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 = 3.2 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?

Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound?

Explanation / Answer

here,

h = 3.2 m

let the speed of 1 before the collison be u

using conservation of energy

m * g * h = 0.5 * m * u^2

9.81 * 3.2 = 0.5 * u^2

solving for u

u = 7.92 m/s

let the common speed be v

using conservation of momentum for collison

m * u = ( 2m + m) * v

7.92 = 3 * v

solving for v

v = 2.64 m/s

the final speed of combination is 2.64 m/s

when the collison is perfectly elastic

let the final speeds be v1 and v2

using conservation of momentum

m * u = m * v1 + 2m * v2

7.92 = v1 + 2 * v2 ....(1)

and

using conservation of kinetic energy

0.5 * m * u^2 = 0.5 * m * v1^2 + 0.5 * 2 * m * v2^2

7.92^2 = v1^2 + 2 * v2^2 ....(2)

from (1) and (2)

v1 = - 2.64 m/s

v2 = 5.28 m/s

the final height gained , h' = v1^2 /( 2g)

h' = 2.64^2 /( 2 * 9.81) m

h' = 0.36 m

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