a) The momentum of the third piece is 19.1 kg m/s
b) The velocity is 6.4 m/s at an angle of [tex]-65^{\circ}[/tex]
Explanation:
a)
We can solve this problem by applying the law of conservation of momentum in 2D. In fact, the total momentum of the board before the explosion must be equal to the total momentum of the 3 pieces after the explosion.
Before the explosion, the total momentum is zero, since the board is at rest. Resolving into the x- and y- components, this means that the two components of the momentum are zero:
[tex]p_x = 0\\p_y = 0[/tex]
After the explosion, we have:
[tex]p_x = (2 kg)(10 m/s)( cos 60^{\circ}) + (1.2 kg)(15 m/s)(cos 180^{\circ})+p_{3x}[/tex]
where [tex]p_{3x}[/tex] is the x-component of the momemtum of the 3rd piece. By solving,
[tex]p_x = -8+p_{3x}[/tex]
And since the x-momentum must be conserved,
[tex]0=-8+p_{3x} \rightarrow p_{3x}=8 kg m/s[/tex]
Similarly, along the y-axis,
[tex]p_y = (2 kg)(10 m/s)( sin 60^{\circ}) + (1.2 kg)(15 m/s)(sin 180^{\circ})+p_{3y}[/tex]
where [tex]p_{3y}[/tex] is the y-component of the momemtum of the 3rd piece. By solving,
[tex]p_y = 17.3+p_{3y}[/tex]
And since the y-momentum must be conserved,
[tex]0=17.3+p_{3y} \rightarrow p_{3y}=-17.3 kg m/s[/tex]
So now we can find the momentum of the third piece:
[tex]p_3 = \sqrt{p_{3x}^2+p_{3y}^2}=\sqrt{8^2+(-17.3)^2}=19.1 kg m/s[/tex]
b)
The momentum of the third piece is given by
[tex]p_3 = m v_3[/tex]
where
m = 3 kg is the mass of the piece
[tex]v_3[/tex] is the velocity
Solving for [tex]v_3[/tex], we find
[tex]v_3 = \frac{p_3}{m}=\frac{19.1}{3}=6.4 m/s[/tex]
We can also find the direction, which is identical to the direction of the momentum, by using the following equation:
[tex]\theta = tan^{-1}(\frac{p_{3y}}{p_{3x}})=tan^{-1}(\frac{-17.3}{8})=-65^{\circ}[/tex]
So, the third piece flies off at 6.4 m/s at an angle of [tex]-65^{\circ}[/tex].
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