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A professional boxer hits his opponent with a 1000 N horizontal blow that lasts 0.180 seconds. The opponent's total body mass is 112 kg, and the blow strikes him near his center of mass while he is motionless in midair.

Answer :

Final answer:

A professional boxer's punch can be analyzed using Newton's second law of motion to understand the concept of impulse and momentum. In this case, the punch results in an impulse of 180 kg.m/s which changes the opponent's momentum, resulting in a final velocity of approximately 1.6 m/s.

Explanation:

The problem at hand is a classic illustration of Newton's second law of motion, which can be interpreted to involve the concept of impulse and momentum. Impulse is force applied over a certain time period which leads to a change in an object's momentum. The momentum, in turn, is just the product of an object's mass and its velocity.

Given that the force applied by the boxer is 1000 N over a period of 0.180 seconds, we can calculate the impulse J = Force x Time, which gives J = 1000 N x 0.180 s = 180 kg.m/s. This impulse will cause a change in the opponent's momentum.

Assuming the opponent was initially at rest (initial momentum = 0), the change in his final momentum will be equal to the impulse itself. Thus, the final momentum of the opponent is 180 kg.m/s. Since momentum (p) is given by the product of mass (m) and velocity (v), i.e., p = m*v, thus we can calculate the final velocity (v) as, v = p/m = 180 kg.m/s / 112 kg ≈ 1.6 m/s.

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