Question

A 4.5 kg dog stands on an 18 kg flatboat at dis-

tance 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 wa-

ter, find how far the dog is then from

the shore.

Answer 1

Since the momentum is conserved:

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

The sum of the displacements is 2.4 m:

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

Solve the system of equations:

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

The displacement of dog is 1.92 m forward, the distance to the shore is 6.1-1.92=4.18 m.