Please help me with the lab.

To fill the table use this link to do the experiment: Physics Simulation: Collisions (physicsclassroom.com)

C Lab 16: Conservation of Momentum Introduction According to the law of conservation of momentum, an impulse generated internal to a system of two or more objects cannot change the total momentum of the system and follows the equation: Par + Pai = Paz + Paz If the total momentum before an impulse is zero, then the total momentum after the impulse must also be zero. The system under test comprises of two mass carts on a frictionless track. An impulse lasting a fraction of a second is created within the system by the release of a compressed spring. The impulses on the mass carts are equal and opposite, and cancel each other out. The mass carts accelerate during the impulse, and coast after. The momentum will be calculated by measuring the velocities of the mass carts while varying the masses of one of the carts. As in all "explosions," the initial velocity and initial momentum of each mass cart are zero. The comparison of the momentum of the two mass carts after the impulse will verify the law of conservation of momentum. Objectives During this laboratory investigation students will: Observe Newton's third law of motion during a collision. Verify the law of conservation of momentum under varying conditions.Pre-lab Questions 1. What does Newton's Third Law tell us about the force experienced by cach mass cart due to an impulse? " Every Action has an equal $ opposite Reaction " 2. The forces must act over the same time interval during an explosion collision (contact force). How are the impulses experienced by each mass cart related. I = AP = FAt Materials Two Vernier mass carts (one with a spring) Vernier frictionless track Two J-kg Vernier attachable masses Stopwatch timer Balance Procedure 1. Obtain all materials listed above. Determine the mass of both mass carts by placing them on a balance. Record their masses in Table 16-1 as Cart A and Cart B. 2. Place the two mass carts on the frictionless track. Adjust the track legs so that the track is level and the carts do not move on their own. Place the track bumpers at the end of both sides of the track (See Figure 16-1). 3. Set the spring on the mass cart and place the two carts together in the middle of the track. Place a piece of tape at point "O", where the mass carts meet. Lightly tap the release mechanism and observe carefully whether the mass carts reach the track bumpers at the same time. 4. Repeat the procedure until the mass carts reach the track bumpers at the same time. When they do, return the carts to their starting points and measure the displacements of each cart from the front of the cart to the bumper. Record these displacement values in the table. Note that one of the displacements will be positive and the other negative as displacement is a vector. 5. Trigger the spring again. This time, use the timer to record the time interval for the carts to reach the bumpers. Record the time in the table.Name 6. Determine the average velocity for each cart by dividing the displacement of the cart during the trial by the time interval. Record your calculation in the table. Note, one velocity will be positive, the other negative. 7. Calculate the momentum of each cart after the impulse using the equation p = my. Add the two values together to determine total momentum after impulse. 8. Secure one of the Vernier masses to mass cart B. Repeat steps 3-7. 9. Add a second Vernier mass to mass cart B, securing with tape. Repeat steps 3-7. 10. Return all materials to designated places and clean up lab table.Data and Calculations Table 16-1 Trial 1 Before Impulse After Impulse Mass Momentum Displacement time interval avg velocity Momentum (kg) (kg.m/s) (m) (5) (m/s [kg:m/s) Cart A MA= dA= PAT Cart B mg 0 Total Error= Trial 2 (+ 500 g ) Before Impulse After Impulse Mass Momentum Displacement time interval ave velocity Momentum (KE) (kg. m/5) (m) (5) (m/5] (kg-m/s] Cart A my= PAI= 0 PAT Cart 8 Pm= 0 Total 0 Error = Trial 3 (6 cool + ) Before Impulse After Impulse Mass Momentum Displacement time interval avg velocity Momentum (ke) (kg.m/s) (m] (5) (m/s) [kg-m/s) Cart A MA= PAI= 0 dA= tm VAJ PAT Cart B my PB= 0 Total 0 Error =Interpretation 1. Determine the percent error for Trials I through 3 using the following formula Errorl % error Ipxal + Ipozl X 100 Trial Percent Error (%) Sample Calculation Space: How does changing the mass of cart B affect the velocity and displacement of cart A? Explain. Which Trial yielded the highest percent error? What do you think caused this error?Name 2. When fired from a cannon, a shell emerges from the barrel with a very high velocity. On the basis of the law of conservation of momentum, explain how the muzzle velocity (the velocity of the shell as it leaves the barrel) of a cannon shell might be measured. 3. The accuracy of a handgun is greatly increased when a silencer is attached to the end of the barrel. Most people believe this is due to the increased barrel length when the silencer is in place. However, it is mainly due to a decrease in the recoil of the gun with the added weight from the silencer. Less recoil equals greater accuracy! Suppose a handgun has a mass of 1.5-kg fires a bullet with mass 0.030-kg at 384 m/s. What is the recoil velocity of the gun when the bullet lelyes? Now add the silencer onto the gun. Suppose the silencer has a mass of 0.50-kg. What is the new recoil velocity of the gun with silencer attached?Before Impulse After Impulse Mass Momentum Displacement time interval avg velocity Momentum (kg) (kg.m/s) (m) (S) (m/5) (kg.m/s) Cart A ma= | Ka PAI= 0 dA= 6 m It= 10 p VAZ= 0. 6 PN= 0.6 Cart B mg Pa:= 0 de- - 5 m VB -0.5 PM -O.S Total x 100 = 10% 0 Error = Total Momentum after = 0. 1 kgms"Before Impulse After Impulse Mass Momentum Displacement time interval avg velocity Momentum (kg) (kg.m/s) (m) (s) (m/5) (kg.m/s) Cart A ma= 1 .5 kg PM= 0 dA = 5 9 8 VAZ= 0.55 PAI 0. 825 Cart B 1.5 Kq PM= 0 de - 4 VB= - 0. 94 PM= - 0.66 0.165 x 100 Total 0 Error = 20 16% Total Momentum after : 0. 165 kams'Before Impulse After Ionpulse Mass Momentum Displacement time interval avg velocity Momentum (kg) (kg.m/s) (m) (s) (m/s) (kg.m/s) Cart A ma = 2 kg PAI= 0 dA= 4 t= 7p VN2= 0.57 1.19 Cart B ms 2 ka Pa:= 0 de - 3 VB - 0 . 42 PM# - 0.84 0.3x100 Total 0 Error = = 30% Total Momentum alty ~ 0.3 kgms