Two-Dimensional Linear Momentum Conservation

• Assume the firework was at rest before the explosion, so the initial momentum is zero.
• The firework explodes into 3 fragments of equal mass, let's call them $m$.

Velocities:

Conservation of momentum:

Horizontal direction: $p_{i_x}=p_{f_x}$

$0=m \ v_2\cos\theta + m\ v_{3,x}$

$0= v_2\cos\theta \ \ \ + \ \ \ \ v_{3,x}$

$v_{3,x}=- v_2\cos\theta$

Therefore, we have the velocity of the 3rd piece in the $x$ direction:

$v_{3,x}=- v_2\cos\theta=-25\cos25\degree=-22.66$ (m/s)

$$ (that is, to the left)

Vertical direction: $p_{i_y}=p_{f_y}$

$0=m\ v_1-m \ v_2\sin\theta + m\ v_{3,y}$

$0= \ \ v_1\ \ -\ \ v_2\sin\theta \ \ + \ \ \ v_{3,y}$

$v_{3,y}=-v_1 \ + \ \ v_2\sin\theta$

Therefore, we have the velocity of the 3rd piece in the $y$ direction:

$v_{3,y}=-v_1 + v_2\sin\theta=-14+25\sin25\degree=-3.43$ (m/s)

$$ (that is, downwards)

Find the resultant velocity:

$v_3=\sqrt{(v_{3,x})^2+(v_{3,y})^2}=\sqrt{22.66^2+3.43^2}=22.92$ (m/s)

Find the direction:

$\alpha=\tan^{-1}\bigg(\dfrac{v_{3,y}}{v_{3.x}}\bigg)=\tan^{-1}\bigg(\dfrac{3.43}{22.92}\bigg)=8.62\degree$

* it should be: tan^-1 (3.43/22.66) $$

(below the horizontal, to the left)