Question

A man with a mass of 95.0 kg is out fishing with his daughter, who has a mass of 25.0 kg. They are initially sitting at opposite ends of their 3.0-m boat, which has a mass of 60.0 kg and is at rest in the middle of a calm lake. The man and the daughter then carefully trade places. We will neglect any resistive forces between the boat and the water.

(a) How far has the boat moved, once the man and his daughter have finished trading places? m

(b) Halfway through the process, the man and the daughter are each traveling with speeds of 1.30 m/s, with the man's velocity directed toward one end of the canoe and the daughter's velocity directed toward the other end of the canoe. That is all as measured by a stationary observer on the shore. As seen by that same stationary observer on the shore, what is the speed of the canoe at this instant?

Answer 1

a).The boat moved

x=1.16m

b). The speed canoe at this instant

v=1.3 m/s

Step-by-step explanation:

Let consider a system of n' particles with respectives velocities and accelerations so for the distance

a)

Let the boat moves behind by x m from center of mass so

`x_(t)=\frac{m_(p)*x'+m_(d)*x'+m_(b)*x'}{m{p}+m_(d)+m_(b)}`

`x_(t)=(95kg*(3m)/(2)'+25kg*(3m)/(2)'+62kg*0')/(95kg+25kg+60kg)`

`=(95kg*(x-1.5)-25kg*(1.5+x)+62kg*(-x))/(95kg+25kg+60kg)`

Resolve to x'

`180x+105=-105`

`x=1.16`

b).

Vcom=0

`v_(t)=\frac{m_(p)*v'+m_(d)*x'+m_(b)*v'}{m{p}+m_(d)+m_(b)}`

`x_(t)=\frac{95*(1.3)'+25*(1.3)'+60*V'}{m{p}+m_(d)+m_(b)}`

Resolve to v'

`78-60v=\\`

`v=(78)/(60)`

`v=1.3 (m)/(s)`