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7. A 50 kg woman standing on a polished floor (with no friction) has a cat jump into her armsand she catches the cat. If the cat has a mass of 5 kg and jumps with a speed of 12 m/sa) How much momentum does the cat have before the woman catches it?b) What happens to the woman/cat combination after the cat is caught?c) What is the total momentum of the woman/cat combination after the cat is caught?d) What is the speed of the woman/cat combination after the cat is caught?

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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