Physics, 21.10.2020 16:01 ian2006huang

A man with a mass of 86.5 kg stands up in a 61-kg canoe of length 4.00 m floating on water. He walks from a point 0.75 m from the back of the canoe to a point 0.75 m from the front of the canoe. Assume negligible friction between the canoe and the water. How far does the canoe move?

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Answer from: hello6975

The displacement of the canoe is 1.46 m

Explanation:

Given that,

Mass of canoe = 61 kg

Mass of man = 86.5 kg

Length = 4 m

Let the the displacement of the canoe is x'

We need to calculate the displacement of the man

Using formula of displacement

$x=x_{2}-x_{1}$

Put the value into the formula

x=4-(0.75+0.75)

$x=2.5\ m$

We need to calculate the displacement of the canoe

Using conservation of momentum

$M_{m}v_{m}=(M_{c}+M_{m})v_{c}$

$M_{m}\dfrac{x}{t}=(M_{c}+M_{m})\dfrac{x'}{t}$

$86.5\times2.5=(61+86.5)\times x'$

$x'=\dfrac{86.5\times2.5}{61+86.5}$

$x'=1.46\ m$

Hence, The displacement of the canoe is 1.46 m

Answer from: Quest

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Answer from: Quest

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A man with a mass of 86.5 kg stands up in a 61-kg canoe of length 4.00 m floating on water. He walks...

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