A small object at rest absorbs a light pulse | vedclass

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(B) The momentum $p$ transferred by a light pulse is given by the relation $p = \frac{E}{c}$,where $E$ is the energy of the pulse and $c$ is the speed

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(B) The momentum $p$ transferred by a light pulse is given by the relation $p = \frac{E}{c}$,where $E$ is the energy of the pulse and $c$ is the speed of light.Energy $E = 	ext{Power} 	imes 	ext{time} = P 	imes t$.Given: $P = 20\,mW = 20 	imes 10^{-3}\,W$,$t = 300\,ns = 300 	imes 10^{-9}\,s$,and $c = 3 	imes 10^8\,m/s$.Substituting the values:$p = \frac{(20 	imes 10^{-3}\,W) 	imes (300 	imes 10^{-9}\,s)}{3 	imes 10^8\,m/s}$$p = \frac{6000 	imes 10^{-12}}{3 	imes 10^8}\,kg\,m/s$$p = 2000 	imes 10^{-20}\,kg\,m/s$$p = 2 	imes 10^{-17}\,kg\,m/s$.Thus,the momentum is $2 	imes 10^{-17}\,kg\,m/s$.

$A$ small object at rest absorbs a light pulse of power $20\,mW$ and duration $300\,ns$. Assuming the speed of light as $3 \times 10^8\,m/s$,the momentum of the object becomes equal to $.........\times 10^{-17}\,kg\,m/s$.

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