Chapter 8, Problem 30P

(a)

To determine

The reason that why the successful tackle constitutes a perfectly inelastic collision.

Answer to Problem 30P

The successful tackle constitutes a perfectly inelastic collision because they are stuck together and the momentum is conserved.

Explanation of Solution

Given info: The mass of the fullback is $90.0\text{kg}$, the speed of the fullback is $5.00\text{m}/\text{s}$, the mass of the opponent is $95.0\text{kg}$ and the speed of the opponent is $3.00\text{m}/\text{s}$.

Momentum is a vector quantity which is always conserved in a closed system while kinetic energy is not a vector quantity. So, it is use the direction and magnitude of the player’s original momenta to figure out how they will be moving post collision. Thus, the successful tackle constitutes a perfectly inelastic collision because they are stuck together and the momentum is conserved.

Conclusion:

Therefore, the successful tackle constitutes a perfectly inelastic collision because they are stuck together and the momentum is conserved.

(b)

To determine

The velocity of the players immediately after the tackle.

Answer to Problem 30P

The velocity of the players immediately after the tackle is $2.9\text{m}/\text{s}$.

Explanation of Solution

Given info: The mass of the fullback is $90.0\text{kg}$, the speed of the fullback is $5.00\text{m}/\text{s}$, the mass of the opponent is $95.0\text{kg}$, the speed of the opponent is $3.00\text{m}/\text{s}$.

Formula to calculate the momentum of the fullback is,

$p_1=m_1{v}_1$

Here,

$p_1$ is the momentum of the fullback.

$m_1$ is the mass of the fullback.

$v_1$ speed of the fullback.

Substitute $90.0\text{kg}$ for $m_1$ and $5.00\text{m}/\text{s}$ for $v_1$ in above equation to find $p_1$.

$\begin{array}{c}{p}_1=\left(90.0\text{kg}\right)\left(5.00\text{m}/\text{s}\right)\\ =450\text{kg-m}/\text{s}\end{array}$

Thus, the momentum of the fullback is $450\text{kg-m}/\text{s}$.

Formula to calculate the momentum of the opponent is,

$p_2=m_2{v}_2$

Here,

$p_2$ is the momentum of the opponent.

$m_2$ is the mass of the opponent.

$v_2$ speed of the opponent.

Substitute $95.0\text{kg}$ for $m_2$ and $3.00\text{m}/\text{s}$ for $v_2$ in above equation to find $p_2$.

$\begin{array}{c}{p}_2=\left(95.0\text{kg}\right)\left(3.00\text{m}/\text{s}\right)\\ =285\text{kg-m}/\text{s}\end{array}$

Thus, the momentum of the opponent is $285\text{kg-m}/\text{s}$.

After the collision, the momentum of the players will be neither be in north nor in east.

It should be in between north and east. So, the addition of vectors is used to calculate the momentum of the players immediately after the tackle.

Formula to calculate the hypotenuse of right triangle formed by placing the vectors is,

$p^2=p_1^2+p_2^2$

Here,

$p$ is the momentum of the players immediately after the tackle.

Substitute $450\text{kg-m}/\text{s}$ for $p_1$ and $285\text{kg-m}/\text{s}$ for $p_2$ in above equation to find $p$.

$\begin{array}{c}{p}^2={\left(450\text{kg-m}/\text{s}\right)}^2+{\left(285\text{kg-m}/\text{s}\right)}^2\\ p=532.65\text{kg-m}/\text{s}\end{array}$

Thus, the momentum of the players immediately after the tackle is $532.65\text{kg-m}/\text{s}$.

Formula to calculate the combined mass of the players is,

$m=m_1+m_2$

Here,

$m$ is the combined mass of the players.

Substitute $90.0\text{kg}$ for $m_1$ and $95.0\text{kg}$ for $m_2$ in above equation to find $m$.

$\begin{array}{c}m=90.0\text{kg}+95.0\text{kg}\\ \text{=185}\text{kg}\end{array}$

Thus, the combined mass of the players is $\text{185}\text{kg}$.

Formula to calculate the velocity of the players immediately after the tackle is,

$\begin{array}{c}p=mv\\ v=\frac{p}{m}\end{array}$

Here,

$v$ is the velocity of the players immediately after the tackle.

