Question

An 80kg quarterback jumps straight up in the air right before throwing a 0.43kg football horizontally at 15m/s . How fast will he be moving backward just after releasing the ball? (THIS IS NOT THE QUESTION THAT IS TO BE ANSWERED)

This is the question (THAT NEEDS TO BE ANSWERED):
Which of the following best describes why you can analyze this event using conservation of momentum?

a. The throwing action is quick enough that external forces may be ignored.

b. External forces don't act on the system during the jump.

c. Conservation of momentum is always the best way to analyze motion.

Answer 1

Concepts and reason

The concept used to solve this problem is conservation of momentum.

Initially, use the conservation of momentum to solve for the final velocity of quarterback.

Finally, analyze the event of jump and throw to identify the reason of using the conservation of momentum.

Fundamentals

The momentum of an object is the product of its mass and velocity. The magnitude of momentum is expressed as,

$p = mv$

Here, $p$ is the momentum, $m$ is the mass and $v$ is the speed.

The conservation of momentum principle says that total momentum of an isolated system is always conserved. This can be expressed as,

${p_{{\rm{initial}}}} = {p_{{\rm{final}}}}$

Here, ${p_{{\rm{initial}}}}$ is the total initial momentum of the system and ${p_{{\rm{final}}}}$ is the total final momentum.

The total final momentum of the quarterback-football system is,

${p_{{\rm{final}}}} = {m_{{\rm{football}}}}{v_{{\rm{football}}}} - {M_{{\rm{quarterback}}}}{v_{{\rm{quarterback}}}}$

Here, ${m_{{\rm{football}}}}$ is mass of football, ${v_{{\rm{football}}}}$ is the velocity of football, ${M_{{\rm{quarterback}}}}$ is the mass of quarterback, and ${v_{{\rm{quarterback}}}}$ is the velocity of the quarterback.

As there is no external force acting on the system then the total initial momentum of the system is equal to zero.

Substitute $0$ for ${p_{{\rm{initial}}}}$ and ${m_{{\rm{football}}}}{v_{{\rm{football}}}} - {M_{{\rm{quarterback}}}}{v_{{\rm{quarterback}}}}$ for ${p_{{\rm{final}}}}$ in the conservation of momentum equation and solve for ${v_{{\rm{quarterback}}}}$.

$\begin{array}{c}\\0 = {m_{{\rm{football}}}}{v_{{\rm{football}}}} - {M_{{\rm{quarterback}}}}{v_{{\rm{quarterback}}}}\\\\{v_{{\rm{quarterback}}}} = \frac{{{m_{{\rm{football}}}}{v_{{\rm{football}}}}}}{{{M_{{\rm{quarterback}}}}}}\\\end{array}$

Substitute $0.43{\rm{ kg}}$ for ${m_{{\rm{football}}}}$, $15{\rm{ m/s}}$ for ${v_{{\rm{football}}}}$, and $80{\rm{ kg}}$ for ${M_{{\rm{quarterback}}}}$ in the equation ${v_{{\rm{quarterback}}}} = \frac{{{m_{{\rm{football}}}}{v_{{\rm{football}}}}}}{{{M_{{\rm{quarterback}}}}}}$ and calculate the velocity of quarterback.

$\begin{array}{c}\\{v_{{\rm{quarterback}}}} = \frac{{\left( {0.43{\rm{ kg}}} \right)\left( {15{\rm{ m/s}}} \right)}}{{\left( {80{\rm{ kg}}} \right)}}\\\\ = 0.081{\rm{ m/s}}\\\end{array}$

The quarterback jumps and throws the ball very swiftly. All this happens in a fraction of second and the effect of other forces on the quarterback are far too less than the effect of throwing the football. The change in the momentum of the quarterback is also large as he throws the ball at large velocity. Hence, the throwing action is quick enough that external forces may be ignored.

Ans:

The velocity of the quarterback backward just after releasing the ball is 0.081 m/s.

The throwing action is quick enough that external forces may be ignored.