The velocity of the third piece is mathematically given as
v=0.88m/s
Generally, the equation for conservation is mathematically given as
[tex]m u=m_{1} v_{1 \mathrm{x}}+m_{2} v_{2 \mathrm{x}}+m_{1} v_{x}[/tex]
Here, is the mass of the grenade, [tex]m_{1}, m_{2}, m_{3}[/tex] the masses of the pieces after an explosion, and [tex]v_{1 \mathrm{x}}, v_{2 \mathrm{x}}[/tex] are velocities of the first and second pieces after the explosion.
[tex]m u_{1} &=m_{1} v_{1 x}+m_{2} v_{2 \mathrm{x}}+m_{3} v_{x} \\0 &=0+(0.2 \mathrm{~kg})((5 \mathrm{~m} / \mathrm{s})\\v_{x} &=\frac{-0.173 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}}{0.3 \mathrm{~kg}} \\&=-0.578 \mathrm{~m} / \mathrm{s}[/tex]
the equation for conservation
[tex]m u=m_{1} v_{1 y}+m_{2} v_{2 y}+m_{3} v_{y}[/tex]
[tex]m u_{1} &=m_{1} v_{1 \mathrm{y}}+m_{2} v_{x y}+m_{3} v_{y} \\0 &=(0.10 \mathrm{~kg})(10 \mathrm{~m} / \mathrm{s})+(0.2 \mathrm{~kg})\left((5 \mathrm{~m} / \mathrm{s}) \cos 10^{\circ}\right)+(0.3 \mathrm{~kg}) v_{y} \\v_{y} &=\frac{-1.98 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}}{0.3 \mathrm{~kg}} \\&=-0.661 \mathrm{~m} / \mathrm{s}[/tex]
In conclusion, the magnitude of the velocity
[tex]v=\sqrt{v_{\mathrm{x}}^{2}+v_{\mathrm{y}}^{2}}[/tex]
[tex]Substitute $-0.578 \mathrm{~m} / \mathrm{s}$ for $v_{\mathrm{s}}$ and $-0.661 \mathrm{~m} / \mathrm{s}$ for $v_{\mathrm{y}}$[/tex]
[tex]v &=\sqrt{v_{x}^{2}+v_{y}^{2}} \\&=\sqrt{(-0.578 \mathrm{~m} / \mathrm{s})^{2}+(-0.661 \mathrm{~m} / \mathrm{s})^{2}} \\&=0.88 \mathrm{~m} / \mathrm{s}[/tex]
v=0.88m/s
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