocity of 3.83m/s at an angle of 40 above the x-axis. What will be the velocity (magnitude and direction) of m 2 after the collision? 7