Lab5

Collision I Lab Report Names:Ting Cheung Tung Section:H5 Date:Oct 29 Purpose To investigate the relationship of impulse to change of momentum Readings You can explore the Wikipedia or your favorite mechanics textbook, Elastic collision, Inelastic collision, Kinematics, kinematics equations, Impulse Theory and concepts in short In a collision it is difficult to use Newton’s Second Law, F = m a because the force varies with time in some generally unknown manner. (Note: a bold letter for F means it is a vector unlike mass, m, which is a scalar.) We can rewrite the second law in a manner more useful for collisions by using the impulse I = ∫ F dt. Integrating both sides of the second law, I = ∫ F dt= ∫ m a dt = ∫ m d v dt dt = ∫ m d v = m v f - m v i where v f and v i are the velocities of m before and after the collision. Notice that F in here is the total force acting on the body. We define the linear momentum of m as P = m v . Then the second law says that the change of momentum of a body during a collision equals the impulse it receives, I = Δ P. Procedure Procedure for Part I When a dropped ball hits something it exerts a force on it (and by Newton’s third law an equal force is exerted on it). We will use the force probe to measure that force. You will drop a play-doh ball and golf ball with the same mass from the same height. The golf ball undergoes a nearly elastic collision while the clay ball has an inelastic collision. Setting Up the Data Collection Program LoggerPro Tung_Lab5_Collision1.nb 1 Printed by Wolfram Mathematica Student Edition

- Load the computer program LoggerPro by double clicking on its icon. - Add the force detector which is facing upward by clicking on the LoggerPro icon , and then clicking on the CH1 ⇒ Choose Sensor ⇒ Force ⇒ Dual Range Force - Gently press on the force probe. You should see a live reading at the bottom-left of the screen that shows the force you are exerting in units of Newtons. - Make sure that the button on the back of force probe is set to ± 50N . Make sure this is also the case on the settings of the sensor in the LoggerPro. If not use Change Sensor Range option to change it to ± 50N . - Zero the force probe by clicking zero icon. (Do this before each measurement, also after the measurement check to see if it is still close to zero and not drifted too much by the bounces.) - Click on the clock icon. Set the data collection rate to 4000 Hz for 0.5s . Tung_Lab5_Collision1.nb 2 Printed by Wolfram Mathematica Student Edition

connected to the same ULI as the force probe). Setting Up the Data Collection Program LoggerPro For this part, you will work together with another lab group. One group measures the force excerpted by the rolling car, the other group measures the velocity of the cart before and after collision using a motion detector. - Add the motion detector by clicking on the LoggerPro icon and then clicking on Dig/SONIC1 ⇒ Choose Sensor ⇒ Motion Detector - Click on the clock icon, and set the data collection rate to 50 Hz . Dialog: Part I. A Golf Ball, A Play-doh Ball, & The Force Sensor Tung_Lab5_Collision1.nb 4 Printed by Wolfram Mathematica Student Edition

When a dropped ball hits something it exerts a force on it (and by Newton’s third law an equal force is exerted on it). We will use the force probe to measure that force. You will drop a play-doh ball and a golf ball with the same mass from the same height . The golf ball undergoes a nearly elastic collision while the clay ball has an inelastic collision. Which do you expect to exert the greater force? Golf ball The contact time of the golf ball is shorter, so I guess it will exert a greater force. Step 1. Measuring the Impulse delivered to golf ball by the force probe - Zero the force probe. Press Collect and drop the golf ball through the plastic tube so that it hits the force probe. - Export your data into a CVS file and then import it into Mathematica using the Import function. (g1 is to label “golf ball’s 1st trial”) Example: fvtg1=Rest[Import[“/home/maryam/Downloads/f1.csv”]] - Now you can plot your force vs time plot using ListPlot function. Example: ListPlot[fvtg1, Joined  True, PlotRange  All, Frame  True, FrameLabel  {“Time (s)”,”Force (N)”}] 0.00 0.05 0.10 0.15 0.20 0.25 0 5 10 15 20 25 Time ( s ) Force ( N ) - From the plot find t0 (* time at which golf ball first hits probe (beginning of the first bounce) *), t1 (* time at which golf ball leaves probe (end of the first bounce) *), t2 (* time at which golf ball hits probe for the second time (beginning of the second bounce) *) and record below. - Measure m using the scale. - Measure h using a ruler. (You may find it convenient to measure the length of the plastic tube and the height of the force probe hook.) Tung_Lab5_Collision1.nb 5 Printed by Wolfram Mathematica Student Edition

