Question

One 110-kg football lineman is running to the right atwhile another 125-kg lineman is running directly toward him at. What are (a) the magnitude and direction of the netmomentum of these two athletes, and (b) their total kinetic energy?

Short answer

(a) The magnitude of net momentum of two athletes is 22.5kg.m/s and its direction is left.

(b) The total kinetic energy of two athletes is 838 J

Step-by-step solution

Given information in the question

Given in the question.

Mass of lineman A,$m_A=110\mathrm{kg}$

Velocity of lineman B,$v_A=2.75\mathrm{m}/\mathrm{s}$

Mass of line man B,$m_B=125\mathrm{kg}$

Velocity of line man B,$v_B=v_{Bx}=-2.60\mathrm{m}/\mathrm{s}$

Formula used in the question

Significance of kinetic energy and momentum

Kinetic Energy

If mass of an object is m and its velocity is v then kinetic energy of object is,

$\mathit{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}$

Momentum

If mass of an object is m and velocity is v then its momentum is

$$\begin{gathered} \overrightarrow{\mathbf{p}}\mathbf{=}\mathit{m}\overrightarrow{\mathbf{v}} \\ \overrightarrow{\mathbf{p}}\mathbf{=}\mathit{m}\overrightarrow{{\mathbf{v}}_{\mathbf{x}}}\mathbf{+}\mathit{m}\overrightarrow{{\mathbf{v}}_{\mathbf{y}}} \end{gathered}$$

${\mathit{p}}_{\mathbf{x}}$is x component of momentum and${\mathit{p}}_{\mathbf{y}}$ is y component of momentum

Finding the magnitude and direction of net momentum for part (a)

Given in the question.

$$\begin{gathered} m_A=110\mathrm{kg} \\ {v}_A=v_{Ax}=2.75\mathrm{m}/\mathrm{s} \\ {m}_B=125\mathrm{kg} \\ {v}_B=v_{Bx}=-2.60\mathrm{m}/\mathrm{s} \end{gathered}$$

As we know the momentum of an object is

$$\begin{gathered} \overrightarrow{p}=m\overrightarrow{v} \\ {p}_x+p_y=m{\overrightarrow{v}}_x+m{\overrightarrow{v}}_y \end{gathered}$$

The x component of the net momentum of the system can be given as

$$\begin{gathered} p_{Ax}+p_{Bx}=m_A{v}_{Ax}+m_B{v}_{Bx} \\ {p}_y=110kg\left(2.75m/s\right)+125kg\left(-2.60m/s\right) \\ {p}_y=-22.5\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

The y component of the net momentum of the system can be given as.

$$\begin{gathered} p_{Ax}+p_{By}=m_A{v}_{Ay}+m_B{v}_{B}y \\ {p}_y=110kg\left(0\right)+125kg\left(0\right) \\ {p}_y=0 \end{gathered}$$

The magnitude of the momentum

$$\begin{gathered} p=\sqrt{p_x^2+p_y^2} \\ =\sqrt{{\left(-22.5\mathrm{kg}.\mathrm{m}/\mathrm{s}\right)}^2+{\left(0\right)}^2} \\ =22.5\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

The Magnitude of net momentum is 22.5kg.m/s

The direction of net momentum

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{p_x}{p_y}\right) \\ ={\tan }^{-1}\left(\frac{0}{-22.5\mathrm{kg}.\mathrm{m}/\mathrm{s}}\right) \\ =0 \end{gathered}$$

Since the x-component is negative the direction is left.

Hence the magnitude of net momentum is 22.5kg.m/s and its direction is left.

Finding the total kinetic energy for part (b)

We know the formula for kinetic energy

$K=\frac{1}{2}m{v}^2$

The total energy is the some of energy of both the athletes.

$$\begin{gathered} K_{total}=K_A+K_B \\ =\frac{1}{2}{\mathrm{m}}_{\mathrm{A}}{\mathrm{v}}_{\mathrm{A}}^2+\frac{1}{2}{\mathrm{m}}_{\mathrm{B}}{\mathrm{v}}_{\mathrm{B}}^2 \\ =\frac{1}{2}\left(110\mathrm{kg}\right){\left(2.75\mathrm{m}/\mathrm{s}\right)}^2+\frac{1}{2}\left(125\mathrm{kg}\right){\left(-2.60\mathrm{m}/\mathrm{s}\right)}^2 \\ =838\mathrm{J} \end{gathered}$$

The total energy of the system of two athletes is 838 J