Question

We can use our results for head-on elastic collisions to analyze the recoil of the Earth when a ball bounces off a wall embedded in the Earth. Suppose that a professional baseball pitcher hurts a baseball$\mathbf{(}\mathit{m}\mathbf{=}\mathbf{155}\mathbf{}\mathbf{\text{g}}\mathbf{)}$with a speed ofrole="math" localid="1668602437434" $\mathbf{100}\mathbf{\text{mih}}\left(v_1=44\text{m}/\text{s}\right)$at a wall that is securely anchored to the Earth, and the ball bounces back with little loss of kinetic energy.

(a) What is the approximate recoil speed of the Earth$\mathbf{(}\mathit{M}\mathbf{=}\mathbf{6}\mathbf{\times }{10}^{24}\text{kg})$?

(b) Calculate the approximate recoil kinetic energy of the Earth and compare it to the kinetic energy of the baseball.

The Earth gets lots of momentum (twice the momentum of the baseball) but very little kinetic energy.

Short answer

(a) The approximate recoil speed of the earth is$2.3\times {10}^{-24}\text{m}/\text{s}$.

(b) The approximate recoil kinetic energy of the earth is $1.6\times {10}^{-23}\text{J}$.

Step-by-step solution

Given data

Ball's mass:$m=155\text{gor}0.155\text{kg}$

Ball's initial speed:$v_{\text{ball},i}=44\text{m}/\text{s}$

Earth's mass:$M=6\times {10}^{24}\text{kg}$

The concept of elastic and inelastic collision

If there is no kinetic energy lost in the impact, the collision is said to be perfectly elastic. A collision is considered to be inelastic if any of the kinetic energy is converted to another kind of energy during the collision.

Determine the speed of the earth

(a)

The net force on the system is zero, so we can apply law of conservation of linear momentum:

$$\begin{gathered} {\overrightarrow{p}}_i={\overrightarrow{p}}_f \\ {\overrightarrow{p}}_{ball,i}={\overrightarrow{p}}_{ball,f}+{\overrightarrow{p}}_{earth,f} \\ {\overrightarrow{p}}_{\text{earth},f}={\overrightarrow{p}}_{\text{ball},i}-{\overrightarrow{p}}_{ball,f} \end{gathered}$$

The ball bounces back, therefore speed of the ball is not changed but its direction is reverse$\left(v_{\text{ball},f}=-v_{\text{ball},i)}\right)$:

$$\begin{gathered} {\overrightarrow{p}}_{\text{earth},f}=2{\overrightarrow{p}}_{\text{ball},i} \\ M{v}_{\text{earth},f}=2m{v}_{\text{ball},i} \\ {v}_{\text{earth},f}=\frac{2m{v}_{\text{ball},i}}{M} \end{gathered}$$

$\begin{array}{rcl}{v}_{\text{earth},f}& =& \frac{2\times 0.155\times 44}{6\times {10}^{24}}\\ & =& 2.3\times {10}^{-24}\text{m}/\text{s}\end{array}$

This value is too small.

Determine the approximate kinetic energy

(b)

Kinetic energy of the ball:

$\begin{array}{rcl}K{E}_{\text{ball},i}& =& \frac{1}{2}m{v}_{\text{ball},i}^2\\ & =& \frac{1}{2}\times 0.155\times (44{)}^2\\ & =& 150\text{J}\end{array}$

Kinetic energy of the Earth:

$K{E}_{\text{earth},f}=\frac{1}{2}M{v}_{\text{earth},f}^2$

$\begin{array}{rcl}K{E}_{\text{earth,f}}& =& \frac{1}{2}\times 6\times {10}^{24}\times {\left(2.3\times {10}^{-24}\right)}^2\\ & =& 1.6\times {10}^{-23}\text{J}\end{array}$

This is very little kinetic energy.