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 a professional baseball pitcher hurls a baseball ( 155 grams) with a speed of 99 miles per hour ( 43.6 m/s) at a wall, and the ball bounces back with little loss of kinetic energy. What is the recoil speed of the Earth ( 6 × 1024 kg)?

Answer 1

Answer:

So recoil speed of the Earth will be

$v = 2.25 \times 10^{-24} m/s$

Explanation:

Here if we assume that during collision if ball will lose very small amount of energy and rebound with same speed

then the impulse given by the ball is

$Impulse = m(v_f - v_i)$

$Impulse = (0.155)(43.6 - (-43.6))$

$Impulse = 13.52 Ns$

so impulse received by the Earth is same as the impulse given by the ball

so here we will have

$Impulse = mv$

$13.52 = (6 \times 10^{24})v$

$v = 2.25 \times 10^{-24} m/s$

Answer 2

The recoil speed of the Earth is $2.25*10^-24 m/s.$

What is collision?

Collision can be regarded as the forceful coming together of bodies.

The impulse of the ball can be calculated as;

$Impulse= MV$

where V=( V1 -Vo)

V1= final velocity=(43m/s)

V0= initial velocity= (-43.6m/s)

m= mass of baseball

Hence,$V= [43.6-(-43.6)]= 87.2m/s$

Then Impulse given by the ball=$( 0.155*87.2)= 13.53Ns.$

We can now calculate the recoil speed of the  earth as;

$Impluse= MVV= Impulse/ mass = 13.52/(6*10^24)$

=2.25*10^-24 m/s

Therefore, the recoil speed of the Earth is 2.25*10^-24 m/s

Learn more about collision at:

https://brainly.com/question/7538238