Question

A dog, with a mass of 10.0 kg, is standing on a flatboat so that he is 22.5 m from the shore. He walks 7.8 m on the boat toward the shore and then stops. The boat has a mass of 46.0 kg. Assuming there is no friction between the boat and the water, how far is the dog from the shore now?

Answer 1

Answer:16.096

Explanation:

Given

mass of dog$\left ( m_d\right )=10kg$

mass of boat$\left ( m_b\right )=46kg$

distance moved by dog relative to ground=$x_d$

distance moved by boat relative to ground=$x_b$

Distance moved by dog relative to boat=7.8m

There no net force on the system therefore centre of mass of system remains at its position

0=$m_d\times x_d+m_b\dot x_b$

0=$10\times x_d+46\dot x_b$

$x_d=-4.6x_b$

i.e. boat will move opposite to the direction of dog

Now

$|x_d|+|x_b|=7.8$

substituting$x_d$value

$5.6|x_b|=7.8$

$|x_b|=1.392m$

$|x_d|=6.4032m$

now the dog is  22.5-6.403=16.096m from shore