Hi, I need help with this Physics Reflection report about Impulse can you please check and add anything that needs to be added thank you

This question deals with uncertainties both in measurements and in calculated quantities. A) Describe how you determined the uncertainty in the measurement of the impulse of the force. (3 marks) To determine the uncertainty in the measurement of the impulse of the force, | followed a systematic approach using PASCO Capstone software and data from the force sensor and motion sensor. First, | collected data from both sensors during the experiment. The force sensor provided the force data, while the motion sensor provided the time data, which was crucial for calculating the impulse using the formula: impulse = force x time. Next, | focused on error analysis to determine the uncertainty in the force measurement from the force sensor. To pl'opagate the uncertainty in the force measurement to the impulse calculation, | applied standard error propagation technigues. This allowed me to quantify the uncertainty in the impulse measurement based on the uncertainty in the force measurement. B) Consider your result for quiz question 1, and describe (if you got it right) or correct (if you got it wrong) your uncertainty calculation for the change in momentum. For this part, you should and justify the key steps in your working through the calculation. (5 marks) Step 1: Understand the given expression for change in momentum, which ism X vl m x v2. Q2 - This expression calculates the change in momentum by subtracting the final momentum (m x v2) from the |nitial momentum ( m x v1). Step 2: Recognize that the uncertainty in a product (or quatient) can be found by adding the relative uncertainties of the components involved, then multiplying by the absolute value of the product (or quotient). The formula for the uncertainty of a product AB (where A and B are measured quantities with uncertainties ud and uB, respectively) is given by u(AB) = A () + (5)" Step 3: Apply the formula for uncertainty in a product to each momentum term separately. For the initial momentum m X v1, the uncertainty is um X v1 + m X uv, since uv is the uncertainty in either velocity ( v1 or v2 ) and um is the uncertainty in the mass m. This simplifies to um X v1 + uv X m using the properties of multiplication. Step 4: Similarly, for the final momentum m X v2, the uncertainty is um X v2 + m X uv, which simplifies to um X v2 + uv X m. Step 5: Recognize that when subtracting quantities, the uncertainties add. Therefore, the total uncertainty in the change in momentum (m X v1 m X v2 ) is the sum of the uncertainties in each term. This means you add the uncertainties found in Steps 3 and 4. Step 6: Combine the expressions for the uncertainties in the initial and final momentum to get the total uncertainty in the change in momentum. This gives us um x vl +uvr xm + um x v2 4+ uv X m. Step 7: Simplify the expression from Step 6. Notice that ur X m appears twice, so it can be combined. The simplified expression for the total uncertainty in the change in momentum is umx (vl +v2) + 2 X ur x m. Step 8: The final expression for the uncertainty in the change in momentum, in terms of the given symbols, is um X (vl + v2) + 2 X uv X m. This formula allows you to calculate the uncertainty in the result of m x v1 m X v2 given the uncertainties in mass (um) and velocity (uv). Interpretation and Discussion of Results - 8 marks Q3 - Synthesizing Knowledge and Critical Thinking - 8 marks In the final part of the laboratory, you explore collisions with the heavy spring and rubber stopper in addition to with the light spring in the earlier parts. Compare and contrast (qualitatively -if you only sketched graphs - or quantitatively - if you took full data on these) the nature of the force vs time graphs and also the velocity results for each case. Specifically: a) If you are doing things qualitatively please provide the sketches of the graphs in question, and if you are doing them more qualitatively please provide any values (with estimated uncertainties). (3 marks) b) For the force vs time graphs/values, compare the maximum force, the duration of the collision, and the impulse that results for each of the three objects. Consider how the two springs compare to one another and how the rubber stopper compares to the springs. (2 marks) c) For the velocity results for the light spring, do the velocity results support the idea that energy is also conserved in the collision? (1 mark) e V) S6. AN () . d) Finally, discuss, or provide an explanation for, the similarities and differences you note. (2 marks) Light Spring Hard Spring 19+ 0.5 18+0.5 0.08 + 0.05 0.05+0.05 0.5004 0.005 0.490+0.005 Rubber Bumper 5810.5 0.025+0.05 0.360+0.005 Maximum Force (N Duration of Collision (s) Impulse (N/s | determined the uncertainty of the maximum force taking half of the interval = 2 (1) = 0.5 and the uncertainty of the duration of collision was also determined in the same way = 2 (0.1) = 0.05. The maximum force exerted by both springs overlaps within a range of 3 uncertainties, indicating a similar maximum force for both. Similarly, the duration of collisions overlaps within 3 uncertainties, suggesting that the time the cart was in contact with each spring was similar. In contrast, the rubber bumper exhibits a higher maximum force due to its lower elasticity, allowing it to absorb more force. The collision duration is also shorter because the bumper does not need to retract before the cart reverses direction. The rubber bumper cannot propel the cart as effectively as the springs due to its lower elasticity, resulting in a smaller impulse. Reflection The maximum force exerted by both springs overlaps within a range of 3 uncertainties, indicating a similar maximum force for both. Similarly, the duration of collisions overlaps within 3 uncertainties, suggesting that the time the cart was in contact with each spring was similar. In contrast, the rubber bumper exhibits a higher maximum force due to its lower elasticity, allowing it to absorb more force. The collision duration is also shorter because the bumper does not need to retract before the cart reverses direction. The rubber bumper cannot propel the cart as effectively as the springs due to its lower elasticity, resulting in a smaller impulse