Question

A 4.5 kg dog stands on an 18 kg flatboat at distance D=6.1 m from the shore. It walks 2.4 m along the boat toward shore and then stops. Assuming no friction between the boat and the water, find how far the dog is then from the shore.

Answer 1

Apply momentum conservation:

$$0=m_bv_b-m_dv_d,\\\space\\ 0=m_b\dfrac{d_b}{t}-m_d\dfrac{d_d}{t},\\\space\\ 0=m_bd_b-m_dd_d.$$

On the other hand, these displacements added to each other give 2.4 m:

$$d_b+d_d=2.4 \text{ m.}$$

Solve the system:

$$d_d=1.92\text{ m},\\ d_d=0.48\text{ m}.$$

So, dog went 1.92 m forward, and the distance to the shore became 6.1-1.92=4.18 m.