course phy 201
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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
You only double the average velocity in order to get final velocity if initial velocity is zero and acceleration is uniform.
During the collision forces vary so acceleration is not uniform; we don't attempt to analyze what happens during the collision. The collision occurs only while the balls are in contact, and so occurs over a very short distance. We neglect that distance in our analysis.
From the instant after collision to the instant of impact with the floor the ball is in free fall and its acceleration is zero. This phase of uniformly accelerated motion (accel = 0 is uniform) begins after the collision, with the balls already in motion. So initial velocity is not zero. Since acceleration is 0, velocity is constant and hence initial, final and average velocities are all the same.
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
68.1 cm is the vertical fall; this is not `dv.
‘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
It isn't clear how you got 0.91 s. The time of fall will be the same as before, assuming distance of fall is the same.
The motion down the ramp, then along the horizontal, is characterized by two different accelerations and cannot be analyzed as a uniform acceleration event. You get the velocity of the first ball from its horizontal range and time of fall.
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
M2/M1 =- 0.56/-0.83= 0.67
0.4 - 1.23 = -0.83. Your result would end up
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
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.
This appears to be very similar to your previous submission, which was posted on 4/4/06.
I've inserted a couple more notes.