Lab 7 - Elasticity

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Dec 6, 2023

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Mohamed Aden 10/13/2023 Lab #7 Elastic and Inelastic collison Q1: What might cause a collision to be other than perfectly elastic or totally inelastic? Collisions can be other than perfectly elasticity or total inelasticity becauseof factors like energy dissipation, deformation, friction, and rotational motion. Real-world collisions mostly involve some level of energy loss as heat or deformation, leading to a reduced elasticity compared to ideal situations. Q2: Modern cars are designed with “crumple zones” (areas that are designed to be deformed during a collision). Considering that this deformation takes energy, is this good or bad design on the part of the engineers? Why or why not? Crumple zones in car is a good design by engineers because thse zones absorb energy during collisions, it is included for safety purposes. These crumple zones reduce the impact transferred to the person in the car which reduces the risk of serious injury. Q3: What quantities are conserved in this collision? Explain your answer. The momentum and kinetic energy are conserved in this collision. Momentum is always conserved if there is no external force acting on the objects, and since the collison is elastic the kinetic energy is conserved. Q4: Assuming that mass is constant, and considering v1 to be an independent variable, Eq. 7.5 describes a linear relationship. Write an expression for the slope of Eq. 7.5. The equation for the slope of the linear relationship is slope = 2𝑚 1 /𝑚 1 + 𝑚 2 Mass of red box = 4kg ( 𝑚 1 ) Mass of blue box = 2kg ( 𝑚 2 )
Elasticity at 1 Trials 𝑣 1 (𝑚𝑠) 𝑣' 2 (𝑚𝑠) 1 10.0 13.3 2 8.0 10.7 3 6.0 8.0 4 4.0 5.3 5 2.0 2.7 Analysis: Q5: Using your values for m1 and m2, what is the numerical value for your theoretical slope? Using our equation from Q4, we’d get the numerical value for our theoretical slope 2(4kg)/4kg + 2kg = 1.3. Q6: Considering your best-fit equation, what is the numerical value for your experimental slope? The slope of our line of best fit is y = 1.33x + 0.02, so the numerical value of our experimental slope is 1.3.
Q7: How does the numerical value for your experimental slope compare to the numerical value for your theoretical slope? Looking at both the theoretical and numerical values of the slopes, they are both the same. There is a connection between the initial velocity of block one and the final velocity of block 2. The final velocity of block two is the initial velocity velocity of block one added to the mass of block one. Q8: Is the total momentum conserved in this collision? Explain your answer. The total momentum is conserved in the collision because there is no external force present. Q9: Assuming mass is constant, and considering v1 to be an independent variable, Eq. 7.6 describes a linear relationship. Write an expression for the slope of Eq. 7.6. The equation for the slope of the linear relationship is slope = . 𝑚 1 /𝑚 1 + 𝑚 2 Elasticity at 0 ( Totally inelastic ) Mass of red box = 4kg ( 𝑚 1 ) Mass of blue box = 2kg ( 𝑚 2 ) Trials 𝑣 1 (𝑚𝑠) 𝑣' 2 (𝑚𝑠) 1 10.0 6.7 2 8.0 5.3 3 6.0 4.0 4 4.0 2.3 5 2.0 1.3 Q10: Using your values for m1 and m2, what is the numerical value for your theoretical slope? Using our equation from Q4, we’d get the numerical value for our theoretical slope 4kg/4kg + 2kg 4/6 = 0.67.
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