Answer
$F_{max} = 986~N$
Work Step by Step
We can let the initial velocity be $-34~m/s$
We can find the change in momentum of the ball:
$\Delta p = p_f-p_i$
$\Delta p = (0.058~kg)(34~m/s)-(0.058~kg)(-34~m/s)$
$\Delta p = 3.944~kg~m/s$
The impulse on the ball from the wall must be equal in magnitude to the ball's change in momentum.
To find the impulse on the ball, we can calculate the area under the force versus time graph.
From $t = 0$ to $t = 6~ms$, the area can be divided into three parts, including a triangle (0 to 2 ms), a rectangle (2 ms to 4 ms), and a triangle (4 ms to 6 ms)
We can find each area separately:
$A_1 = \frac{1}{2}(F_{max})(2.0\times 10^{-3}~s) = (F_{max})(1.0\times 10^{-3}~s)$
$A_2 = (F_{max})(2.0\times 10^{-3}~s)$
$A_3 = \frac{1}{2}(F_{max})(2.0\times 10^{-3}~s) = (F_{max})(1.0\times 10^{-3}~s)$
We can find an expression for the impulse from $t = 0$ to $t = 6~ms$:
$J = (F_{max})(1.0\times 10^{-3}~s)+(F_{max})(2.0\times 10^{-3}~s)+(F_{max})(1.0\times 10^{-3}~s)$
$J = (F_{max})(4.0\times 10^{-3}~s)$
We can find the value of $F_{max}$:
$J = 3.944~kg~m/s$
$(F_{max})(4.0\times 10^{-3}~s) = 3.944~kg~m/s$
$F_{max} = \frac{3.944~kg~m/s}{4.0\times 10^{-3}~s}$
$F_{max} = 986~N$
