Question

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?

Answer 1

The displacement of the canoe is 1.46 m

Step-by-step 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)(x)/(t)=(M_(c)+M_(m))(x')/(t)`

`86.5*2.5=(61+86.5)* x'`

`x'=(86.5*2.5)/(61+86.5)`

`x'=1.46\ m`

Hence, The displacement of the canoe is 1.46 m