## College Physics (4th Edition)

(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.
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.