Part 2 of 3 - Categorize We use the momentum version of the isolated system model to analyze the collision, and the ener

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Part 2 Of 3 Categorize We Use The Momentum Version Of The Isolated System Model To Analyze The Collision And The Ener 1 (65.02 KiB) Viewed 26 times

Part 2 of 3 - Categorize We use the momentum version of the isolated system model to analyze the collision, and the energy version of the isolated system model to analyze the subsequent sliding process. Part 3 of 3-Analyze The collision, for which figures and are before and after pictures, is perfectly inelastic, and momenturn is conserved for the system of clay and block. We have M₁V₁ (M₁M₂)V₂ my 1 Gy 22 7.50 m In the sliding process occurring between figures and the original kinetic energy of the surface, block, and clay is equal to the increase in internal energy of the system due to friction. +₂₂² - Substituting the expression for in terms of the total mass and friction coefficient, we have m₂ +₂₂² = {m₂ + m₂}gl Solving for v₂ gives v2 = √2129 V1 = √2(0.65 = 9.77 ✔m/s. Now from the momentum conservation equation, we have the following. 07. v₂ ✔7.50 ✔ 0.108 ✔kg 0.095 x Your response differs from the correct answer by more than 100 % . kg Submit Skip (you cannot come back! | m)(9.80 m/s²) 9.77✔ m/s) Enter a number differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s.