CALC? Starting at t = 0, a horizontal net force is applied to a box that has an initial momentum What is the momentum of the box at t = 2.00 s?

Solution 16E Momentum ( p ) and variable force ( F) are related by the following equation, t2 p = Fdt t1 Given that, net force F = (0.280 N/s) t i + ( 0.450 N/s )t j Initial momentum p = ( 3.00 kg. m/s) i + (4.00 kg.m/s) j Let us now calculate the x-component first. 2 p x (0280 N/s)t i dt 0 t2 2 2 p x 0.280 N/s i ( 2 s0 p = 0.560 N.s i x Let us now calculate the y-component. 2 p y (0.450 N/s )t jdt 0 2 t3 2 p y 0.450 N/s j ( ) 3 0 p = 0.150 N/s j × 8 s 3 y p y 1.2 N.s j Therefore, the net x-component of momentum = 3.00 kg.m/s i + 0.560 kg.m/s i = 2.44 kg.m/s i The net y-component of momentum = 4.00 kg.m/s j 1.2 kg.m/s j = 2.8 kg.m/s j Therefore, omentum at t = 2.00 s ( 2.44 kg.m/s i + 2.8 kg.m/s j)