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 = 2.0 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. (Figure 1) 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. 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.

Explanation / Answer

for block of mass "m"

using conservation of energy

Kinetic energy at bottom = Potential energy at the top

(0.5) m v2 = mgh

v = sqrt(2gh)

v = sqrt(2 x 9.8 x 2) = 6.26 m /s                 (speed of block "m" just before collision with block "2m")

V = speed of the combination after the collision

using conservation of momentum

mv = (m + 2m) V

mv = 2mV

V = v/2 = 6.26 / 2 = 3.1 m/s

b)

v1 = speed of block "m" after collision

v2 = speed of block (2m) after collision

using conservation of momentum

mv = m v1 + 2m v2

v1 + 2 v2 = 6.26                               

v1 = 6.26 - 2 v2                                           eq-1

for perfectly elastic collision

Velocity of saperation = velocity of approach

v2 - v1 = v - 0

v2 = v1 + v

Using eq-1

v2 = 6.26 - 2 v2   + 6.26

v2 = 4.2 m/s

v1 = 6.26 - 2 v2 = 6.26 - 2 (4.2) = - 2.14 m/s

height gained = h1 = v12 /(2g) = (- 2.14)2 / (2 x 9.8) = 0.23 m = 23 cm

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