Answer
(a) The force exerts an impulse of 2.5 kg m/s.
(b) (i) The speed is 6.25 m/s to the right.
(ii) The speed is 3.75 m/s to the right.
Work Step by Step
(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.
