2_2Lab

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Feb 20, 2024

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2.2 Lab: Energy, Momentum, Conservation Laws Part 1 Recoil Before a gun fires a bullet, the gun & bullet make up a system with zero momentum. After the bullet fires, the recoil based on Newton’s Third Law causes the gun to move away from the bullet. If we know the masses of the gun, the bullet, and the velocity of the bullet, we can predict the recoil velocity of the gun. Simulation: oPhysics Momentum & Energy: Explosive "Collision" Open the simulation. Follow directions carefully for full credit. Settings: ● Set Initial Velocity Boxes (m/s) slider to 0. ● Set Mass of red box (kg) to your choice of one of these values: 1.2, 1.3, 1.4, 1.5 The instructor will use 1.9 kg. ● Set Mass of blue box (kg) to your choice of one of these values: 4.1, 4.2, 4.3, 4.4, 4.5 kg. The instructor will use 4.9 kg. ● Leave the Explosive energy at 50 J. For this simulation, the red box will represent a projectile such as a bullet from a gun moving in the negative direction. The blue box will be the recoiling object, such as a gun, moving in the positive direction. Record your initial masses in the table below. Click “Run” and watch. After a few seconds, the boxes will be pushed apart by 50 J of energy applying equal and opposite forces. Pause the simulation once you can clearly see the information on the motion. The final velocity, momentum ( p ), and kinetic energy ( KE ) of each box is given by the sim. Record the values in the table below. Remember to include units with all numbers. Vectors that point to the left should be given negative values. Answer the questions that follow. Object Mass final velocity momentum kinetic energy Red box (bullet) 1.3 kg -7.664 m/s -9.964 kgm/s 38.182 J Blue box (gun) 4.2 kg 2.372 m/s 9.964 kgm/s 11.818 J What is the total momentum of the boxes before the explosion? 0 kgm/s What is the total momentum of the boxes after the explosion? 0 kgm/s What is the total kinetic energy of the boxes after the explosion? 50 J Paste Screenshot here. 1

Unit 2 Practice Problem Example: A 0.02 kilogram bullet is launched with a muzzle velocity of 864.28 meters per second. It is fired from a rifle with a mass of 2.89 kilograms. What is the recoil velocity of the weapon in meters per second? Round your final answer to two decimals. The Law of Conservation of Momentum states: the total linear momentum of an isolated system is constant. So the equation for the gun recoil problem can be stated as (total mv before firing) = (total mv after firing). Remember that momentum is calculated as the mass × velocity . The total mv means adding the momentums of both objects (as vectors). ( m g v g + m b v b ) before = ( m g v g + m b v b ) after Since neither gun nor bullet has velocity before firing, the total momentum before firing is zero. After firing we must add the mv of both objects. 0 = m g v g + m b v b In the problem, the unknown is v g , so solve for that variable. v g = − m b v b m g Since in our simulation, the bullet is represented by the red box and the gun by the blue box, you can change the subscripts as follows: v ¿ = − m ¿ v ¿ m ¿ Use the equation to calculate the velocity of the gun (the blue box in our simulation). Round your predicted velocity to 3 places after the decimal. Predicted velocity of blue box (gun) recoil -0.596 m/s Simulation value for the blue box velocity -0.596 m/s The two values should be close if not exactly the same 2

Provide a screen shot of your calculation. Screen Shot of calculator (not the simulation). THOUGHT PROCES: CALCULATOR: An example of the calculation is shown below. v b = − m ¿ v ¿ m ¿ = − ( 1.1 kg ) ( − 8.616 m s ) 4.9 kg = 1.934 m s Note the negative sign in front of the fraction changes the negative velocity to give a positive velocity for the answer, indicating the gun recoils in the positive direction, opposite direction from the bullet. This exactly matches the result given by the simulation Applying the same equation to the practice problem example we get v 2 = − m bullet v bullet m gun = − ( 0.02 kg ) ( − 864.28 m s ) 2.89 kg = 5.98 m s Part 2 Inelastic Collision Open the PhET Collision Lab Simulation . 3

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Settings ● Click on the “Intro” tab. ● Check the boxes: Velocity, Values. ● Change the Elasticity Slider to 0% for a totally inelastic collision. ● Check the “More Data” box. ● Change the mass of ball 1 (on left) to the maximum value of 3.0 kg. ● Change the mass of ball 2 (on right) to a value between 1.31 and 1.99 kg. The instructor will use 1.10 kg. ● Ignore the position data, so long as ball 1 is on the left of ball 2. ● Change the velocity of ball 1 to any value between 2.31 and 2.99 m/s. The instructor will use 2.10 m/s. ● Change the velocity of ball 2 to zero (0.00). ● Record your initial values in the table below. ● Calculate the initial momentum of the system. Formula momentum = p = mv . ● Calculate the initial total kinetic energy of the system given by the simulation also. Formula: KE = 1 2 m v 2 . ● Round all these values to 3 places after the decimal. ● Paste Screenshot here. THOUGHT PROCESS: CALCULATOR: 4

