(5) A package of mass m is released from rest at a warehouse loading dock and sl

Question

(5) A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0 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 as shown in the figure. 3.0 m a. Suppose the packages stick together. What is their common speed after the collision? b. Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound?

Explanation / Answer

Given,

h = 3 m ; mass1 = m ; mass2 = 2m ;

a)The box at the top will have a speed of:

v = sqrt (2 g h) when it collides

its momentum, p1 = m sqrt(2gh)

final momentum will be:

pf = (m + 2m)vf

pf = 3 m vf

from conservation of momentum, p1 = pf

m sqrt(2gh) = 3m vf

vf = sqrt(2 gh)/3

vf = sqrt (2 x 9.81 x 3)/3 = 2.56 m/s

Hence, vf = 2.56 m/s

b)v1' = (m1 - m2)/(m1 + m2) v1

v1' = -m/3m v1 = -1/3sqrt (2gh)

from conservation of energy:

1/2 m v1^2 = m g h'

1/2 (-1/3 sqrt(2gh)^2) = g h'

1/2 x 1/9 x 2 gh = gh'

h' = h/9 = 3/9 = 1/3 = 0.33 m

Hence, h' = 0.33 m = 33 cm

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.