Answer
The magnitude of the force is $471.2N$ and its direction is opposite from the direction of the ball before contact.
Work Step by Step
First, let's calculate the acceleration of the ball during contact.
We have $v_0=2.1m/s$, $v=-2.0m/s$ and $t=3.3\times10^{-3}s$, so
$$v=v_0+at$$ $$a=\frac{v-v_0}{t}=\frac{-4.1m/s}{3.3\times10^{-3}s}=-1.24\times10^3m/s^2$$
Next, applying Newton's 2nd Law of Motion, we can find the average net force exerted on the ball:
$$\sum F=m_{ball}a=0.38kg\times(-1.24\times10^3m/s^2)$$ $$\sum F=-471.2N$$
So the magnitude of the force is $471.2N$ and its direction is opposite from the direction of the ball before contact.
