In a slow-pitch softball game, a 0.200 kg softball crossed the plate at 18.00 m/
ID: 1485931 • Letter: I
Question
In a slow-pitch softball game, a 0.200 kg softball crossed the plate at 18.00 m/s at an angle of 45.0° below the horizontal. The batter hits the ball toward center field, giving it a velocity of 42.0 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 to zero linearly in another 4.00 ms, what is the maximum force on the ball?
N (magnitude)
° (above the horizontal)
Explanation / Answer
Lets take the components of monentum
The initial momentum is Px1 = mVx1 = m(18cos-45°) and Py1 = mVy1 = m(18sin-45°)
The final momentum is Px2 = mVx2 = m(-42cos30°) and Py2 = m(42sin30°)
So, the Px is Px2-Px1 = 0.2*(-42cos30°-18cos-45°) = -9.8201 kg.m/s and
Py = 0.2*(42sin30°-(18sin-45°)) = 1.6544kg.m/s
The total impulse is then F*t = ((-9.8201)² + 1.6544²) = 9.958 kg.m/s
The angle above the horizontal is arctan (6.429/8.908) = 9.56 °
b) Fmax = 9.958/.024 = 414.91 N. The angle above the horizontal is arctan (6.429/8.908) = 9.56 °