In[11]:= mg1 = 0.0458 (* mass of golf ball *) hg1 = 0.115 (* height that golf ball falls to hit probe *) t0g1 = 0.002 (* time at which golf ball first hits probe ( beginning of the first bounce ) *) t1g1 = 0.015 (* time at which golf ball leaves probe ( end of the first bounce ) *) t2g1 = 0.076 fg1 = Interpolation [ fvtg1 ] ; (* time at which golf ball hits probe for the second time ( beginning of the second bounce ) *) Ipg1 = NIntegrate [ fg1 [ t ] , { t, t0g1, t1g1 }] (* I probe impulse that force probe delivers to golf ball during the first bounce; that is, ∫ t0 t1 F probe on ball ( t ) dt *) Out[11]= 0.0458 Out[12]= 0.115 Out[13]= 0.002 Out[14]= 0.015 Out[15]= 0.076 Out[17]= 0.0343058 In[18]:= mg2 = 0.0458 (* mass of golf ball *) hg2 = 0.115 (* height that golf ball falls to hit probe *) t0g2 = 0.001 (* time at which golf ball first hits probe ( beginning of the first bounce ) *) t1g2 = 0.011 (* time at which golf ball leaves probe ( end of the first bounce ) *) t2g2 = 0.101 fg2 = Interpolation [ fvtg2 ] ; (* time at which golf ball hits probe for the second time ( beginning of the second bounce ) *) Ipg2 = NIntegrate [ fg2 [ t ] , { t, t0g2, t1g2 }] (* I probe impulse that force probe delivers to golf ball during the first bounce; that is, ∫ t0 t1 F probe on ball ( t ) dt *) Out[18]= 0.0458 Out[19]= 0.115 Out[20]= 0.001 Out[21]= 0.011 Out[22]= 0.101 Out[24]= 0.0705709 Tung_Lab5_Collision1.nb 7 Printed by Wolfram Mathematica Student Edition

In[25]:= mg3 = 0.0458 (* mass of golf ball *) hg3 = 0.115 (* height that golf ball falls to hit probe *) t0g3 = 0.002 (* time at which golf ball first hits probe ( beginning of the first bounce ) *) t1g3 = 0.013 (* time at which golf ball leaves probe ( end of the first bounce ) *) t2g3 = 0.081 fg3 = Interpolation [ fvtg3 ] ; (* time at which golf ball hits probe for the second time ( beginning of the second bounce ) *) Ipg3 = NIntegrate [ fg3 [ t ] , { t, t0g3, t1g3 }] (* I probe impulse that force probe delivers to golf ball during the first bounce; that is, ∫ t0 t1 F probe on ball ( t ) dt *) Out[25]= 0.0458 Out[26]= 0.115 Out[27]= 0.002 Out[28]= 0.013 Out[29]= 0.081 Out[31]= 0.0553816 You can find the impulse that force probe delivers to golf ball during the first bounce by integrating force between t0 and t1 using the following two commands, fg1=Interpolation[fvtg1]; Ipg1=NIntegrate[fg1[t], {t, t0g1, t1g1}] Question 1. - What force(s) are acting on the ball before it hits the probe? Force of Gravity - What force(s) are acting on the ball during the first collision, i.e. between t0 and t1? What is their relative sign? Force of Gravity(acting downwards) and the Impulse Force(Acting upwards) So the force of gravity is negative, the impulse force is positive Step 2. Calculating the total impulse experienced by the golf ball - Use kinematic equations to calculate v0 and v1 from information collected inStep 1. Remember that when you drop the golf ball, it starts at rest. Be careful with signs: an upward velocity is positive, a down- ward velocity is negative. In[32]:= v0g1 = - Sqrt [ 2 ( 9.8 ) hg1 ] (* velocity of golf ball at time t0 *) v1g1 = ( 9.8 ) ( t2g1 - t1g1 ) / 2 (* velocity of golf ball at time t1 *) p0g1 = mg1 ( v0g1 ) (* momentum of golf ball at time t0 *) p1g1 = mg1 ( v1g1 ) (* momentum of golf ball at time t1 *) Tung_Lab5_Collision1.nb 8 Printed by Wolfram Mathematica Student Edition

