Fall22 Conservation of Momentum Online 8.16.22

Conservation of Momentum Lab Online Purpose The purpose of this activity is to examine the relationship between the momentum masses that undergoes elastic and inelastic collisions, and the effects the two different types of collisions have on the total energies of the masses. Theory Newton’s Third Law tells us that when mass 1 (m 1 ) exerts a force on mass 2 (m 2 ) then m 2 must exert a force on m 1 of equal magnitude but opposite in direction. This can be written as a simple algebraic equation; F 12 =− F 21 Since Newton’s second law tells us that all forces can be written as F = ma , where m is the object’s mass and a is its current acceleration, we can substitute that in giving us; m 1 a 1 =− m 2 a 2 The average acceleration is the change in an object’s velocity per unit time, a avg = ∆v ∆t , so we can also substitute this in for the two accelerations giving us; m 1 ∆ v 1 ∆t =− m 2 ∆v 2 ∆t You should notice that there are no subscripts on the ∆t . The reason there is no subscript on the ∆t is because the two masses are exerting forces on each other over the exact same time period. m 1 can’t touch m 2 without m 2 touching m 1 , and vice versa. This means we can multiply this equation by ∆t to remove it from the equation all together. m 1 ∆v 1 =− m 2 ∆v 2 Now we can expand the deltas, then distribute the masses and the negative sign, giving us. m 1 v 1 − m 1 v 1 i =− m 2 v 2 + m 2 v 2 i Finally, if we now regroup with the initial velocities on left side of the equation and the final velocities on the right side of the equation, we get: 1

m 1 v 1 i + m 2 v 2 i = m 1 v 1 + m 2 v 2 This equation is The Law of Conservation of Momentum for an elastic collision , and as you have just seen, we can get it directly from Newton’s Third Law. The product of a mass and its velocity is called the mass’s momentum ( p = mv ) , and in the SI system it has the units of kilograms·meters/seconds (kg·m/s). The Law of conservation of Momentum tells us that the sum of the momentums of the two masses before their collision is equal to the sum of their momentums after their collision. This law can be extended for any number of masses interacting with each other. In ideal elastic collision not only is the momentum of the masses conserved but so is the mechanical energy of the masses. Assuming a perfectly horizontal collision (so we can ignore gravitational potential energy) the conservation of mechanical energy equation for a perfectly elastic collision would be the following: 1 2 m 1 v 1 i 2 + 1 2 m 2 v 2 i 2 = 1 2 m 1 v 1 2 + 1 2 m 2 v 2 2 Using a very similar argument as stated above to obtain the law of Conservation of Momentum for an elastic collision we can also obtain The Law of Conservation of Momentum for an Inelastic Collision. m 1 v 1 i + m 2 v 2 i = ( m 1 + m 2 ) v In a perfectly inelastic collision, the masses become stuck together once they collide and therefore move with the same velocity after the collision, as opposed to a perfectly elastic collision where the two masses bounce off each other after the collision and fly off with different final velocities. Also, while mechanical energy is conserved in a perfectly elastic collision, it is NOT conserved in a perfectly inelastic collision. Among some other reasons why the system of masses loses some energy in an inelastic collision, one can think of some of the energy being ‘used’ to cause the two masses to become stuck together. So, the sum of the final mechanical energies of the two masses is always less than the sum of their initial mechanical energies. 1 2 m 1 v 1 i 2 + 1 2 m 2 v 2 i 2 > 1 2 m 1 v 1 2 + 1 2 m 2 v 2 2 Finally, it should be pointed out that momentum is a vector quantity, and therefore when dealing with motion in more than one direction, one needs to write separate conservation of momentum equations for each direction involved. Such as the following: m 1 v 1 ix + m 2 v 2 ix = m 1 v 1 x + m 2 v 2 x 2

7. In the grey box near the top right of your screen, make the following selections. a. Check Momentum. Velocity should already be checked. If not, check it. b. Uncheck Reflecting Border c. Elasticity is set to 100% Procedure Run 1 1. Check the box near the bottom left of your screen that reads ‘More Data’. This will cause Position, Velocity and Momentum to appear. 2. In the grey data box, set the mass of the first ball to 1.50 kg. Record this value in the Table under Run 1 for m 1 . a. Set the velocity of the first mass to 1.00 m/s. b. Record this value in the Table under Run 1 for v 1i . c. Record the given initial momentum for mass 1, p 1i , in the Table under Run 1. 3. In the grey data box, set the mass of the second ball to 1.50 kg. Record this value in the Table under Run 1 for m 2 . a. Set the velocity of the second mass to 0.00 m/s. b. Record this value in the Table under Run 1 for v 2i . c. Record the given initial momentum for mass 2, p 2i , in the Table under Run 1. 4. To start the experiment, click the play button in the center of your screen. a. Once you click the play button it will turn into a pause button, which you will click to stop the experiment a second or two after the two masses collide. b. The final velocities and momentums for the masses are now given in the grey data box. Record them in the Table under Run 1 for v 1 , v 2 , p 1 , p 2 . 4

