91Ó°ÊÓ

The mass of the baseball is \({m_1} = 144\;{\rm{g}} = 0.144\;{\rm{kg}}\).

The initial speed of the baseball is \({u_1} = 28.0\;{\rm{m/s}}\).

The initial speed of the brick is \({u_2} = 0\).

The mass of the brick is \({m_2} = 5.25\;{\rm{kg}}\).

After the collision, the speed of the brick is \({v_2} = 1.10\;{\rm{m/s}}\).

Let \({v_1}\) be the speed of the baseball after the collision.

The total momentum of the baseball and the brick after the collision is equal to the total momentum of the baseball before the collision as the brick was at rest initially.

Part (a)

The total momentum before the collision is \(\left( {{m_1}{u_1} + {m_2}{u_2}} \right)\)

The total momentum after the collision is \(\left( {{m_1}{v_1} + {m_2}{v_2}} \right)\).

Now, from the concept of momentum conservation,

\(\begin{array}{c}{m_1}{v_1} + {m_2}{v_2} = {m_1}{u_1} + {m_2}{u_2}\\{m_1}{v_1} = {m_1}{u_1} + {m_2}{u_2} - {m_2}{v_2}\\{v_1} = \frac{{{m_1}{u_1} + {m_2}{u_2} - {m_2}{v_2}}}{{{m_1}}}\end{array}\)

Now, substituting the values in the above equation, you get:

\(\begin{array}{c}{v_1} = \frac{{\left[ {\left( {0.144\;{\rm{kg}}} \right) \times \left( {28.0\;{\rm{m/s}}} \right)} \right] + \left[ {\left( {5.25\;{\rm{kg}}} \right) \times 0} \right] - \left[ {\left( {5.25\;{\rm{kg}}} \right) \times \left( {1.10\;{\rm{m/s}}} \right)} \right]}}{{0.144\;{\rm{kg}}}}\\ = - 12.10\;{\rm{m/s}}\end{array}\)

The negative sign in the expression of speed suggests that the baseball's speed is in the direction opposite to its initial speed.

Hence, after the collision, the baseball's speed is \(12.10\;{\rm{m/s}}\).

Part (b)

The total kinetic energy before the collision is:

\(\begin{array}{c}{K_{\rm{i}}} = \frac{1}{2}{m_1}u_1^2 + \frac{1}{2}{m_2}u_2^2\\ = \left[ {\frac{1}{2} \times \left( {0.144\;{\rm{kg}}} \right) \times {{\left( {28.0\;{\rm{m/s}}} \right)}^2}} \right] + \left[ {\frac{1}{2} \times \left( {5.25\;{\rm{kg}}} \right) \times {{\left( 0 \right)}^2}} \right]\\ = 56.4\;{\rm{J}}\end{array}\)

The total kinetic energy after the collision is:

\(\begin{array}{c}{K_{\rm{f}}} = \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_2}v_2^2\\ = \left[ {\frac{1}{2} \times \left( {0.144\;{\rm{kg}}} \right) \times {{\left( { - 12.10\;{\rm{m/s}}} \right)}^2}} \right] + \left[ {\frac{1}{2} \times \left( {5.25\;{\rm{kg}}} \right) \times {{\left( {1.10\;{\rm{m/s}}} \right)}^2}} \right]\\ = 13.7\;{\rm{J}}\end{array}\)

Hence, the kinetic energy before the collision is 56.4 J, and the final kinetic energy after the collision is 13.7 J.