(I) A 110-kg tackler moving at 2.5 m/s meets head-on (and holds on to) an 82-kg halfback moving at 5.0 m/s. What will be their mutual speed immediately after the collision?

The mass of the tackler is \({m_1} = 110\;{\rm{kg}}\).

The mass of the halfback is \({m_2} = 82\;{\rm{kg}}\).

The initial speed of the halfback is \({v_2} = 5.0\;{\rm{m/s}}\) (towards the tackler).

Assume that the direction of the motion of the halfback is positive, then the direction of the motion of the tackler is negative.

Also, the initial speed of the tackler is \({v_1} = - 2.5\;{\rm{m/s}}\) (towards the halfback).

Let \(v\) be their mutual speed immediately after the collision.

The momentum of the tackler and the halfback is the same before and after the tackle. There is no external force if you consider the tackler and the halfback as a system.

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

And the total momentum after the tackle is \(\left( {{m_1} + {m_2}} \right)v\).

Using the concept of momentum conservation, you get

\(\begin{array}{c}\left( {{m_1} + {m_2}} \right)v = \left( {{m_1}{v_1} + {m_2}{v_2}} \right)\\v = \frac{{{m_1}{v_1} + {m_2}{v_2}}}{{{m_1} + {m_2}}}\\v = \frac{{\left[ {\left( {110\;{\rm{kg}}} \right) \times \left( { - 2.5\;{\rm{m/s}}} \right)} \right] + \left[ {\left( {82\;{\rm{kg}}} \right) \times \left( {5.0\;{\rm{m/s}}} \right)} \right]}}{{\left( {110\;{\rm{kg}}} \right) + \left( {82\;{\rm{kg}}} \right)}}\\v = 0.70\;{\rm{m/s}}{\rm{.}}\end{array}\)

Hence, their mutual speed is \(0.70\;{\rm{m/s}}\) immediately after the collision.