Acatcher catches a 150g baseball traveling horizontally at 25 m/s. if the ball takes 20.ms to stop once it is in contact with the catcher's glove, what average force did the ball exert on the catcher?

The magnitude of the force exerted by the ball on the catcher is 1.9 × 10²  N

Explanation:

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Let´s find the acceleration of the ball that makes it stop when caught by the catcher. The acceleration can be calculated from the equation of velocity considering that it is constant:

v = v0 + a · t

We know that initially the ball was traveling at 25 m/s, so, if we consider the position of the catcher as the origin of the frame of reference, then, v0 = -25 m/s. We also know that it takes the ball 20 ms (0.02 s) to stop (i.e. to reach a velocity of 0). Then using the equation of velocity:

v = v0 + a · t

0 m/s = -25 m/s + a · 0.020 s

25 m/s/ 0.020 s = a

Now, using the second law of Newton, we can calculate the force exerted by the catcher on the ball:

F = m · a

Where:

F = force.

m = mass of the ball.

a = acceleration.

F = 0.150 kg · (25 m/s/ 0.020 s) = 1.9 × 10²  N

According to Newton´s third law, the force exerted by the ball on the catcher will be of equal magnitude but opposite direction. Then, the force exerted by the ball on the catcher will have a magnitude of 1.9 × 10²  N.

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