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An airline employee tosses two suitcases in rapid succession with a horizontal velocity of 7.2 ft/s onto a 50-lb baggage carrier which is initially at rest. Problem 14.003.a Conservation of momentum: two colliding suitcases Knowing that the final velocity of the baggage carrier is 4.8 ft/s and that the first suitcase the employee tosses onto the carrier has a weight of 30 lb, determine the weight of the other suitcase. (You must provide an answer before moving on to the next part.) The weight of the other suitcase is lb.

Answer :

hamzaahmeds

Answer:

m₁ = 70 lb

Explanation:

Here we will use the law of conservation of momentum:

m₁u₁ + m₂u₂ + m₃u₃ = m₁v₁ + m₂v₂ + m₃v₃

where,

m₁ = mass of first suitcase = ?

m₂ = mass of second suitcase = 30 lb

m₃ = mass of baggage carrier = 50 lb

u₁ = initial speed of first suitcase = 7.2 ft/s

u₂ = initial speed of second suitcase = 7.2 ft/s

u₃ = initial speed of baggage carrier = 0 ft/s

v₁ = Final speed of first suitcase = 4.8 ft/s

v₂ = Final speed of second suitcase = 4.8 ft/s

v₃ = Final speed of baggage carrier = 4.8 ft/s

because after collision all three will have same speed

Therefore,

(m₁)(7.2 ft/s) + (30 lb)(7.2 ft/s) + (50 lb)(0 ft/s) = (m₁)(4.8 ft/s) + (30 lb)(4.8 ft/s) + (50 lb)(4.8 ft/s)

(m₁)(7.2 ft/s) + (216 lb ft/s) + (0 lb ft/s) = (m₁)(4.8 ft/s) + (144 lb ft/s) + (240 lb ft/s)

(m₁)(7.2 ft/s) - (m₁)(4.8 ft/s) = 168 lb ft/s

m₁ = (168 lb ft/s)/(2.4 ft/s)

m₁ = 70 lb

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