91Ó°ÊÓ

Given Data:

The initial speed of the bullet is ${\mathrm{v}}_0$

The vertical height risen by the block is $\mathrm{h}=0.80 \mathrm{m}$

The mass of the bullet is: $\mathrm{m}=0.012 \mathrm{kg}$

The mass of the block of wood is: $\mathrm{M}=0.800 \mathrm{kg}$

The length of the cord is $\mathrm{l}=1.6 \mathrm{m}$

The final speed of a combined object is calculated by equating the net force on the combined object to the centripetal force of the combined object for that position.

The final speed of the combined object is given as:

$\mathrm{v}=\sqrt{\frac{\mathrm{T}â‹\dots \mathrm{l}-(\mathrm{m}+\mathrm{M})\mathrm{gh}}{\mathrm{m}+\mathrm{M}}}$

Here, g is the gravitational acceleration and its value is .

Substitute all the values in the above equation.

$$\begin{gathered} \mathrm{v}=\sqrt{\frac{(4.80 \mathrm{N})(1.6 \mathrm{m})-(0.012 \mathrm{kg}+0.8 \mathrm{kg})(9.8 \mathrm{m}/{\mathrm{s}}^2)(0.8 \mathrm{m})}{(0.012 \mathrm{kg}+0.8 \mathrm{kg})}} \\ \mathrm{v}=1.27 \mathrm{m}/\mathrm{s} \end{gathered}$$

The initial speed of the bullet is found by using the third equation of motion between initial position of hanged wooden block and position of wooden block after rising some height.

The initial speed of bullet is given as:

${\mathrm{v}}_0=\sqrt{{\mathrm{v}}^2+2\mathrm{gh}}$

Substitute all the values in the above equation.

$$\begin{gathered} {\mathrm{v}}_0=\sqrt{{(1.27 \mathrm{m}/\mathrm{s})}^2+2(9.8 \mathrm{m}/{\mathrm{s}}^2)(0.8 \mathrm{m})} \\ {\mathrm{v}}_0=4.16 \mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the initial speed of bullet is$4.16 \mathrm{m}/\mathrm{s}$.