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Sagot :
Answer:
a) The momentum of the cat before the woman catches it = 60 kgm/s
b) After the cat is caught, the woman and the cat move with an equal velocity
c) Total momentum of the woman/cat combination after the cat is caught =
60 kgm/s
d) The speed of the woman/cat combination after the cat is caught = 1.09 m/s
Explanation:Mass of the woman, m₁ = 50 kg
The woman was standing before she catches the cat. This means that she was at rest and the initial velocity is 0
u₁ = 0 m/s
The mass of the cat, m₂ = 5 kg
The cat's speed, u₂ = 12 m/s
Momentum = Mass x Velocity
a) The momentum of the cat before the woman catches it = m₂u₂
The momentum of the cat before the woman catches it = 5(12)
The momentum of the cat before the woman catches it = 60 kgm/s
b) After the cat is caught, the woman and the cat move with an equal velocity
c)
Total momentum of the woman/cat combination before the cat is caught= Total momentum of the woman/cat combination after the cat is caught
The initial momentum of the woman = m₁u₁
The initial momentum of the woman = 50(0) = 0 kgm/s
Total momentum of the woman/cat combination after the cat is caught = 0 + 60
Total momentum of the woman/cat combination after the cat is caught =
60 kgm/s
d) The speed of the woman/cat combination after the cat is caught
m₁u₁ + m₂u₂ = (m₁ + m₂)v
50(0) + 5(12) = (50+5)v
60 = 55v
v = 60/55
v = 1.09 m/s
The speed of the woman/cat combination after the cat is caught = 1.09 m/s
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