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)
<u>m₁ = 70 lb</u>
