In a slow-pitch softball game, a 0.200 kg softball crossed the plate at 11.0 m/s
ID: 1978953 • Letter: I
Question
In a slow-pitch softball game, a 0.200 kg softball crossed the plate at 11.0 m/s at an angle of 35.0° below the horizontal. The batter hits the ball toward center field, giving it a velocity of 40.0 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
the initial horizontal momentum component is subtracted
from the final horizontal momentum component
Take direction up and towards center field as positive
a)Horizontal pi = 0.200*-11.0*cos(-35.0º)= -1.802
Horizontal pf = 0.200*40.0*cos30.0º = 6.92
Subtract the first from the second (observe the signs):
p horizontal = 6.92-(-1.802) = 8.722 N-s -----------Horizontal impulse
Veritical pi = 0.200*-11*sin(-35º) = 1.261 Ns
vertical pf = 0.200*40*sin30º = 4 Ns
Again subtract the first from the second
p vertical = 2.739 N-s vertical impules
Total impulse = [8.722^2 + 2.739^2] = 9.141 N-s
The direction is arctan(2.739/8.722) = 17.434º.
The angle is positive, so it is above the horizontal.
b) Impulse = F*dt.
integral = 0.5*F*4.00 + F*20.0 + 0.5*F*4.00 = 24.0*F N-ms = 24*10^-3*F N-s.
This must equal the total impulse of 9.141
so 24.0*10^-3*F = 9.141 and
F =380.875 N
The direction is the same as the impulse, 17.434º above horizontal