Answer
a) $\rm 1.0\; m/s$
b) $\rm 90\;N$
c) $\rm -90\;N$
Work Step by Step
a)
We know that the initial velocity of the clay ball just before it hits the block is 10 m/s and after 10 ms its speed (while it is stuck with the block and moving as one unit) is $v_{xf}$. Noting that this final velocity is constant since the block slides on a frictionless surface as the author told us.
According to Newton's third law, we know that the force exerted by the clay ball on the block is equal in magnitude to the force exerted by the block on the clay ball but in the opposite direction. We call these two forces $F$ but we should be careful with the direction.
Thus, the force exerted on the block is to the right while the force exerted on the clay is to the left.
$$\sum F_{x,block}=F=m_{block}a_{x,block}$$
$$\sum F_{x,clay}=-F= m_{clay}a_{x,clay}$$
From the last two formulas;
$$m_{block}a_{x,block}=-m_{clay}a_{x,clay} $$
Plugging the known;
$$900 a_{x,block}=-100a_{x,clay} $$
$$9 a_{x,block}=- a_{x,clay}\tag 1 $$
Applying the kinematic formulas of velocity on both objects;
$$v_{fx}=v_{ix,clay}+a_{x,clay}t\tag 2$$
$$v_{fx}=\overbrace{v_{ix,block}}^{0}+a_{x,block}t\tag 3$$
From these two equations;
$$a_{x,block}t=v_{ix,clay}+a_{x,clay}t$$
Plugging from (1);
$$a_{x,block}t=v_{ix,clay}+(-9a_{x,block})t$$
$$a_{x,block}t=v_{ix,clay} -9a_{x,block}t$$
$$10a_{x,block}t=v_{ix,clay} $$
Thus,
$$a_{x,block}=\dfrac{v_{ix,clay}}{10t}$$
Plugging the known;
$$a_{x,block}=\dfrac{10}{10\times10\times 10^{-3}}=\bf 100\;\rm m/s^2$$
Plugging into (3) to find the final velocity of the system;
$$v_{fx}= 100t=100\times 10\times 10^{-3}$$
$$v_{fx} =\color{red}{\bf 1}\;\rm m/s$$
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b)
The force exerted by the clay on the block is given by the very first equation above;
$$F_{\rm clay\;on\;block}=m_{block}a_{block}=0.9\times 100=\color{red}{\bf 90}\;\rm N$$
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c)
And the force exerted by the block on the clay is the same as the previous force but in the opposite direction.
$$F_{\rm block\;on\;clay}=\color{red}{\bf -90}\;\rm N$$
