## University Physics with Modern Physics (14th Edition)

(a) $J = F~t = (2500~N)(0.0010~s)$ $J = 2.5~kg~m/s$ The force exerts an impulse of 2.5 kg m/s. (b) (i) $p_2-p_1 = J$ $mv_2 = J+mv_1$ $v_2 = \frac{J+mv_1}{m}$ $v_2 = \frac{(2.5~kg~m/s)+(2.00~kg)(5.00~m/s)}{2.00~kg}$ $v_2 = 6.25~m/s$ The speed is 6.25 m/s to the right. (ii) $p_2-p_1 = J$ $mv_2 = J+mv_1$ $v_2 = \frac{J+mv_1}{m}$ $v_2 = \frac{(-2.5~kg~m/s)+(2.00~kg)(5.00~m/s)}{2.00~kg}$ $v_2 = 3.75~m/s$ The speed is 3.75 m/s to the right.