A particle of mass 2, kg is initially at rest | vedclass

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(D) The impulse imparted to the particle is equal to the area under the $F-t$ graph, which is equal to the change in linear momentum of the particle.I

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$50$

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(D) The impulse imparted to the particle is equal to the area under the $F-t$ graph, which is equal to the change in linear momentum of the particle.Impulse $J = \int F dt = 	ext{Area under } F-t 	ext{ graph}$.Area $= 	ext{Area of triangle (0 to 2 s)} + 	ext{Area of rectangle (2 to 4 s)} + 	ext{Area of trapezoid (4 to 6 s)} + 	ext{Area of triangle (6 to 10 s)}$.Area $= (\frac{1}{2} 	imes 2 	imes 10) + (2 	imes 10) + (\frac{1}{2} 	imes (10 + 20) 	imes 2) + (\frac{1}{2} 	imes 4 	imes 20)$.Area $= 10 + 20 + 30 + 40 = 100 \, Ns$.Change in momentum $\Delta p = m(v - u) = 100$.Given $m = 2 \, kg$ and initial velocity $u = 0$.$2(v - 0) = 100$.$2v = 100$.$v = 50 \, ms^{-1}$.

$A$ particle of mass $2\, kg$ is initially at rest. $A$ force acts on it whose magnitude changes with time. The force-time graph is shown below. The velocity of the particle after $10\, s$ is ......... $ms^{-1}$.

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