A firecracker is thrown with a velocity of 30 | vedclass

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(D) The explosion force that splits the firecracker is internal to the system,so the path of the centre of mass of the system remains unchanged.The ra

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$63 \, m$ or $117 \, m$

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(D) The explosion force that splits the firecracker is internal to the system,so the path of the centre of mass of the system remains unchanged.The range of the centre of mass of the system is given by $R = \frac{u^2 \sin 2	heta}{g}$.Here,$u = 30 \, ms^{-1}$ and the angle with the horizontal is $	heta = 90^{\circ} - 75^{\circ} = 15^{\circ}$.Therefore,$R = \frac{30 	imes 30 	imes \sin(2 	imes 15^{\circ})}{10} = 90 	imes \sin(30^{\circ}) = 90 	imes 0.5 = 45 \, m$.So,the position ($x$-coordinate) of the centre of mass is at $45 \, m$ distance from the origin.Using the formula for the centre of mass,$X_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$,where $m_1 = m_2 = m$ and $X_{CM} = 45 \, m$:$45 = \frac{m(27) + m(x)}{m + m}$$45 = \frac{27 + x}{2}$$90 = 27 + x \Rightarrow x = 63 \, m$.However,the first piece could also fall at $-27 \, m$ (behind the origin) if the explosion is strong enough to reverse its direction:$45 = \frac{m(-27) + m(x)}{2m}$$90 = -27 + x \Rightarrow x = 117 \, m$.Thus,the other piece will fall at $63 \, m$ or $117 \, m$ from the shooting point.

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