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(D) The change in momentum $\Delta \vec{p}$ is equal to the impulse $\vec{J} = \int \vec{F} dt$. This is equal to the area under the force-time graph

Vedclass · Accepted Answer

$5\sqrt{2}\,kg\,m/s$

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(D) The change in momentum $\Delta \vec{p}$ is equal to the impulse $\vec{J} = \int \vec{F} dt$. This is equal to the area under the force-time graph for each component.For the $x$-component: $\Delta p_x = 	ext{Area} = 30\,N 	imes 0.1\,s = 3\,N\cdot s$.For the $y$-component: $\Delta p_y = 	ext{Area} = \frac{1}{2} 	imes 80\,N 	imes 0.1\,s = 4\,N\cdot s$.For the $z$-component: $\Delta p_z = 	ext{Area} = -50\,N 	imes 0.1\,s = -5\,N\cdot s$.The magnitude of the total change in momentum is $\Delta p = \sqrt{(\Delta p_x)^2 + (\Delta p_y)^2 + (\Delta p_z)^2}$.$\Delta p = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}\,kg\,m/s$.

$A$ particle is acted upon by a force whose components' variations with time are shown in the diagrams. The magnitude of the change in momentum of the particle in $0.1\,s$ is:

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