In a slow-pitch softball game, a 0.200 kg softball crossed the plate at 11.0 m/s
ID: 2019608 • 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 42.0m/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)
thanks!
Explanation / Answer
(a) We'll arbitrarily choose up and towards center field as positive directions.
Initial momentum:
pix = mvix = (0.200 kg)(-11.0 m/s)(cos 35.0°) = -1.80 Ns
piy = mviy = (0.200 kg)(-11.0 m/s)(sin 35.0°) = -1.26 Ns
Final momentum:
pfx = mvfx = (0.200 kg)(42.0 m/s)(cos 30.0°) = 7.27 Ns
pfy = mvfy = (0.200 kg)(42.0 m/s)(sin 30.0°) = 4.20 Ns
Find the magnitude of the impulse:
Jx = pfx = 7.27 Ns - (-1.80 Ns) = 9.07 Ns
Jy = pfy = 4.20 Ns - (-1.26 Ns) = 5.46 Ns
vector addition to find Jtotal
magnitude of Jtotal = [(9.07 Ns)2 + (5.46 Ns)2] = 10.6 Ns
Find the angle of the impulse:
tan = Jy / Jx
= tan-1(5.46 Ns / 9.07 Ns) = 31.0°
(b) Its easier to solve this problem graphically. Let's picture the situation.
- The force increases from 0 to Fmax in 0.004 s
- The force then remains constant at Fmax for 0.020 s
- The force then decreases from Fmax to 0 in 0.004 s
- Shape of the graph is essentially a trapezoid
Find the magnitude of Fmax:
Area of graph = Area leftTriangle + Area rectangle + Area rightTriangle
A = 0.5(0.004s)(Fmax) + (0.020s)(Fmax) + 0.5(0.004s)(Fmax)
A = (0.024s)Fmax
J = Ft = Area under the curve of the force vs time graph
J = (0.024 s) Fmax
Fmax = J / 0.024 s
Fmax = 10.6 Ns / 0.024 s = 442 N
Find the angle of Fmax:
Fmax = J / 0.024 s
Fmax-x = Jx / 0.024 s = 9.07 Ns / 0.024s = 378 N
Fmax-y = Jy / 0.024 s = 5.46 Ns / 0.024s = 228 N
tan = Fmax-y / Fmax-x
= tan-1(228 N / 378 N) = 31.1°
Hope that helps!