91Ó°ÊÓ

In this case, the block or bullet moves vertically, an inelastic collision occurs, and momentum is conserved. Also, after the collision, the energy is conserved in the rising of the block and bullet.

Given data:

The mass of the block,\(m = 1.40\;{\rm{kg}}\).

The mass of the bullet,\(m' = 25\;{\rm{g}}\).

The speed of the bullet, \(v' = 230\;{\rm{m/s}}\).

The relation to calculate the height can be written as follows:

\(v' = \frac{{m + m'}}{m}\sqrt {2gh} \)

Here,\(g\)is the gravitational acceleration, and h is the required height.

Plugging values in the above equation,

\(\begin{array}{c}230\;{\rm{m/s}} = \left( {\frac{{1.40\;{\rm{kg}} + \left( {25\;{\rm{g}} \times \frac{{1\;{\rm{g}}}}{{1000\;{\rm{g}}}}} \right)}}{{1.40\;{\rm{kg}}}}} \right)\sqrt {2\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right)h} \\h = 0.83\;{\rm{m}}\end{array}\)

Thus, \(h = 0.83\;{\rm{m}}\) is the height risen by the block.