In a slow-pitch softball game, a 0.200 kg softball crossed the plate at 19.0 m/s

Question

In a slow-pitch softball game, a 0.200 kg softball crossed the plate at 19.0 m/s at an angle of 40.0° below the horizontal. The batter hits the ball toward center field, giving it a velocity of 50.0 m/s at 30.0° above the horizontal.
(a) Determine the impulse delivered to the ball.
N·s (magnitude)
° (above the horizontal)

(b) If the force on the ball increases linearly for 4.00 ms, holds constant for 20.0 ms, then decreases linearly to zero in another 4.00 ms, what is the maximum force on the ball?
N (magnitude)
° (above the horizontal)

Explanation / Answer

Mass of the soft Ball = M = 0.2 kg Intia velocity (with respect to Horizontial) : U cos = 19 cos 40   = 14.55 m/s final velocity = V cos 30 = 50 cos30 = 43.3 m/s thus, change of momentum : P = M (V - U )     = (0.2)(43.3 - 14.55)     = 5.75 Kgm/s (a) impulse delivered to the ball: = change of momentum = 5.75 m/s (b) Time of contact : t = 20 ms = 20x10-3 s Note : Among the given timings force holds with the time should be considered . so, thus, Impulse = I = F x t F = Impulse / t     = 5.75 / 20 x10-3        = 287.5 N

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