We can use the textbook 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 (m = 155 grams = 0.155 kg) with a speed of 103 miles per hour (vball = 45.3 m/s) at a wall, and the ball bounces back with little loss of kinetic energy.(a) What is the recoil speed of the Earth (M = 6multiply1024 kg)?

Recoil velocity; v2 = 1.17 x 10^(-24) m/s

Explanation:

We are given;

Mass of baseball; m1 = 0.155 kg

Velocity of baseball; v1 = 45.3 m/s

Mass of earth; m2 = 6 x 10^(24) kg

Lets the recoil velocity be v2.

From conservation of momentum, initial momentum is equal to final momentum.

Thus,

m1•v1 = m2v2

Thus, v2 = m1•v1/m2

Plugging in the relevant values to get ;

v2 = 0.155 x 45.3/(6 x 10^(24))

v2 = 1.17 x 10^(-24) m/s

the answer would be 32*f