This is a momentum problem, so let's find the velocity of the ball initially. It can be given by:
[tex]\vec{F} = m\vec{a}[/tex]
Let's plug in values and solve for a:
[tex]\vec{a} = \frac{2000}{450} = 4.44 m/s^2[/tex]
Now, we can multiply by time to get velocity:
[tex]\vec{v} = \vec{a}t = 4.44(5) = 22.2m/s[/tex]
We have all the information we need, so let's set up an equation to solve for the velocity of the wall after the collision:
[tex]mv_{ib}= mv_{fb} + mv_{w}[/tex]
Solve for velocity of wall finally:
[tex]v_w = \frac{mv_{ib}-mv_{fb}}{m} [/tex]
Plug in all values:
[tex]v_w = \frac{450(22.2)-450(10)}{500} = 10.98 \frac{m}{s} [/tex]
So, the final velocity of the wall will be 10.98 m/s.
The KE in this problem initially was 110889 J (1/2mv^2). After the collision it was 52640.1 so, the KE decreased.
Since there was no mashing of objects together, this was an elastic collision.