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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)?

Respuesta :

Answer:

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

kendrich

Answer:

The explaination and attachment better portrays the answer. kindly check explaination.

Explanation:

To be able to solve for the kinetic energy and the recoil speed. Let have an understanding of those words.

Kinetic energy is the energy of motion. If an object is moving, it is said to have kinetic energy.

Please kindly check attachment for the detailed answer.

Ver imagen kendrich

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