A package of mass is released from rest at a warehouse loading dock and slides d

Question

A package of mass is released from rest at a warehouse loading dock and slides down the = 3.4- high, frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass , from the bottom of the chute. (Figure 1) /Users/cpi/Desktop/10.P42.png A) Suppose the packages stick together. What is their common speed after the collision? Express your answer to two significant figures and include the appropriate units. B)Suppose the packages stick together. What is their common speed after the collision? Express your answer to two significant figures and include the appropriate units.

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 3.4] =>v = 8.16 m/s 1) By the law of momentum conservation:- =>m1u1+m2u2 = (m1+m2) x v =>m x 8.16+ 0 = 3m x v =>v = 2.72 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 = -8.16-------------(i) BY the law of momentum conservation:- =>m1u1 + m2u2 = m1v1 + m2v2 =>m x 8.16+ 0 = mv1 + 2mv2 =>v1+2v2 = 8.16---------------(ii) By 2 x (i) + (ii):- =>3v1 = -8.16 =>v1 = -2.72 m/s Thus again by the law of energy conservation,Let the package of mass m gain h meter due the velocity of 2.72 m/s =>PE(top) = KE(bottom) =>mgh = 1/2mv1^2 =>h = v1^2/2g =>h = (2.72)^2/(2 x 9.8) =>h = 0.377 m

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