Question

To practice problem-solving strategy 9.1 a strategy for conservation of momentum problems. an 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . how fast will he be moving backward just after releasing the ball?

Answer 1

Refer to the diagram shown below.

By definition momentum =  mass *  velocity.

Before throwing the ball:
The initial momentum is
P₁ = 0.

After throwing the ball:
Let u = the backward velocity of the quarterback.
The momentum is 
P₂ = (0.43 kg)*(15 m/s) + (80 kg)*(- u m/s)

Conservation of momentum requires that
P₂ = P₁
6.45 - 80u = 0
u = 6.45/80 = 0.0806 m/s

Answer: 0.08 m/s backward

Answer 2

The quarterback will be moving with a velocity of $\boxed{0.08\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ in backward direction.

Further Explanation:

Given:

The mass of the quarterback is $80\,{\text{kg}}$.

The mass of the basketball is $0.43\,{\text{kg}}$.

The velocity of the basketball in the forward direction is $15\,{{\text{m}} \mathord{\left/{\vphantom{{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$.

Concept:

The quarterback throws the basket ball in the forward direction. It means that the momentum of the quarterback in the horizontal direction will change but the overall momentum of the ball and the man system will remain conserved in the horizontal direction.

The momentum of the body is defined as the product of the mass and velocity of the body. It is expressed as:

$\boxed{p=m \times v}$

The initial momentum of the system in horizontal direction is zero because as the quarterback jumps up, the ball and the man do not have any velocity in horizontal direction but as he throws the ball in forward direction, he will gain some momentum in order to make the momentum of the system conserved.

So, the expression for momentum conservation of the system is:

${p_i} = {p_b} - {p_m}$  

Here, ${p_i}$ is the initial momentum of system, ${p_b}$ is the momentum of the ball and ${p_m}$ is the momentum of the man.

The momentum of the man is negative because he gains the speed in backward direction.

Substitute the values in the equation:

$\begin{aligned}0&=\left({{m_b}\times {v_b}}\right)-\left( {{m_m}\times {v_m}} \right) \\{v_m}&= \frac{{0.43 \times 15}}{{80}}\\&=\frac{{6.45}}{{80}}\\&= 0.08\,{{\text{m}}\mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}$  

Thus, the quarterback will be moving with a velocity of $\boxed{0.08\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ in backward direction.

Learn More:

1. During soccer practice, Maya kicked a soccer ball 37° off the ground at 25 m/s brainly.com/question/11023695

2. A 700-kg car, driving at 29 m/s, hits a brick wall and rebounds with a speed of 4.5 m/s brainly.com/question/9484203

3. What is the magnitude fcont of the force that the car exerts on the truck? Use the coordinate system shown in part b brainly.com/question/2235246

Answer Details:

Grade: College

Subject: Physics

Chapter: Momentum Conservation

Keywords:  Conservation of momentum, 80 kg quarterback, jumps straight up, football, initial momentum, gains momentum, backward direction.