A 5.00 kg chunk of ice is sliding at 12.0 m/s on thefloor of an ice-covered vall
ID: 1744889 • Letter: A
Question
A 5.00 kg chunk of ice is sliding at 12.0 m/s on thefloor of an ice-covered valley when it collides with and sticks toanother 5.00 kg chunk of ice that is initially at rest. (Seethe figure below (Intro 1 figure) .)Since the valley is icy, there is no friction. a)After the collision, how high above the valleyfloor will the combined chunks go? (Hint: Break thisproblem into two parts-the collision and the behavior after thecollision-and apply the appropriate conservation law to eachpart.) H= m A 5.00 kg chunk of ice is sliding at 12.0 m/s on thefloor of an ice-covered valley when it collides with and sticks toanother 5.00 kg chunk of ice that is initially at rest. (Seethe figure below (Intro 1 figure) .)Since the valley is icy, there is no friction. a)After the collision, how high above the valleyfloor will the combined chunks go? (Hint: Break thisproblem into two parts-the collision and the behavior after thecollision-and apply the appropriate conservation law to eachpart.) H= m a)After the collision, how high above the valleyfloor will the combined chunks go? (Hint: Break thisproblem into two parts-the collision and the behavior after thecollision-and apply the appropriate conservation law to eachpart.) H= mExplanation / Answer
Given : m1 = 5.00 kg u1 = 12.0 m /s m2 = 5.00 kg u2 = 0 m /s As the particles stick together after colision , Finalvelocity after collision is : v = ( m1 u1 + m2 u2 ) / (m1 +m2) = ( 5* 12.0 + 0 ) / ( 5 + 5) = 60 / 10 = 6 m/s We know that : h = v2 / 2g = 36 / ( 2* 9.8) = 1.837 m Hope this helps u!Related Questions
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