Substitute $\text{185}\text{kg}$ for $m$ and $532.65\text{kg-m}/\text{s}$ for $p$ in above equation to find $v$.

$\begin{array}{c}v=\frac{532.65\text{kg-m}/\text{s}}{\text{185}\text{kg}}\\ =2.87\text{m}/\text{s}\\ \approx 2.9\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the velocity of the players immediately after the tackle is $2.9\text{m}/\text{s}$.

(c)

To determine

The decrease in the mechanical energy as a result of the collision.

Answer to Problem 30P

The decrease in the mechanical energy as a result of the collision is $774.5\text{J}$.

Explanation of Solution

Given info: The mass of the fullback is $90.0\text{kg}$, the speed of the fullback is $5.00\text{m}/\text{s}$, the mass of the opponent is $95.0\text{kg}$, the speed of the opponent is $3.00\text{m}/\text{s}$.

Formula to calculate the kinetic energy of the fullback is,

$E_1=\frac{1}{2}{m}_1{v}_1^2$

Here,

$E_1$ is the kinetic energy of the fullback.

Substitute $90.0\text{kg}$ for $m_1$ and $5.00\text{m}/\text{s}$ for $v_1$ in above equation to find $E_1$.

$\begin{array}{c}{E}_1=\frac{1}{2}\left(90.0\text{kg}\right){\left(5.00\text{m}/\text{s}\right)}^2\\ =1125\text{J}\end{array}$

Thus, the kinetic energy of the fullback is $1125\text{J}$.

Formula to calculate the kinetic energy of the opponent is,

$E_2=\frac{1}{2}{m}_2{v}_2^2$

Here,

$E_2$ is the kinetic energy of the opponent.

Substitute $95.0\text{kg}$ for $m_2$ and $3.00\text{m}/\text{s}$ for $v_2$ in above equation to find $E_2$.

$\begin{array}{c}{E}_2=\frac{1}{2}\left(95.0\text{kg}\right){\left(3.00\text{m}/\text{s}\right)}^2\\ =427.5\text{J}\end{array}$

Thus, the kinetic energy of the opponent  is $427.5\text{J}$.

Formula to calculate the total kinetic energy of the player prior to the collsion is,

$E=E_1+E_2$

Here,

$E$ is the total kinetic energy of the player prior to the collsion.

Substitute $1125\text{J}$ for $E_1$ and $427.5\text{J}$ for $E_2$ in above equation to find $E$.

$\begin{array}{c}E=1125\text{J}+427.5\text{J}\\ =1552.5\text{J}\end{array}$

Thus, the total kinetic energy of the player prior to the collsion is $1552.5\text{J}$.

Formula to calculate the total kinetic energy followed by the collision is,

$E^{\prime }=\frac{1}{2}m{v}^2$

Here,

$E^{\prime }$ is the total kinetic energy followed by the collision.

Substitute $\text{185}\text{kg}$ for $m$ and $2.9\text{m}/\text{s}$ for $v$ in above equation to find $E^{\prime }$.

$\begin{array}{c}{E}^{\prime }=\frac{1}{2}\left(\text{185}\text{kg}\right){\left(2.9\text{m}/\text{s}\right)}^2\\ =777.92\text{J}\\ \approx 778\text{J}\end{array}$

Thus, the total kinetic energy followed by the collision is $778\text{J}$.

Formula to calculate the decrease in the mechanical energy as a result of the collision is,

$E_{\text{mechanical}}=E-E^{\prime }$

Here,

$E_{\text{mechanical}}$ is the decrease in the mechanical energy as a result of the collision.

Substitute $1552.5\text{J}$ for $E$ and $778\text{J}$ for $E^{\prime }$ in above equation to find $E_{\text{mechanical}}$.

$\begin{array}{c}{E}_{\text{mechanical}}=1552.5\text{J}-778\text{J}\\ =774.5\text{J}\end{array}$

This decrease in the mechanical energy as a result of the collision is converted into heat energy to tackle.

Conclusion:

Therefore, the decrease in the mechanical energy as a result of the collision is $774.5\text{J}$.