Run 2 1. Click on the orange refresh button in the bottom right of your screen to reset the experiment. After clicking the refresh button you will need to redo the following in the grey box near the top right of your screen. a. Check Momentum. Velocity should already be checked. If not, check it. b. Uncheck Reflecting Border. c. Elasticity is set to 100% 2. Check ‘More Data’ again, and in the grey data box set the initial values to the following, and then record them in the Table under Run 2: a. Mass 1, m 1 = 1.0 kg b. Mass 2, m 2 = 2.0 kg c. Velocity 1, v 1i = 2.0 m/s d. Velocity 2, v 2i = -0.5 m/s e. Also, record the initial momentum values in the Table under Run 2. 3. Now repeat Step 4 in Run 1, and record the final values in the Table under Run 2. Run 3 1. Click on the orange refresh button in the bottom right of your screen to reset the experiment. After clicking the refresh button you will need to redo the following in the grey box near the top right of your screen. a. Check Momentum. Velocity should already be checked. If not, check it. b. Uncheck Reflecting Border. c. Elasticity is set to 0 % 2. Repeat experiment with the same initial settings as Run 1. a. Record all values in the Table under Run 3. Run 4 3. Click on the orange refresh button in the bottom right of your screen to reset the experiment. After clicking the refresh button you will need to redo the following in the grey box near the top right of your screen. a. Check Momentum. Velocity should already be checked. If not, check it. b. Uncheck Reflecting Border. c. Elasticity is set to 0 % 4. Repeat experiment with the same initial settings as Run 2. d. Record all values in the Table under Run 4. 5

( 1.50 kg )( 1.00 m s )+( 1.50 kg )( 0 )= ( 3.00 kg ) ( 0.5 m s ) 1.50 kg· m s = 1.50 kg·m / s Run 4: ( 1.00 kg )( 2.00 m s )+ ( 2.00 kg ) ( − 0.50 m s ) = ( 3.00 kg ) ( 0.33 m s ) 1.00 kg· m s = 1.00 kg·m / s 2. For which of the four runs was energy conserved? Show Calculations. (20 points) Run 1: 1 2 m 1 v 1 i 2 + 1 2 m 2 v 2 i 2 = 1 2 m 1 v 1 2 + 1 2 m 2 v 2 2 1 2 ( 1.50 kg )( 1.00 m s ) 2 + 1 2 ( 1.50 kg )( 0 ) 2 = 1 2 ( 1.50 kg )( 0 ) 2 + 1 2 ( 1.50 kg )( 1.00 m s ) 2 0.75 J = 0.75 J Run 2: 1 2 ( 1.00 kg ) ( 2.00 m s ) 2 + 1 2 ( 2.00 kg ) ( − 0.50 m s ) 2 = 1 2 ( 1.00 kg ) ( − 1.33 m s ) 2 + 1 2 ( 2.00 kg )( 1.17 m s ) 2 2.25 J = 2.25 J Run 3: 1 2 m 1 v 1 i 2 + 1 2 m 2 v 2 i 2 > 1 2 m 1 v 1 2 + 1 2 m 2 v 2 2 1 2 ( 1.50 kg ) ( 1.00 m s ) 2 + 1 2 ( 1.50 kg ) ( 0 ) 2 > 1 2 ( 1.50 kg ) ( 0.5 m s ) 2 + 1 2 ( 1.50 kg )( 0.50 m s ) 2 0.75 J > 0.375 J Run 4: 1 2 ( 1.00 kg ) ( 2.00 m s ) 2 + 1 2 ( 2.00 kg ) ( − 0.50 m s ) 2 > 1 2 ( 1.00 kg ) ( 0.33 m s ) 2 + 1 2 ( 2.00 kg )( 0.33 m s ) 2 2.25 J > 0.163 J 7

3. Remember that energy is never created nor destroyed, but simply changes form. So, besides some of the energy being used to cause the two masses to stick together in a perfectly inelastic collision, what else could account for some of the loss of the mechanical energy? (20 points) Some potential factors that could account for this loss of mechanical energy include friction and air resistance. 8