● Click Play (optional: set the motion setting to “slow”) and pause after the collision takes place. If the objects don’t stick together, set the elasticity to zero, hit the blue reset button ( ), and repeat. ● Paste Screenshot here. ● Record the velocity of the two balls (now stuck together) after the collision. ● Calculate the initial and final momentum values and the total momentum values for each ball. Reminder: the variable p stands for momentum. ● Calculate the final kinetic energy of the objects after the collision. ● Answer the questions that follow the table. Object Mass (kg) v i (m/s) v f (m/s) p i (kg m/s) p f (kg m/s Ball 1 3.00 2.55 1.70 7.65 5.10 Ball 2 1.50 0.00 0.00 2.55 Total p before collision 7.65 kg m/s 5

Total p after collision 8.925 kg m/s Kinetic energy initial 9.1875 J Kinetic energy final 4.845 J Was the total kinetic energy of the system conserved in the collision? a Yes x b No Was the total momentum of the system conserved in the collision? x a Yes b No Practice Problem Example: A 2,626.51 kilogram truck runs into the rear of a 1,078.24 kilogram stationary car. The truck and car are locked together after the collision and move with speed 6.16 meters per second. What was the speed (in meters per second) of the truck before the collision? Round your answer to two decimal places. We will use our simulation masses and final velocity to calculate the initial velocity of the ball that was moving before the collision, similar to the truck car collision problem. Conservation of Momentum m 1 v 1 + m 2 v 2 = ( m 1 + m 2 ) v f Since v 2 = 0, this simplifies to m 1 v 1 = ( m 1 + m 2 ) v f Solving this for v 1 , we get v 1 = [ m 1 + m 2 ] m 1 v f Use this equation to calculate the initial velocity of Ball 1 using your recorded data. An example of the instructor’s data is included below. m 1 = Mass of ball 1 m 2 = Massof ball 2 v f = velocityof balls aftercollision v 1 = velocity of ball 1 beforethe collision 6

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Calculated velocity of ball 1 before collision 4.37 m/s. Simulation value of ball 1 before collision 6.16 m/s These should be close if not exactly the same. Make a screen shot of your calculator and paste it here. Screen shot of Calculator (not the sim). THOUGHT PROCESS: CALCULATOR: Instructor example: v 1 = [ m 1 + m 2 ] m 1 v f = [ ( 3.0 kg ) + ( 1.10 kg ) ] 3.0 kg ( 1.54 m s ) = 2.105 m s Simulation setting: 2.10 m/s Applying this formula to the Practice Problem example, we get 7

v t = ( m t + m c ) v t + c m t = [ ( 2626.51 kg ) + ( 1078.24 kg ) ] ( 6.16 m s ) 2626.51 kg = 8.69 m s 8

Part 3 Conservation of Potential and Kinetic Energy during Free Fall Open the PhET Projectile Motion Simulation . Settings ● Click on the “Lab” tab. ● Rotate the cannon to point straight up (90 ᵒ ). ● Set the initial speed to anything between 11 m/s and 25 m/s. The instructor will use 30 m/s. ● Change the gravity setting to 9.80 m/s 2 . ● Fire the cannon and wait for the cannonball to return to the ground. ● If the cannonball travels above the view of the simulation, click the magnifying glass with the minus sign in it to shrink the scale until you can see the maximum position of the cannonball. ● Drag the measuring tool (blue box with the crosshairs on the left side) and position the crosshairs on the green dot at the top of the trajectory. Record the maximum height, the initial velocity, and the gravity value in the table below. ● Use the formula presented in class (also see below the data table) to calculate what it predicts for the maximum height of the projectile. Initial upward velocity 15 m/s Maximum height 11.48m Acceleration of gravity 9.80ms Calculated maximum height 11.48m Paste a screenshot showing your maximum height measurement and your initial velocity. Paste Screen Shot here. 9

