Question

As part of an interview for a summer job with the Coast Guard, you are asked to help determine the search area for two sunken ships by calculating their velocity just after they collided. According to the last radio transmission from the 40,000-ton luxury liner, the Hedonist, it was going due west at a speed of 20 knots in calm seas through a rare fog just before it was struck broadside by the 60,000- ton freighter, the Ironhorse, which was traveling north at 10 knots. The transmission also noted that when the freighter's bow pierced the hull of the liner, the two ships stuck together and sank together. To determine the search area find the resulting components of the velocity.

Answer 1

Answer:

The resultant velocity is  $v_t=10 knots$

Explanation:

Apply the law of conservation of momentum

     $M_L *v_L + M_f * V_f = (M_L + M_f) v_t$

Where $M_L$ is the mass of the Luxury Liner = 40,000 ton

            $v_L$ is the velocity of Luxury Liner = 20 knots due west

            $M_f$ mass of freighter = 60,000

           $v_f$ is the velocity of freighter = 10 knots due north

Apply the law of conservation of momentum toward the the west direction

         $v_f = 0 \ knots$

So the equation would be

              $M_L *v_L = (M_L + M_f) v_t$

Substituting values

            $40000*20 = (40000+ 60000)v_t_w$

Where $v_t_w$ the final velocity due west

Making $v_t_w$ the subject

          $v_t_w = \frac{40,000* 20}{(40000 + 60000)}$

                $= 8 \ knots$

Apply the law of conservation of momentum toward the the north direction          

          $v_L = 0 \ knots$

So the equation would be

           $M_f *v_f = (M_L + M_f) v_t_n$

Where $v_t_n$ the final velocity due north

     Making $v_t_n$ the subject

          $v_t_n = \frac{60,000* 10}{(40000 + 60000)}$

                $= 6 \ knots$

The resultant velocity is

       $v_t = \sqrt{v_t_w^2 + v_t_n^2}$

            $= \sqrt{8^2 +6^2}$

           $v_t=10 knots$