Question

A 500 g rubber ball is dropped from a height of 10 m and undergoes a perfectly elastic
collision with the earth.
(a) What is the earth’s velocity after the collision? Assume the earth was at rest just before
the collision
(b) How many years would it take the earth to move 1.0 mm at this speed?

Answer 1

Answer on Question#45365, Physics, Mechanics | Kinematics | Dynamics

Let us denote:

$v$ - the speed of the ball just before the collision with Earth, $v_1$ - speed of the ball after collision, $v_2$ - speed of the Earth after collision, $h$ - height of ball above the Earth, $m$ - mass of the ball, $M$ - mass of the Earth.

a) First, let us find the speed of the ball just before the collision with Earth. Using equations of motion, $h = h_0 - \frac{g t^2}{2}$ and $v = g t$, obtain $v = \sqrt{2 h g} \approx 14 \frac{m}{s}$ - the speed of the ball just before the collision with Earth.

In order to find the velocity of Earth after collision, let us use the law of conservation of linear momentum and law of conservation of energy:

$$m v = m v_1 + M v_2$$ $$m v^2 = m v_1^2 + M v_2^2$$

Solving this system of equation for $v_1, v_2$, obtain $v_1 = \frac{(m - M)v}{m + M}$, $v_2 = \frac{2mv}{m + M}$.

Using given data $m = 0.5 \, \text{kg}$, $M = 5.98 \cdot 10^{24} \, \text{kg}$ and velocity $v = 14 \, \frac{\text{m}}{\text{s}}$ found above, obtain $v_2 = 2.34 \cdot 10^{-24} \, \frac{\text{m}}{\text{s}}$ - the velocity of Earth after collision.

b) The time to move $1 \, \text{mm}$ is equal $t = \frac{1 \cdot 10^{-3} \, \text{m}}{2.34 \cdot 10^{-24} \, \frac{\text{m}}{\text{s}}} = 4.27 \cdot 10^{20} \, \text{s} \approx 1.35 \cdot 10^{13} \, \text{years}$.

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