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 speed $v_o$. The third piece has twice the mass as the other two. What are the speed and direction of the third piece? Give the direction as an angle east of north

Short answer

The speed of the third piece is $\frac{v_o}{\sqrt{2}}$at${45}^o$

Step-by-step solution

Step 1. Given information

A firecracker in a coconut blows the coconut into three pieces.

To find the speed of the third piece.

Step 2. Analyze and apply

We can conserve the momentum on both axis,

For x-axis:

$$\begin{gathered} m_1{v}_{1,x}+m_2{v}_{2,x}+m_3{v}_{3,x}=0 \\ m(-v_o)+0+2m{v}_{3,x}=0 \\ {v}_{3,x}=\frac{v_0}{2} \end{gathered}$$

Step 3. Evaluating at y-axis

For y-component:

$$\begin{gathered} m_1{v}_{1,y}+m_2{v}_{2,y}+m_3{v}_{3,y}=0 \\ m\left(0\right)+m(-v_o)+2m{v}_{3,y}=0 \\ {v}_{3,y}=\frac{v_o}{2} \end{gathered}$$

The final speed, $v=\sqrt{{v_x}^2+{v_y}^2}=\frac{v_0}{\sqrt{2}}$

The direction will be,$$\begin{gathered} \tan \theta =\frac{v_{3,y}}{v_{3,x}}=1 \\ \theta ={45}^o \end{gathered}$$