Conservation of Momentum

By setting up head-on collisions of various spherical objects and allowing them immediately after collision to fall a known distance under the influence of gravity, we can from the horizontal ranges of their falls determine their velocities immediately after collision. If one object is stationary prior to collision, and if the velocity of the other immediately before collision is determined, we can then compare total momentum before collision to total momentum after. This comparison constitutes a test of the Law of Conservation of Momentum for two objects. See CD EPS01 for Lab Kit Experiment 19. Note that the setup here has been adapted somewhat from the version on the video clip; however the differences and similarities should be easy to understand. In this experiment we will allow the larger marble in the kit to roll from rest down the binding indentation of a hardcover book inclined at an angle, then onto the binding indentation of another book lying flat on a horizontal tabletop. This setup replaces the curved-end incline seen in the video clips. The setup of the books is shown in the figure below. Note that the binding indentation in recent printings of the book might not be deep enough to keep the marble 'in the groove'. If this is the case for your book you can make the following adaptation: • Crease a strip of paper, with the crease about 1/2 inch from its edge. • If started from the middle of the 'high end' of the book the ball will roll toward the lower end, and probably will also tend to roll toward the binding edge of the book. If not you can tilt the book very slightly toward the binding edge to ensure that this happens. • If you place the creased edge of the paper parallel to the indentation but on the flat surface of the book, that will be sufficient to keep the marble rolling in a straight line. The strip of paper should lie flat on the book, with the plane of the 1/2-inch crease in the upward direction and sloping slightly away from the path of the marble. • Do this for both books. • The right end of the 'blue' book is elevated so the marble will achieve a significant velocity while rolling down the binding indentation. If the book is not moved we can expect the marble to achieve very nearly the same velocity with every trial. • The books will be aligned so that the marble rolls from the binding indentation of the 'blue' book directly and without interruption onto the binding indentation of the 'green' book. The 'green' book will not quite be as thick as the 'blue' book so that the marble will drop smoothly from the first book to the second. • The marble will roll with significant velocity off of the second book, which will be lying flat on a horizontal table. The velocity of the marble as it exits the second book is expected to be very nearly the same from trial to trial. The setup is illustrated below: The setup with creased strips of paper: End view of book with creased paper and marble: The marble, referred to hereafter as the 'ball', will roll down the two books and strike another marble (also called a ball) head-on, after which both balls will fall as projectiles to the floor. We will obtain data to determine the velocities of the balls after impact and the velocity of the first ball before impact, from which we can make various tests of the conservation of momentum. You will collide two balls of unequal mass, with the larger ball rolling down the incline and the smaller set up as a stationary target. • Using a section of a straw, as in the video clip, set up the smaller ball at the edge of