In[48]:= Ip2 = ( p1g2 - p0g2 ) ; NumberForm [ Ip2, 2 ] Error2 = ( 100 ( Ip2 - Ipg2 ) / Ip2 ) ; NumberForm [ Error2, 3 ] Out[49]//NumberForm= 0.089 Out[51]//NumberForm= 20.7 In[52]:= Ip3 = ( p1g3 - p0g3 ) ; NumberForm [ Ip3, 2 ] Error3 = ( 100 ( Ip3 - Ipg3 ) / Ip3 ) ; NumberForm [ Error3, 3 ] Out[53]//NumberForm= 0.084 Out[55]//NumberForm= 34.1 - If your percentage error is equal or less than %15 record your data in the table below. Replace “?” with corresponding variable from above. If your percentage error is large, try to find out where the error (systematic error) lies and discuss it. Remember that Coefficient Of Restitution (COR) is defined as COR = |normal velocity right after the colli- sion|/|normal velocity right before the collision|. - Repeat the above steps at least two more times and report the results again on the same table below ( you can re-use the code you developed above. ). In other word, fill out the second and third row on this table too. The percentage error of my data is larger than 15%. I guess the ball might not be hit the force sensor accu- rately. It may have hit the sides of the tube and lose enemy and momentum. Therefore, the error would be higher than 15% In[112]:= Grid { Text [ "Golf ball, elastic collision" ] , SpanFromLeft } ,  "data set", "m, kg", "t0, s", "t1, s", "t2, s", "h, m", "v0, m / s", "v1, m / s", "COR", "p1 - p0, N * s", "  t0 t1 F ( t ) dt, N * s", " p1 - p0 - ∫ t0 t1 F ( t ) dt p1 - p0 * 100 % "  , { "fvg1", "0.0458", "0.002", "0.015", "0.076", "0.115", " - 1.501", "0.298", "0.199", "0.082", "0.0343", "58.4" } , { "fvg2", "0.0458`", "0.001", "0.011", "0.101", "0.115", " - 1.501", "0.441", "0.294", "0.089", "0.0705", "20.7" } , { "fvg3", "0.0458", "0.002", "0.013", "0.081", "0.115", " - 1.501", "0.333", "0.222", "0.084", "0.0553", "34.1" } , Frame  All  Tung_Lab5_Collision1.nb 10 Printed by Wolfram Mathematica Student Edition

Out[112]= Golf ball, elastic collision data set m, kg t0, s t1, s t2, s h, m v0, m / s v1, m / s COR p1 - p0, N * s ∫ t0 t1 F ( t ) dt , N * s  p1 - p0 - ∫ t0 t1 F ( t ) dt   ( p1 - p0 )* 100 % fvg1 0.0458 0.002 0.015 0.076 0.115 - 1.501 0.298 0.199 0.082 0.0343 58.4 fvg2 0.0458` 0.001 0.011 0.101 0.115 - 1.501 0.441 0.294 0.089 0.0705 20.7 fvg3 0.0458 0.002 0.013 0.081 0.115 - 1.501 0.333 0.222 0.084 0.0553 34.1 Step 3. Inelastic collision by play-doh ball Make a play-doh ball of the same mass as the golf ball and repeat steps one and two above and record the results on the table below. Again, you are doing all the steps at least three times, i.e. each trial is a row on the table below. In[56]:= fvtp1 = Rest [ Import [ "C:\\Users\\User\\Desktop\\Mathematica\\Tung _ Lab5 _ Collision1\\p1.csv" ]] ; ListPlot [ fvtp1, Joined  True, PlotRange  All, Frame  True, FrameLabel  { "Time ( s ) ", "Force ( N ) " }] fvtp2 = Rest [ Import [ "C:\\Users\\User\\Desktop\\Mathematica\\Tung _ Lab5 _ Collision1\\p2.csv" ]] ; ListPlot [ fvtp2, Joined  True, PlotRange  All, Frame  True, FrameLabel  { "Time ( s ) ", "Force ( N ) " }] fvtp3 = Rest [ Import [ "C:\\Users\\User\\Desktop\\Mathematica\\Tung _ Lab5 _ Collision1\\p3.csv" ]] ; ListPlot [ fvtp3, Joined  True, PlotRange  All, Frame  True, FrameLabel  { "Time ( s ) ", "Force ( N ) " }] Out[57]= 0.0 0.1 0.2 0.3 0.4 0 5 10 15 Time ( s ) Force ( N ) Tung_Lab5_Collision1.nb 11 Printed by Wolfram Mathematica Student Edition

