A particle of mass 2 kg moves with an initial velocity of (4i + 2j) m/s on the x-y plane. A force F = (2i – 8j) N acts on the particle. The initial position of the particle is (2m, 3 m). 3 m). A particle of mass 2 kg moves with an initial velocity of (4i + 2j) m/s on the x-y plane. A force F = (2i - 8j) N acts on the particle. The initial position of the particle is (2m September 29, 2020 Category: Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) , Facebook Messenger WhatsApp Share this:TwitterFacebook Related