a table just past the end of the ramp (i.e., the two-book system described above). Position the ball also that the collision will be head-on in both a horizontal and a vertical plane. • Position sheets of paper overlaid with carbon paper to detect the positions at which the balls strike the floor. If carbon paper is not available devise another method for determining with reasonable accuracy the positions at which the balls strike the floor. • Release the larger ball from the end of the ramp and allow it to collide with the 'target' ball, after which the two balls will fall to the floor and, if carbon paper is available, leave marks indicating the positions at which they struck. • Take any other data you will need to determine the velocities of the balls immediately after collision. M2= v0=0m/s a= 9.8m/s^2 ‘dsv = 68.1cm = 0.681m ‘dsh= 10.51cm = 0.17.51m ‘dt = srt(2*’dsv/a) = srt(2*0.681m/9.8m/s^2) =0.373 vAve = ‘ds/’dt = 0.1051m/0.373 s = 0.28m/s*2 = 0.56m/s = vf M1 = v0= 0m/s a = 9.8m/s^2 ‘dsv = 68.1cm = 0.681m ‘dsh= 7.5 cm = 0.075m ‘dt = srt(2*’dsv/a) = srt(2*0.681m/9.8m/s^2) =0.373s vAve = ‘ds/’dt = 0.075m/0.373s = 0.20m/s * 2 = 0.4m/s=vf • Using the same procedures as in previous experiments, determine the horizontal velocities of the falling balls. [These procedures are based on the projectile properties of each ball; measuring the horizontal range of the ball and the distance of fall you analyze the vertical motion (which should have initial velocity zero if the balls collide as instructed) to determine the time of fall and then from the range determine the horizontal velocity.] • Allow the ball rolling down the incline to fall freely without colliding with the second ball, and collect the data you will need to determine the velocity with which it left the ramp. • Determine the horizontal velocity of the ball as it falls. ‘dsh = 16.98cm = 0.1698m ‘dv= 68.1cm = 0.681m a = 9.8m/s^2 ‘ds ramp =56.1cm =0.561m ‘dt ramp = 0.91s vAve = ‘ds/’dt = 0.561m/0.91s = 0.62m/s * 2 =1.23 m/s = vf • Letting m1 stand for the mass of the larger ball and m2 for the mass of the smaller, write expressions for the total momentum of the two balls before collision and after collision. • The momentum of the first ball immediately before collision, using the velocity you reported above (the velocity based on the mean range and distance of fall). Be sure to use both the numerical value of the velocity and its units. This will be reported in the first line below. • The momentum of the first ball immediately after collision, using the velocity you reported above. This will be reported in the second line below. • The momentum of the second ball immediately after collision, using the velocity you reported above. This will be reported in the third line below. • The total momentum of the two balls immediately before collision. This will be reported in the fourth line below. • The total momentum of the two balls immediately after collision. This will be reported in the fifth line below M1*1.23/s = -M2* 0.0m/s M1*0.4m/s = -M2* 0.56m/s • Set the two expressions equal to obtain an equation expressing momentum conservation. • The resulting equation will have m1 and m2 as unknowns. M1*1.23m/s + M2* 0.00m/s =M1*0.40m/s + M2* 0.56m/s • Using simple algebra, rearrange the equation to get only the ratio m2 / m1 on the left-hand side. The other side will reduce to a single number, which will be the ratio m2 / m1 of the mass of the smaller to the larger ball. M1*1.23m/s + M2* 0.00m/s =M1*0.4m/s + M2* 0.56m/s M2*0.00m/s-M2*0.56m/s = M1*0.4m/s – M1*1.23m/s M2*-0.56m/s = M1*-0.58m/s

