Question

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 105 miles per hour ( Vball = 46.2 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= 6 × 10²⁴ kg)?

Vearth=

(b) Calculate the recoil kinetic energy of the Earth and compare to the kinetic energy of the baseball. The Earth gets lots of momentum (twice the momentum of the baseball) but very little kinetic energy.

Kearth =

Kbaseball =

Answer 1

by momentum conservation

M = mass of earth = 6*10^24 kg  

little loss of kinetic energy so we can assume thaat after bouncing ball has same ans opposite velocity

mu = mv+ MVearth

so Vearth = m ( u-v) /M = 0.155* ( 46.2+ 46.2) / ( 6*10^24) = 2.387*10^-24 m/s

b) Calculate the recoil kinetic energy​

KE earth = 1/2*M*Vearth^2 = 0.5*6*10^24*( 2.387*10^-24)^2 =1.7093307e-23 Joule

KE base ball = 0.5*0.155*46.2^2 = 165.4191 J

KE/KEearth = 165.4191 / (1.7093307e-23) =9.67741935e24

let me know in a comment if there is any problem or doubt