91Ó°ÊÓ

The impulse exerted on an object is the product of the applied force and the elapsed time. It varies directly with the applied force.

The mass of the fullback is \({m_1} = 95\;{\rm{kg}}\).

The initial speed of the fullback is \({\vec v_{1,{\rm{i}}}} = 3.0\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}\).

The time interval is \(\Delta {t_1} = 0.85\;{\rm{s}}\).

Since at the final position, the fullback is stopped, his final velocity is \({\vec v_{1,{\rm{f}}}} = 0\).

(a)The original momentum of the fullback can be calculated as shown below:

\(\begin{array}{c}{{\vec P}_1} = {m_1}{{\vec v}_{1,{\rm{i}}}}\\{{\vec P}_1} = \left( {95\;{\rm{kg}}} \right)\left( {3.0\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}} \right)\\{{\vec P}_1} = 285\;{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/{\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {\rm{s}}}} \right.\\} {\rm{s}}}\end{array}\)

Thus, the original momentum of the fullback is \(285\;{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/{\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {\rm{s}}}} \right.\\} {\rm{s}}}\) towards the east.

(b)

The impulse exerted on the fullback can be calculated as shown below:

\(\begin{array}{c}{{\vec F}_1}\Delta t = {m_1}\left( {{{\vec v}_{1,{\rm{f}}}} - {{\vec v}_{1,{\rm{i}}}}} \right)\\{{\vec F}_1}\Delta t = \left( {95\;{\rm{kg}}} \right)\left[ {\left( 0 \right) - \left( {3.0\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}} \right)} \right]\\{{\vec F}_1}\Delta t = - 285\;{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/{\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {\rm{s}}}} \right.\\} {\rm{s}}}\end{array}\)

Thus, the impulse exerted on the fullback is \( - 285\;{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/{\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {\rm{s}}}} \right.\\} {\rm{s}}}\) towards the west.

(c)The impulse exerted on the tackler is equal in magnitude to that exerted on the fullback but in the opposite direction. That is, \({\vec F_2}\Delta t = 285\;{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/{\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {\rm{s}}}} \right.\\} {\rm{s}}}\).

Thus, the impulse exerted on the tackler is \(285\;{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/{\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {\rm{s}}}} \right.\\} {\rm{s}}}\).

(d)The average force exerted on the tackler can be calculated as shown below:

\(\begin{array}{c}{{\vec F}_2} = \frac{{{m_2}\left( {{v_{2,{\rm{f}}}} - {v_{2,{\rm{i}}}}} \right)}}{{\Delta {t_1}}}\\{{\vec F}_2} = \frac{{{{\vec F}_2}\Delta t}}{{\Delta {t_1}}}\\{{\vec F}_2} = \frac{{\left( {285\;{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/{\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {\rm{s}}}} \right.\\} {\rm{s}}}} \right)}}{{\left( {0.85\;{\rm{s}}} \right)}}\\{{\vec F}_2} = 335.3\;{\rm{N}}\end{array}\)

Thus, the average force exerted on the tackler is \(335.3\;{\rm{N}}\).