Do you mean M2*-0.56m/s = M1*-0.83m/s ?

M2/M1 =- 0.56/-0.83= 0.67

If the equation is as I suggested above, then M2 / M1 = -0.83 m/s / (-.56 m/s) = 1.5 approx.

• What do you get for the ratio? 0.67 • Measure the diameters of the two balls. Is the ratio of the masses equal to the cube of the ratio of the diameters? If the balls are made of the same material, why would we expect that the ratios would behave in this manner? M1= 2.5 M2 = 1.7 M2/M1 =0.68

m2 / m1 would be 1.7^3 / 2.5^3, not 1.7 / 2.5. Volume and hence mass is proportional to the cube of the radius.

Mine isn’t but hey should be because if they are made of the same thing then the only difference is how large they are so they the larger the diameter the more mass they would have Using the program MOMSIM (available on the 164.106.222.236 homepage under Simulations), analyze the first collision from a variety of reference frames. • Enter the velocities and the mass ratio, as requested, for the balls in the first collision. • If the collision is viewed from a vehicle which is moving smoothly along with the moving ball just before collision, and which continues moving smoothly at this velocity, then to an occupant of the vehicle it will appear that the this ball is initially standing still and that the second ball approaches and strikes the first. After collision the first ball will appear to move 'backwards'. • Using the simulation, give your vehicle a velocity equal to that of the moving ball just before collision, and observe the collision from this frame of reference. Describe what you see and why what you see makes good sense as a collision. The second ball once hit moves very fast while the first ball that made the contact slows down considerably. Some of the energy is being displaced from the first ball into the second ball when they make contact which makes the second ball go faster and the first to slow down. v1=-0.83 v2=-0.67 M1*1.23/s = -M2* 0.0m/s M1*0.83m/s = -M2* 0.67m/s M1*1.23m/s + M2* 0.00m/s =M1*0.83m/s + M2* 0.67m/s M2*0.00m/s-M2*0.67m/s = M1*0.83m/s – M1*1.23m/s M2*-0.67m/s = M1*-0.4m/s M2/M1 =- 0.67/-0.4= 0.67=1.675 • Now use the simulation to view the collision from a vehicle which is moving at half the before-collision velocity of the first ball. Describe what you see and why this collision makes good sense. The green ball moves very little once it is hit. Still in the direction that it was moving V1=-0.22 V2=-0.06 once they hit each other they slow down so there velocity is slower M1*1.23/s = -M2* 0.0m/s M1*0.22m/s = -M2* 0.06m/s M1*1.23m/s + M2* 0.00m/s =M1*0.22m/s + M2* 0.06m/s M2*0.00m/s-M2*0.06m/s = M1*0.22 m/s – M1*1.23m/s M2*-0.06m/s = M1*-1.01m/s M2/M1 =- 1.01/-0.06 =16.8 • Repeat using a frame of reference of the second ball after collision. The green ball stopped all together. If the red ball was moving at the same speed as the green ball well when they hit the green ball is smaller and so it ends up stopping which the red ball continues but slower V1=-0.16 because it hits the car and then goes back in the same direction it was coming and is slowing down V2 = 0 velcoity is zero because the car stopped M1*1.23/s = -M2* 0.0m/s M1*0.16m/s = -M2* 0.0m/s M1*1.23m/s + M2* 0.00m/s =M1*0.16m/s + M2* 0.0m/s M2*0.00m/s-M2*0.0m/s = M1*0.16 m/s – M1*1.23m/s M2*-0.0m/s = M1*-1.07m/s M2/M1 =- 0/-1.07 =0 • Verify in detail that the conservation of momentum is validated in each of these frames of reference. Explain why the velocities are as indicated on the simulation, and calculate total momentum before and after collision for each frame of reference. Analyze the collision from the center-of-mass frame, using various coefficients of restitution. • The center of mass of the system at any given instant is the position relative to which the two balls would balance if their positions were frozen and they were placed on a beam rotating about the center of mass. • As the balls move toward or away from collision, their center of mass moves in almost every reference frame. The one frame in which the center of mass does not move is called the center-of-mass frame. • The total momentum in the center-of-mass frame is 0. • If the balls were to stick together after collision, they would be moving with the velocity of the center-of-mass frame. This velocity is easily found to be vCM = (m1 v1 + m2 v2) / (m1 + m2). • Determine vCM from the velocities of the balls before collision, and observe the collision from this frame. vCM = (m1 1.23m/s + m2 0m/s) / (m1 + m2). • Now let the computer calculate and represent the actual velocities to be expected after collision, based on the velocities before collision. First run the simulation, then make this choice afterwards. How do you do this with the Program???? Can’t do the rest because don’t know how to do this part • • You will need to select a coefficient of restitution, which is the ratio of the magnitude of the relative velocity of the balls after collision to that before. Begin by selecting 0 and 1, and observe how the white dot, which is at the center of each circle before collision, tends to move differently after collision. These white dots represent the positions of the balls according to the computer's calculations, which are based on the initial velocities you provided.. • See if you can find a coefficient of resitution which best models your observations by keeping the white dots as close to the positions of the balls as possible. • How closely were you able to model your observed velocities? How nearly 'true' then were your conclusions? How does this compare with the error in your experiment (the degree to which your results failed to verify momentum conservation)? Analysis of errors • Discuss possible sources of error in this experiment. Didn’t measure correctly when they hit the paper. Balls might not have had direct contact. • Estimate the possible error ranges in your data, and determine whether within the resulting ranges of observed momenta we can conclude that momentum is conserved. "

Your work on this experiment is good overall. Your analysis is generally accurate, but be sure to see my notes on a couple of points. I don't see anything wrong with your data, but it does lead to a contradictory conclusion about the ratio of masses.