Collision Question We can use the textbook results for head-on elastic collision

Question

Collision 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 = 6x10^24 kg)? V Earth = m/s (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. K earth = J K baseball = J

Explanation / Answer

Mass of Ball = 0.155 Kg
Speed of Ball = 46.2 m/s

After Collision -
Speed of Ball = - 46.2 m/s
Recoil speed of earth = vearth

Initial Momentum = m*v = 0.155 * 46.2 Kgm/s
Initial Momentum = 7.161 Kg m/s

Final Momentum = - 0.155 * 46.2 + (6 * 10^24)*vearth

Initial Momentum = Final Momentum
7.161 Kg m/s = - 0.155 * 46.2 + (6 * 10^24)*vearth
vearth =  (2 * 7.161 ) / (6*10^24)
vearth = 2.387 * 10^-24 m/s

K.E Earth = 0.5 * 6.4*10^24 * (2.387 * 10^-24)^2 J
K.E Earth = 1.82 * 10^-23 J

K.E Baseball = 0.5 * 0.155 * 46.2^2 J
K.E Baseball = 165.42 J

Comparing Kinetic Energy -
K.E Baseball / K.E Earth = 165.42/(1.82 * 10^-23) = 9.1 * 10^24

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