A particle of mass 2 kg moves with an initial velocity of \(\vec{v} = 4 \hat{i} + 4 \hat{j}\) m/s. A constant force of \(\vec{F} = - 20 \hat{j}\) N is applied on the particle. Initially, the particle was at (0, 0). the x-coordinate of the particle when its y-coordinate again becomes zero is given by :
Answer & explanation
Correct answer: option 3
\(\vec{a} = \frac{\vec{F}}{m} = - 10 \hat{j}\) (m/s)2
Displacement in y-direction : y = ut + \(\frac{1}{2}\) at2
\(\Rightarrow 0 = 4*t*\frac{-1}{2}*10*t^2\)
\(t = \frac{4}{5} \text{ s}\)
\(\Rightarrow x = 4 t\)
\(\Rightarrow x = 4*\frac{4}{5}\)
\(\Rightarrow x = 3.2 \text{ m}\)