Please help?
2017-12-08 2:07 pm
Sandy is sitting at rest in a canoe 20 meters from the shore with no paddle. Her mass together with that of everything in the canoe and the canoe itself is 125kg. She finds an old chemistry book at the bottom of the canoe and throws the book out of the canoe into the water at 14m/s away from sea shore. She reaches the shore 25 seconds later,traveling at constant speed. What is the mass of the chemistry book?
回答 (1)
2017-12-08 2:38 pm
Let y kg be the mass of the chemistry book.
Mass of the canoe (with every in it other than the chemistry book), m₁ = (125 - y) kg
Mass of the book, m₂ = y kg
Speed of the canoe after Sandy throws the chemical book
= (20 m) / (25 s)
= 0.8 m/s
Before Candy throws the chemistry book :
Velocity of the canoe, u₁ = 0 m/s
Velocity of the book, u₂ = 0 m/s
Just after Candy throws the chemistry book :
Velocity of the canoe, v₁ = 0.8 m/s
Velocity of the book, v₂ = -14 m/s
(Velocity of the book is negative because its direction is opposite to that of the canoe.)
By the law of conservation of momentum :
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0 = (125 - y) (0.8) + y (-14)
0 = 100 - 0.8y - 14y
14.8y = 100
y = 6.8 (to 2 sig. fig.)
Mass of the chemistry book = 6.8 kg
Mass of the canoe (with every in it other than the chemistry book), m₁ = (125 - y) kg
Mass of the book, m₂ = y kg
Speed of the canoe after Sandy throws the chemical book
= (20 m) / (25 s)
= 0.8 m/s
Before Candy throws the chemistry book :
Velocity of the canoe, u₁ = 0 m/s
Velocity of the book, u₂ = 0 m/s
Just after Candy throws the chemistry book :
Velocity of the canoe, v₁ = 0.8 m/s
Velocity of the book, v₂ = -14 m/s
(Velocity of the book is negative because its direction is opposite to that of the canoe.)
By the law of conservation of momentum :
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0 = (125 - y) (0.8) + y (-14)
0 = 100 - 0.8y - 14y
14.8y = 100
y = 6.8 (to 2 sig. fig.)
Mass of the chemistry book = 6.8 kg
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