Answer
After the collision, the speed of the target body is $2.8~m/s$
Work Step by Step
By conservation of momentum, the final momentum of the system is equal to the initial momentum of the system in both the horizontal direction and the vertical direction.
Let's assume the projectile was initially moving in the +x-direction and was then deflected in the +y-direction.
We can find the x-component $p_x$ of the target body's final momentum:
$p_x + (2.0~kg)(0) = (2.0~kg)(8.0~m/s)$
$p_x = 16.0~kg~m/s$
We can find the y-component $p_y$ of the target body's final momentum:
$p_y + (2.0~kg)(6.0~m/s) = (2.0~kg)(0)$
$p_y = -12.0~kg~m/s$
We can find the magnitude of the target body's final momentum:
$p = \sqrt{p_x^2+p_y^2}$
$p = \sqrt{(16.0~kg~m/s)^2+(-12.0~kg~m/s)^2}$
$p = 20.0~kg~m/s$
Let $M$ be the mass of the target body. Let $v$ be the final speed of the target body. We can find an expression for $M$:
$M~v = p$
$M~v = 20.0~kg~m/s$
$M = \frac{20.0~kg~m/s}{v}$
Since the collision was perfectly elastic, the final kinetic energy is equal to the initial kinetic energy:
$KE_f = KE_0$
$\frac{1}{2}Mv^2+\frac{1}{2}(2.0~kg)(6.0~m/s)^2 = \frac{1}{2}(2.0~kg)(8.0~m/s)^2$
$Mv^2 = (2.0~kg)(8.0~m/s)^2-(2.0~kg)(6.0~m/s)^2$
$Mv^2 = 56.0~J$
$(\frac{20.0~kg~m/s}{v})~(v^2) = 56.0~J$
$v = \frac{56.0~J}{20.0~kg~m/s}$
$v = 2.8~m/s$
After the collision, the speed of the target body is $2.8~m/s$.
