Answer
$60\;\rm N$
Work Step by Step
According to Newton's second law, the force is given by
$$F=ma$$
So, to find the force we need to draw a graph of acceleration $a$ versus $m^{-1}$.
The slope of this graph at any point will be
$${\rm slpoe}=\dfrac{a}{\frac{1}{m}}=ma$$
which is the force magnitude.
We got the masses of the balls and now we need to find their acceleration.
We know that the ball, each ball, starts from rest and we know its final velocity, so we can use the kinematic formula of
$$v^2=v_i^2+2ad$$
whereas $d$ is the distance traveled, which is 15 cm for each ball as well.
Thus,
$$a=\dfrac{v^2}{2d}=\dfrac{v^2}{2\cdot 0.15}$$
Thus,
$$\boxed{a=\dfrac{v^2}{0.3}}$$
This is the formula that we will use to get the acceleration of each ball, as you see in the table below.
\begin{array}{|c|c|c|}
\hline
{\rm Mass\;(kg)}& {\rm speed\;(m/s)}& \rm Acceleration \;(m/s^2)\\
\hline
0.200 & 9.4 & 294.53 \\
\hline
0.400& 6.3 & 132.3 \\
\hline
0.600 & 5.2& 90.13 \\
\hline
0.800& 4.9& 80.03 \\
\hline
1.000 & 4& 53.33 \\
\hline
\end{array}
Now we can draw the graph of $a$ versus $\dfrac{1}{m}$ by using the table below.
\begin{array}{|c|c|c|}
\hline
{\rm \dfrac{1}{Mass}\;(kg^{-1})} & \rm Acceleration \;(m/s^2)\\
\hline
5& 294.53 \\
\hline
2.5& 132.3 \\
\hline
(5/3)=1.667& 90.13 \\
\hline
1.25& 80.03 \\
\hline
1.0& 53.33 \\
\hline
\end{array}
From the figure below, it is obvious that the force of the piston is
$$F\approx \color{red}{\bf 60}\;\rm N$$
