91Ó°ÊÓ

According to the law of conservation of linear momentum, the momentum of a closed, isolated system remains conserved when no external force acts on the system.

For example, consider a system of two objects that collide head-on with each other. When there is no external force acting on the system, the total momentum of the system before the collision is equal to its total momentum after the collision, i.e., the total momentum of the system remains conserved.

Mass of the bullet,\({m_1} = 22\;{\rm{g}} = 22 \times {10^{ - 3}}\;{\rm{kg}}\).

Mass of wooden block,\({m_2} = 2.0\;{\rm{kg}}\).

The initial velocity of the bullet,\({u_1} = 240\;{\rm{m/s}}\).

The final velocity of the bullet,\({v_1} = 150\;{\rm{m/s}}\).

Since the wooden block is held stationary initially, the initial velocity of the wooden block,\({u_2} = 0\;{\rm{m/s}}\).

Let the final velocity of the wooden block is \({v_2}\).

Consider a bullet and block system. Since no external force is acting on the system, the law of conservation of momentum can be applied to the bullet and block system.

According to the law of conservation of momentum, the initial momentum of the system is equal to the final momentum of the system, i.e.,

\(\begin{array}{c}{p_{\rm{i}}} = {p_{\rm{f}}}\\{m_1}{u_1} + {m_1}{v_1} = {m_2}{u_2} + {m_2}{v_2}\\\left( {22 \times {{10}^{ - 3}}\;{\rm{kg}}} \right)\left( {240\;{\rm{m/s}}} \right) + \left( {2.0\;{\rm{kg}}} \right)\left( {0\;{\rm{m/s}}} \right) = \left( {22 \times {{10}^{ - 3}}\;{\rm{kg}}} \right)\left( {150\;{\rm{m/s}}} \right) + \left( {2.0\;{\rm{kg}}} \right){v_2}\\{v_2} = \frac{{\left( {22 \times {{10}^{ - 3}}\;{\rm{kg}}} \right)\left( {240\;{\rm{m/s}}} \right) - \left( {22 \times {{10}^{ - 3}}\;{\rm{kg}}} \right)\left( {150\;{\rm{m/s}}} \right)}}{{\left( {2.0\;{\rm{kg}}} \right)}}\\ = 0.99\;{\rm{m/s}}\end{array}\)

Thus, the block will move with a velocity of 0.99 m/s after the bullet emerges from it.