Question

A firecracker in a coconut blows the coconut into three pieces. Two pieces of equal mass fly off south and west, perpendicular to each other, at 21 m/s . The third piece has twice the mass as the other two.
what is the speed of the third piece?

what is the direction of the third piece? (degrees north of east)?

Answer 1

Explanation:

Let us assume that piece 1 is $m_{1}$ is facing west side. And, piece 2 is $m_{2}$ facing south side.

Let $m_{1} \text{and} m_{2}$ = m and $m_{3}$ = 2m. Hence,

        $2mv_{3y} = m(21)$

          $v_{3y} = \frac{21}{2}$

                     = 10.5 m/s

$2mv_{3x} = m(21)$

          $v_{3x} = \frac{21}{2}$

                     = 10.5 m/s

  $v_{3} = \sqrt{v^{2}_{3x} + v^{2}_{3y}}$

              = $\sqrt{(10.5)^{2} + (10.5)^{2}}$

              = 14.84 m/s

or,          = 15 m/s

Hence, the speed of the third piece is 15 m/s.

Now, we will find the angle as follows.

          $\theta = tan^{-1} (\frac{v_{3y}}{v_{3x}})$

                      = $tan^{-1}(\frac{10.5}{10.5})$

                      = $tan^{-1} (45^{o})$

                      = $45^{o}$

Therefore, the direction of the third piece (degrees north of east) is north of east.