Secure 874528818offset next Hanework Problem 11.25-Enhanced - with Feedback prev

Question

Secure 874528818offset next Hanework Problem 11.25-Enhanced - with Feedback previous8 Problem 11.25-Enhanced-with Feedback Part A Apackage of mass m is released tom rent a warehouse loning dock and sldes down te h 38 m . , frid nless chute to a wašing truck. Unfortunately, the truck driver went on a break without harving removed the peevious package, of mass 2m Suppose the packages stick together What is their common speed afer the colision? Express your answer to two significant figures and include the appropriate units. from the botiom of the chute. Eigure 1) You may want to review (D274-276 Correct Here we learm how to use the laws of conservation of momertum and energy to calcuilate the speed of a body after an inelasse collision Part B Suppose the collision between the packages is perlectly elastic. To what height does the package of mass m rebound? Express your answer to two significant figures and include the appropriate units. Value Units Incorrect; Try Again; 5 attempts remaining Contin

Explanation / Answer

Let the velocity of the package of m kg is v m/s as it reaches the bottom, By the law of energy conservation:-
=>PE(top) = KE(bottom)
=>mgh = 1/2mv^2
=>v = sqrt[2gh]
=>v = sqrt[2 x 9.8 x 3.8]
=>v = 8.63 m/s
1) By the law of momentum conservation:-
=>m1u1+m2u2 = (m1+m2) x v
=>m x 8.63 + 0 = 3m x v
=>v = 2.9 m/s

2) Let the velocity of the m is v1 and the velocity of the 2m is v2 after the collision,by the energy conservation:-
=>v1 - v2 = u2 - u1
=>v1 - v2 = -8.63 -------------(i)
BY the law of momentum conservation:-
=>m1u1 + m2u2 = m1v1 + m2v2
=>m x 8.63 + 0 = mv1 + 2mv2
=>v1+2v2 = 8.63 ---------------(ii)
By 2 x (i) + (ii):-
=>3v1 = -8.63
=>v1 = -2.9 m/s
Thus again by the law of energy conservation,Let the package of mass m gain h meter due the velocity of 2.9 m/s
=>PE(top) = KE(bottom)
=>mgh = 1/2mv1^2
=>h = v1^2/2g
=>h = (2.9)^2/(2 x 9.8)
=>h = 0.43 m

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