In[69]:= mp2 = 0.0458 hp2 = 0.115 t0p2 = 0.001 t1p2 = 0.011 t2p2 = 0.038 fp2 = Interpolation [ fvtp2 ] ; Ipp2 = NIntegrate [ fp2 [ t ] , { t, t0p2, t1p2 }] Out[69]= 0.0458 Out[70]= 0.115 Out[71]= 0.001 Out[72]= 0.011 Out[73]= 0.038 Out[75]= 0.0544819 In[76]:= mp3 = 0.0458 hp3 = 0.115 t0p3 = 0.001 t1p3 = 0.013 t2p3 = 0.041 fp3 = Interpolation [ fvtp3 ] ; Ipp3 = NIntegrate [ fp3 [ t ] , { t, t0p3, t1p3 }] Out[76]= 0.0458 Out[77]= 0.115 Out[78]= 0.001 Out[79]= 0.013 Out[80]= 0.041 Out[82]= 0.0585862 In[83]:= v0p1 = - Sqrt [ 2 ( 9.8 ) hp1 ] v1p1 = ( 9.8 ) ( t2p1 - t1p1 ) / 2 p0p1 = mg1 ( v0p1 ) p1p1 = mg1 ( v1p1 ) Out[83]= - 1.50133 Out[84]= 0.2793 Out[85]= - 0.068761 Out[86]= 0.0127919 Tung_Lab5_Collision1.nb 13 Printed by Wolfram Mathematica Student Edition

In[87]:= v0p2 = - Sqrt [ 2 ( 9.8 ) hp2 ] v1p2 = ( 9.8 ) ( t2p2 - t1p2 ) / 2 p0p2 = mg1 ( v0p2 ) p1p2 = mg1 ( v1p2 ) Out[87]= - 1.50133 Out[88]= 0.1323 Out[89]= - 0.068761 Out[90]= 0.00605934 In[91]:= v0p3 = - Sqrt [ 2 ( 9.8 ) hp3 ] v1p3 = ( 9.8 ) ( t2p3 - t1p3 ) / 2 p0p3 = mg1 ( v0p3 ) p1p3 = mg1 ( v1p3 ) Out[91]= - 1.50133 Out[92]= 0.1372 Out[93]= - 0.068761 Out[94]= 0.00628376 In[95]:= Ip4 = ( p1p1 - p0p1 ) ; NumberForm [ Ip4, 2 ] Error4 = ( 100 ( Ip4 - Ipp1 ) / Ip4 ) ; NumberForm [ Error4, 3 ] Out[96]//NumberForm= 0.082 Out[98]//NumberForm= 55.5 In[99]:= Ip5 = ( p1p2 - p0p2 ) ; NumberForm [ Ip5, 2 ] Error5 = ( 100 ( Ip5 - Ipp2 ) / Ip5 ) ; NumberForm [ Error5, 3 ] Out[100]//NumberForm= 0.075 Out[102]//NumberForm= 27.2 Tung_Lab5_Collision1.nb 14 Printed by Wolfram Mathematica Student Edition

In[110]:= COR [ v1g2, v0g2 ] Out[110]= 0.293739 In[109]:= COR [ v1g3, v0g3 ] Out[109]= 0.221936 In[113]:= COR [ v1p1, v0p1 ] COR [ v1p2, v0p2 ] COR [ v1p3, v0p3 ] Out[113]= 0.186035 Out[114]= 0.0881217 Out[115]= 0.0913855 Question 4. - Discuss why the following two statements are incorrect : A. The clay ball will exert the greater force because it has to be stopped, while the golf ball retains most of its speed. The golf ball will exert a greater force than the clay ball. As the impulse is the integral of the force. The clay ball and the golf ball fall from the same height and they experience the same acceleration(gravity). Therefore, they will have the same final velocity. The velocity when they hit the probe is the same. However, the clay touches the ground longer so the force act on it is less than the golf ball. The change in momentum of golf ball is higher. B. The balls will exert the same force because they have the same mass and are moving at the same speed. The impulse would be equal to the final momentum- initial momentum. Although they have the same initial momentum, their final momentum is different. Therefore, the force exert of both balls are not the same.2 Part II. The Cart, The Motion Sensor, & The Force Sensor In the second part we will make a similar measurement with a rolling cart hitting the force probe. For this part, you will work together with another lab group. One group measures the force exerted by the rolling car, the other group measures the velocity of the cart before and after collision using a motion detector. Members of both group have to save the data from the force probe and motion detector in their Sakai Tung_Lab5_Collision1.nb 16 Printed by Wolfram Mathematica Student Edition

