#### Answer

(a) The horizontal component of velocity of the third piece is $-0.133~m/s$
The vertical component of velocity of the third piece is $-4.07~m/s$
(b) After the explosion, the center of mass is stationary at the origin.

#### Work Step by Step

The initial momentum of the system is zero. By conservation of momentum, the momentum of the system is zero after the explosion.
We can find the x-component of the third piece:
$(\frac{M}{4})(5.0~m/s)~cos~37^{\circ}-(\frac{M}{3})(4.0~m/s)~cos~45^{\circ}+(\frac{5M}{12})v_x = 0$
$(\frac{5M}{12})v_x = (\frac{M}{3})(4.0~m/s)~cos~45^{\circ}-(\frac{M}{4})(5.0~m/s)~cos~37^{\circ}$
$5~v_x =(4)(4.0~m/s)~cos~45^{\circ} -(3)(5.0~m/s)~cos~37^{\circ}$
$5~v_x = -0.6658~m/s$
$v_x = -0.133~m/s$
The horizontal component of velocity of the third piece is $-0.133~m/s$
We can find the y-component of the third piece:
$(\frac{M}{4})(5.0~m/s)~sin~37^{\circ}+(\frac{M}{3})(4.0~m/s)~sin~45^{\circ}+(\frac{5M}{12})v_y = 0$
$(\frac{5M}{12})v_y = -(\frac{M}{3})(4.0~m/s)~sin~45^{\circ}-(\frac{M}{4})(5.0~m/s)~sin~37^{\circ}$
$5~v_y =-(4)(4.0~m/s)~sin~45^{\circ} -(3)(5.0~m/s)~sin~37^{\circ}$
$5~v_y = -20.34~m/s$
$v_y = -4.07~m/s$
The vertical component of velocity of the third piece is $-4.07~m/s$
(b) Before the explosion, the center of mass was stationary at the origin. By conservation of momentum, after the explosion, the center of mass is stationary at the origin.