A particle of mass 500 text{ g} is at rest. I | vedclass

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(C) The area under the $P-t$ graph is equal to the work done, which is equal to the change in kinetic energy $(\Delta K)$.Given mass $m = 500 	ext{ g

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$5 	ext{ N-s}$

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(C) The area under the $P-t$ graph is equal to the work done, which is equal to the change in kinetic energy $(\Delta K)$.Given mass $m = 500 	ext{ g} = 0.5 	ext{ kg}$.The area of the trapezoid $OABC$ is given by:$	ext{Area} = \frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes (	ext{height})$$	ext{Area} = \frac{1}{2} 	imes (2 + 8) 	imes 5 = \frac{1}{2} 	imes 10 	imes 5 = 25 	ext{ J}$.Since the particle starts from rest, the initial kinetic energy is $0$. Thus, the final kinetic energy at $t = 5 	ext{ s}$ is $K = 25 	ext{ J}$.We know that kinetic energy $K = \frac{p^2}{2m}$, where $p$ is the momentum.$25 = \frac{p^2}{2 	imes 0.5}$$25 = \frac{p^2}{1}$$p^2 = 25$$p = 5 	ext{ kg m/s} = 5 	ext{ N-s}$.

$A$ particle of mass $500 \text{ g}$ is at rest. It is free to move along a straight line. The power delivered to the particle varies with time according to the following graph. The momentum of the particle at $t = 5 \text{ s}$ is

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