A cricket ball of mass 150 g has an initial velocity u=(3i^+4j^)ms-1 and a final velocity v=-(3i^+4j^)ms-1, after being hit. The change in momentum (final momentum - initial momentum) is (in kg-ms-1)

Given, u=(3i^+4j^)ms-1 and v=-(3i^+4j^)ms-1Mass of the ball = 150 g = 0.15 kg ∴ ∆p= Change in momentum            = Final momentum - Initial momentum = mv - mu           =m(v-u)=(0.15)[-(3i^+4j^)-(3i^+4j^)]           =(0.15)[-6i^-8j^)]=-[0.15×6i^+0.15×8j^)]           =-(0.9i^+1.20j^) Hence, ∆p=-(0.9i^+1.2j^)