Question

In a slow-pitch softball game, a 0.200 kg softball crossed the plate at 20.0 m/s at an angle of 45.0° below the horizontal. The batter hits the ball toward centerfield, giving it a velocity of 44.0 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 forceon the ball?
N (magnitude)
° (above the horizontal)



Please help! I have no idea how to do this

Answer 1

We'll arbitrarily choose up and towards center field as positive directions.

(a) Find the impulse (magnitude and angle)

Initial momentum:
p_ix = mv_ix = (0.200 kg)(-19.0 m/s)(cos 40.0°) = -2.91 Ns
p_iy = mv_iy = (0.200 kg)(-19.0 m/s)(sin 40.0°) = -2.44 Ns

Final momentum:
p_fx = mv_fx = (0.200 kg)(50.0 m/s)(cos 30.0°) = 8.66 Ns
p_fy = mv_fy = (0.200 kg)(50.0 m/s)(sin 30.0°) = 5.00 Ns

Find the magnitude of the impulse:
I_x = Δp_fx = 8.66 Ns - (-2.91 Ns) = 11.57 Ns
I_y = Δp_fy = 5.00 Ns - (-2.44 Ns) = 7.44 Ns
I_Total = (11.57 Ns)i + (7.44 Ns)j
magnitude of I_Total = √[(11.57 Ns)² + (7.44 Ns)²] = 13.76 Ns

Find the angle of the impulse:
tan θ = I_y / I_x
θ = tan^-1(7.44 Ns / 11.57 Ns) = 32.7°

(b) Find the maximum force (magnitude and angle)

Its easier to solve this problem graphically. Sorry...wanted to insert a graph but the "custom diagram" feature refuses to work for me T_T. Draw on paper if needed. Let's picture the situation.

- The force increases from 0 to F_max in 0.0.004 s
- The force then remains constant at F_max for 0.020 s
- The force then decreases from F_max 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)(F_max) + (0.020s)(F_max) + 0.5(0.004s)(F_max)
A = (0.024s)F_max

I = ∫Fdt = Area under the curve of the force vs time graph
I = (0.024 s) F_max
F_max = I / 0.024 s
F_max = 13.76 Ns / 0.024 s = 573 N

Find the angle of Fmax:
F_max = I / 0.024 s
F_max = [(11.57 Ns)i + (7.44 Ns)j] / 0.024s = (482 N)i + (310 N)j
tan θ = (F_max)_y / (F_max)_x
θ = tan^-1(310 N / 482 N) = 32.7°

Hope that helps!