question 25 25. I, A packagc of mass ,n is \'eleased from rest at a warc ing doc

Question

question 25

25. I, A packagc of mass ,n is 'eleased from rest at a warc ing dock and slides down the 3.0-m-high, frictionless chute of FIGURE EX11.2s to a waiting truck. Unfortunately. went on a break without having removed the previous package.,O mass 2m, from the bottom of the chute. the truck driver a. Suppose the packages stick together. What is their common speed after the collision? b. Suppose the collision between the packages is perfectly clastic. To what height does the package of mass m reboun tIt 3.0 m FIGURE EX11.25

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]
=>v = 7.67 m/s
1) By the law of momentum conservation:-
=>m1u1+m2u2 = (m1+m2) x v
=>m x 7.67 + 0 = 3m x v
=>v = 2.56 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 = -7.67 -------------(i)
BY the law of momentum conservation:-
=>m1u1 + m2u2 = m1v1 + m2v2
=>m x 7.67 + 0 = mv1 + 2mv2
=>v1+2v2 = 7.67 ---------------(ii)
By 2 x (i) + (ii):-
=>3v1 = -7.67
=>v1 = -2.56 m/s
Thus again by the law of energy conservation,Let the package of mass m gain h meter due the velocity of 2.56 m/s
=>PE(top) = KE(bottom)
=>mgh = 1/2mv1^2
=>h = v1^2/2g
=>h = (2.56)^2/(2 x 9.8)
=>h = 0.33m

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