Question

21 Q. A 12.0 kg shopping cart rolls due south at 1.50 m's. After striking the buasper of a car.i travels at 0.80 m/s, 30 E of S. What is the magnitude of the change in momentum sustaine by the shopping cart? A. 8.4 kg m/s B. 9.7 kg m/s C. 11 kg m/s D. 27 kg m/s

Answer 1

Given things m = 12 kg v_i = 1.50 m/s v_f = 0.80 m/s theta = 30 degree initial momentum is y direction o_i = m v_i = 12 kg times 1.50 m/s = 18 kg m/s final momentum has x and y component so p_fy = m v_f cos theta = 12 times 0.80 times cos 30 = 8.31 p_fx = m v_f sin theta = 12 times 0.80 times sin 30 = 4.8 change in momentum of case in its initial direction is (y direction) \r\n delta p_xy = 18 - 8.31 = 9.69 kg m/s delta p_x = 0 - 4.8 = -4.8 kg m/s (initially p_ix = 0) delta p = squareroot delta p_y^2 + delta p_x^2 = squareroot (9.69)^2 + (-4.8)^2 = 10.8 kg m/s delta p almostequalto 11 kg m/s