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

Question

A package of mass m is released from rest at a warehouse loading dock and slides down the h = 2.4 m - high, frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass 2m, from the bottom of the chute. 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 collision between the packages is perfectly elastic. To what height does the package of mass m rebound? Express your answer to two significant figures and include the appropriate units. Figure: http://session.masteringphysics.com/problemAsset/1563635/6/10.P42.png

Explanation / Answer

A ) u1 = sqrt(2*g*h) = (2*9.81*2.4)^.5 = 6.86 m/s

initial momentum Pi = m1*u1+m2*u2

final momentum Pf = m1*V1+m2*V2

KEi = 0.5*m1*u1^2 + 0.5*m2*u2^2

KEf = 0.5*m1*V1^2 + 0.5*m2*V2^2

Pf = Pi

m1*(u-V1) = m2*(V2-u2)........1

KEf = KEi

m1*(u1^2-V1^2) = m2*(V2^2-u2^2)...............2

u1 - V1 = V2 - u2

u1 -u2 = V2 -V1

V2 = u1 - u2 + v1....(3)

using 3 in 1

V1 = (m1-m2)*u1 / (m1+m2) + (2*m2*u2)/(m1+m2) = (m-2m)*6.86/(m+2m) = 2.29 m/s

V2 = (m2-m1)*u2 / (m1+m2) + (2*m1*u1)/(m1+m2) = (2*m*6.86)/(m+2m)) = 4.57 m/s

A) perfectly in-elastic

Pf = Pi

m*u1 = (m+2m)*V

V = u1/3 = 2.29 m/s

B) elastic

V1 = (m1-m2)*u1 / (m1+m2) + (2*m2*u2)/(m1+m2) = (m-2m)*6.86/(m+2m) = 2.29 m/s

V2 = (m2-m1)*u2 / (m1+m2) + (2*m1*u1)/(m1+m2) = (2*m*6.86)/(m+2m)) = 4.57 m/s

m*g*h1 = 0.5*m*V1^2

h1 = V1^2/2g = (2.29^2)/(2*9.81) = 0.27 m

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