Question

Man weighs 100kg hurries forward at 0.500 m/sec. the boat he is in weighs 155kg moves backward at some other velocity. WHat is that velocity?

Answer 1

Task. Man weighs $m_{1}=100$ kg hurries forward at $v_{1}=0.500$ m/sec. The boat he is in weighs $m_{2}=155$ kg moves backward at some other velocity, $v_{2}$. What is that velocity?

Solution. Assume that the system “man-boat” is closed and its total impulse is zero. In particular, there is no water resistance.

Then the total impulse of the system is given by the formula:

$\vec{p}=m_{1}\vec{v_{1}}+m_{2}\vec{v_{2}}=0.$

Notice that $\vec{v_{1}}$ and $\vec{v_{2}}$ have opposite directions, so

$p=m_{1}v_{1}-m_{2}v_{2}=0.$

Hence

$m_{1}v_{1}=m_{2}v_{2}\qquad\Rightarrow\qquad v_{2}=\frac{m_{1}v_{1}}{m_{2}}.$

Substituting values we obtain

$v_{2}=\frac{m_{1}v_{1}}{m_{2}}=\frac{100*0.5}{155}=0.32\ m/s.$

Answer. $v_{2}=0.32$ m/s.