In a baseball game, a batter hits the 0.150-kg ball straight back at the pitcher at 180 km/h . If the ball is traveling at 160 km/h just before it reaches the bat, what is the magnitude of the average force exerted by the bat on it if the collision lasts 5.0 ms?

Solution:

Newton's second law of motion states that the rate of change of momentum of a body over time is directly proportional to the force applied, and occurs in the same direction as the applied force.

Using the formula

F = m (v - u)/ t

From the question we know that

V = 180 km/hr = (180 × 1000)/ 3600 = 50 m/s

u = 160 km/hr = (160 × 1000)/ 3600 = 44.44 m/s

t = 5 s

Substituting these values

F = 0.150 (50 - 44.44)/ 5

By further calculation

F = (0.150 × 5.56)/5

F = 0.835/5

So we get

F = 0.17 N when rounded to two decimal places.

Therefore, the magnitude of the average force is 0.17N.

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In a baseball game, a batter hits the 0.150-kg ball straight back at the pitcher at 180 km/h . If the ball is traveling at 160 km/h just before it reaches the bat, what is the magnitude of the average force exerted by the bat on it if the collision lasts 5.0 ms?

Summary:

In a baseball game, a batter hits the 0.150-kg ball straight back at the pitcher at 180 km/h . If the ball is traveling at 160 km/h just before it reaches the bat, the magnitude of the average force exerted by the bat on it if the collision lasts 5.0 ms is 0.17N.