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The formula is based on the conservation of energy. It assumes the total energy at the start, all kinetic energy and no potential energy is equal to the total energy at the end, all potential energy and no kinetic energy. 1 2 m v 2 = mgd Solving for the height we get d = v 2 ( 2 g ) An example of the calculation is shown using the instructor’s data. d = v 2 ( 2 g ) = ( 30 m s ) 2 [ ( 2 ) ( 9.80 m s 2 ) ] = 45.92 m This matches the simulation height exactly. Part 4: Conservation of Potential and Kinetic Energy on a Ramp If there is time, open the PhET Energy Skate Park Simulation . Settings ● Click on the “Measure” Tab. ● Check the “Grid” and “Speed” check boxes. ● Check the second track shape on the left, a ramp going down the to the right. ● Track should look like this. ● Optional: Click the red dot at the top and extend the ramp up and to the left. ● Pause the simulation. ● Set the skateboarder anywhere on the ramp between 4 and 6 meters. ● Use the tape measure to measure the vertical distance from the red dot under the skateboard and the ground. Record this value in the table. The instructor will use 7.0 meters. ● Leave the friction setting at the default of none. 10

● Note that the skateboarder’s initial speed is zero (rest). Optional: check the slow setting. Click Play and pause the simulation once the skateboarder is on the ground. Record the final speed of the skateboarder. Initial height of skateboarder 5.0 m Final speed of skateboarder at bottom 9.9 m/s Acceleration of gravity 9.8m/s/s Calculated final speed 9.90 m/s. The formula is based on the same assumption as in part 3 above. Solving the conservation of energy equation for the final speed, we get v = √ 2 gd Use your starting height and the acceleration of gravity to calculate the final speed and record it in the table. Round your result to 2 places after the decimal. It should be close to if not exactly the same as the simulation speed. The calculation using the instructor’s data is shown below. v = √ 2 gd = √ ( 2 ) ( 9.8 m s 2 ) ( 7.0 m ) = 11.7 m s ● With the simulation paused, click the “Restart Skater” button and make a screenshot of your initial position. Paste Screenshot here. CALCULATIONS: 11

● Click play and pause when the skater is on the ground and make a second screenshot of your final position. Paste Screenshot here. This formula works for any case where gravity is the sole accelerating force and where air resistance and friction can be ignored, such as free fall of a dropped object. Instructor Example Screenshots: 12

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Note that the same speed is attained with any mass object. You can repeat the run with different skaters with different masses and the final speed is always the same. Problem Example: A ball is at rest 75.86 meters above the ground. If it is allowed to fall under the influence of gravity only (ignore air resistance), how fast (in meters per second) is it moving just before hitting the ground? Round your final answer to two decimal places. Solution: v = √ 2 gd = √ ( 2 ) ( 9.8 m s 2 ) ( 75.86 m ) = 38.56 m s If there is time, try the following experiment. This will not be graded. ● Pause the simulation. ● Start with the skateboarder’s red dot at the 8 meter mark to the right side of the track (not on the track). ● Set the simulation to run in slow motion. ● Click play but pause again before the skateboarder reaches the ground. ● Use the step button to the right of the play button to advance the skateboarder to just above the ground, but not yet on the ground. ● Note the speed on the meter. ● Now calculate the speed using the formula for 8 m height. ● This shows the path does not matter, so long as there is no friction or air resistance. ● Position the energy crosshair device over the various dots that track with the freefalling skateboarder. Note the total energy does not change. What is lost in PE is gained in KE. 13

● Finally calculate the GPE at the top using the formula GPE = mgh and the KE at the bottom using the formula KE = 0.5 m v 2 . Both values should be close to the same. solution: To calculate the ball's speed just before impact, we will use the following data: Height above ground (h) equals 75.86 meters Gravitational acceleration (g) equals 9.8 m/s2 Formula for calculating speed using the principle of energy conservation is: v = √(2 * g * h) Now, let us plug the values into the formula and determine the ball's speed (v) just before it hits the ground. v = √(2 * 9.8 m/s² * 75.86 m) v = 1480.4288 m2/s2 and v 38.56 m/s Approximately 38.56 meters per second, rounded to two decimal places, is the speed of the ball just before it hits the earth. In accordance with the principle of energy conservation, the skateboarder's experiment demonstrates that the ultimate speed is independent of the path taken as long as there is no friction or air resistance. Throughout the motion, the total energy remains constant, and any potential energy lost is converted to kinetic energy. Using the formula GPE = mgh, we can calculate the gravitational potential energy (GPE) at the summit using the same height (h = 75.86 m) and the mass of the ball. However, the mass of the object is not provided in the data, so this calculation cannot be completed without it. To calculate the kinetic energy (KE) at the bottom using the formula KE = 0.5mv2, we would also need the ball's mass and velocity at the bottom, which are not provided in the data. If you know the mass of the ball and its speed at the bottom, you can verify that the GPE and KE values are near to one another, demonstrating the conservation of energy. 14

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