Question

A force F→=(2i^+j^+3k^)N acts on a particle of mass 1 kg for 2 s. If initial velocity of particle is u→=(2i^+j^)m/s, then speed of particle at the end of 2 s will be

Answer 1

V = 6i+3j+3k is the answer

Answer 2

by impulse momentum equation

$\int\limits {f} \, dt=m_1v_1-m_2v_2$

on putting values you get 6i+3j+6k