Question

A fisherman on a stationary boat jumps off onto a jetty with a velocity of 1.5 ms-1, causing the boat to move backwards.

(i) If the mass of the fisherman and the boat are 65 kg and 450 kg respectively, determine the velocity of the boat. (ii) If the fisherman’s feet are in contact with the jetty for 10 ms, determine the magnitude of the average force exerted on his feet.

Answer 1

(i) the law of conservation of momentum says

$$0=m_fv_f-m_bv_b$$

Hence, the velocity of the boat

$$v_b=v_f\frac{m_f}{m_b}=1.5*\frac{65}{450}=0.22 \:\rm m/s$$

(ii) the magnitude of the average force

$$F=\frac{m_f\Delta v_f}{\Delta t}=\frac{65*1.5}{10*10^{-3}}=9750\:\rm N$$