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A pitched 140 g baseball, in horizontal flight with a speed of 39 m/s, is struck by a batter. after leaving the bat, the ball travels in the opposite direction with speed also 39 m/s. what impulse acts on the ball while it is in contact with the bat? (give the magnitude.)

User Moonchild
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Final answer:

The impulse exerted on the baseball by the bat is calculated by finding the change in momentum. With a mass of 140 g and initial and final speeds of 39 m/s in opposite directions, the magnitude of the impulse is 21.84 kg m/s.

Step-by-step explanation:

The student has asked about the impulse that acts on a baseball when it is struck by a bat and changes direction. To answer this, we must calculate the change in momentum, which is the definition of impulse (Impulse = change in momentum). The mass of the baseball is given as 140 g (0.14 kg), and the initial and final velocities are given as 39 m/s in opposite directions.

The initial momentum (pi) of the baseball is the product of its mass and initial velocity, pi = m * vi = 0.14 kg * (-39 m/s). The negative sign indicates that the initial direction is to the left (we can assume left is the negative x-direction).

The final momentum (pf) is pf = m * vf = 0.14 kg * 39 m/s, as the ball is now moving to the right (positive x-direction).

The change in momentum, or the impulse (J), is therefore J = pf - pi = (0.14 kg * 39 m/s) - (0.14 kg * -39 m/s) = 10.92 kg m/s - (-5.46 kg m/s) = 21.84 kg m/s. This is the magnitude of the impulse exerted on the baseball by the bat. Note that the direction of the force is not requested, only its magnitude.

User Jscott
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