91影视

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:$M=6脳{10}^{24}\text{kg}$

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.

(a)

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

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

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

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

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

This value is too small.

(b)

Kinetic energy of the ball:

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

Kinetic energy of the Earth:

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

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

This is very little kinetic energy.