#### Answer

$p_x$ of the third fragment is -1 N s
$p_y$ of the third fragment is -2 N s

#### Work Step by Step

We can use conservation of momentum to find the components of the third fragment's momentum. Since the initial momentum of the object is zero, the final momentum is also zero.
Let's consider the horizontal components of momentum.
$p_{1x}+p_{2x}+p_{3x} = 0$
$p_{3x} = -p_{1x}-p_{2x}$
$p_{3x} = -(-2~N~s)-(3~N~s)$
$p_{3x} = -1~N~s$
$p_x$ of the third fragment is -1 N s.
Let's consider the vertical components of momentum.
$p_{1y}+p_{2y}+p_{3y} = 0$
$p_{3y} = -p_{1y}-p_{2y}$
$p_{3y} = -(2~N~s)-0$
$p_{3y} = -2~N~s$
$p_y$ of the third fragment is -2 N s.