Drop Boxes. You will use this data for your next lab prerequisite assignment. Step 1. Elastic Collision, Collecting Data - Put the cart on the track in between the force probe and the motion detector. The motion detector and the force probe should both be facing towards the cart. - Move the force detector such that it is parallel to the track at a position such that it will hit the cart close to the center. - Set the motion detector of the group at the adjacent table on the track. - Make a few test runs and be sure the motion detector follows the cart until it hits the force probe. BE GENTLE! Do not push the car too quickly. - When the cart hits the probe, one group member should be holding the probe tightly so that it does not recoil. - Zero the force probe, press Collect on both computer, and set the cart in motion so that it hits the force probe and bounces back. - Repeat until you get smooth a curve for the position of the cart. - Export your data from the force probe and motion detector in CVS format. - Save the data files in your Drop Box for your next lab prerequisite assignment, see below. Step 2. Elastic Collision, Measuring Impulse, Using Force Data & Using Velocity Data Prerequisite Assignment for Collision II lab Please submit together with Collision I lab report. Calculating the Impulse - Import the force detector data from .csv file. - Plot the F(t) data showing the collision. - Determine the time before and after collision. - Calculate the impulse Ipc delivered by probe on the cart by integrating force vs time over the collision time period. You can use example for the golf ball. In[124]:= Force = Rest [ Import [ "C:\\Users\\User\\Desktop\\Mathematica\\Tung _ Lab5 _ Collision1\\Force.csv" ]] ; ListPlot [ Force, Joined  True, PlotRange  All, Frame  True, FrameLabel  { "Time ( s ) ", "Force ( N ) " }] Tung_Lab5_Collision1.nb 17 Printed by Wolfram Mathematica Student Edition

Out[  ]= 1 2 3 4 5 0.2 0.4 0.6 0.8 In[128]:= t0 = 1.59; t1 = 3.17; t2 = 4.91; line1 = Fit [ Select [ x, t0 < First [#] < t1 - .05 & ] , { 1, t, t^2 } , t ] ; (* makes a parabolic fit to the data between t0 and t1 - 0.05 s *) line2 = Fit [ Select [ x, t1 + .05 < First [#] < t2 & ] , { 1, t, t^2 } , t ] ; vi = ∂ t line1 / . t  t1; (* calculates velocity at the moment t1 *) vf = ∂ t line2 / . t  t1; Show [ ListPlot [ x ] , Plot [ line1, { t, 0, t1 } , PlotStyle  Red, PlotLegends  { NumberForm [ line1, 2 ] , "" }] , Plot [ line2, { t, t1, t2 } , PlotStyle  Magenta, PlotLegends  { NumberForm [ line2, 2 ] , "" }] , PlotLabel  "collision time" t1 , Frame  True, FrameLabel  { "Time ( s ) ", "Displacement ( m ) " }] ListPlot [ v, Frame  True, FrameLabel  { "Time ( s ) ", "Velocity ( m / s ) " }] Grid [{{ Text [ "collision time" t1 ] , SpanFromLeft } , { "vi, m / s", "vf, m / s" } , { NumberForm [ vi, 3 ] , NumberForm [ vf, 3 ]}} , Frame  All ] Out[135]= 0 1 2 3 4 5 0.0 0.2 0.4 0.6 0.8 Time ( s ) Displacement ( m ) 3.17 collision time - 0.058 + 0.25 t + 0.011 t 2 1.7 - 0.35 t + 0.022 t 2 Tung_Lab5_Collision1.nb 19 Printed by Wolfram Mathematica Student Edition

Out[136]= 0 1 2 3 4 5 - 1.0 - 0.5 0.0 0.5 1.0 1.5 Time ( s ) Velocity ( m / s ) Out[137]= 3.17 collision time vi, m / s vf, m / s 0.318 - 0.215 Testing (the change in momentum) = (impulse) relation Calculate and record the results in the table below. In[142]:= Grid { Text [ "Cart, Elastic collision" ] , SpanFromLeft } ,  "m, kg", "vi, m / s", "vf, m / s", "COR", "pf - pi, N * s", "Ipc =  t1 - 0.1 t1 + 0.1 F ( t ) dt, N * s", " pf - pi - Ipc pf - pi * 100 % "  , { "0.5", "0.318", " - 0.215", "0.678", " - 0.267", " - 2.209", "2.89" } , Frame  All  Out[142]= Cart, Elastic collision m, kg vi, m / s vf, m / s COR pf - pi, N * s Ipc =∫ t1 - 0.1 t1 + 0.1 F ( t ) dt, N * s pf - pi - Ipc pf - pi * 100 % 0.5 0.318 - 0.215 0.678 - 0.267 - 2.209 2.89 In[118]:= pf = 0.5 * - 0.215 pi = 0.5 * - 0.318 Out[118]= - 0.1075 Out[119]= - 0.159 In[  ]:= pf - pi Out[  ]= 0.0515 In[4]:= COR [- 0.215, 0.317 ] Out[4]= 0.678233 In[140]:= FFF = Interpolation [ Force ] ; In[141]:= Ipc = NIntegrate [ FFF [ t ] , { t, 3.27, 3.07 }] Out[141]= - 241 620. Tung_Lab5_Collision1.nb 20 Printed by Wolfram Mathematica Student Edition