JEE MainPhysicsModerateNewton's Laws of Motion

The particle of mass 1 kg is acted upon by a
force $\overset{\rightarrow}{F} = F_{x}\overset{\hat{}}{i} + F_{y}\overset{\hat{}}{j}$
from t = 0 to t = 4 sec. Find the velocityof the particle at t=4 sec.,
if its initial velocity is
$\overset{\rightarrow}{u} = - \overset{\hat{}}{i} + \overset{\hat{}}{j}$ and
the variation of applied force with time is as shown in the graphs

Show answer and solution

\int_{0}^{4}{F_{x}dt = \Delta P_{x} = m(v_{x} - u_{x})}

\Rightarrow 0 = m(v_{x} - u_{x})

\Rightarrow V_{x} = u_{x} = - 1

\int_{0}^{4}{F_{y}dt = \Delta P_{y} = m(v_{y} - u_{y})}

\Rightarrow \frac{1}{2} \times 4 \times 5 = 1(v_{y} - 1) \Rightarrow v_{y} = 11

\overset{\rightarrow}{v} = - \overset{\hat{}}{i} + 11\overset{\hat{}}{j}

Source: Grass-G / P-16 Physics paper 25 Nov First.docx

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