The position-time graph for a particle of mas | vedclass

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(A) The mass of the particle is $m = 100 \ g = 0.1 \ kg$.From the position-time graph,the velocity $v$ is the slope of the line.For $t < 5 \ s$,the in

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$0.3$

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(A) The mass of the particle is $m = 100 \ g = 0.1 \ kg$.From the position-time graph,the velocity $v$ is the slope of the line.For $t $v_i = \frac{10 - 20}{5 - 0} = \frac{-10}{5} = -2 \ m/s$.For $t > 5 \ s$,the final velocity $v_f$ is the slope of the line from $(5, 10)$ to $(15, 20)$:$v_f = \frac{20 - 10}{15 - 5} = \frac{10}{10} = 1 \ m/s$.Impulse $I$ is equal to the change in momentum:$I = \Delta p = m(v_f - v_i) = 0.1 	imes [1 - (-2)] = 0.1 	imes 3 = 0.3 \ N-s$.

The position-time graph for a particle of mass $100 \ g$ is shown in the figure. The impulse acting on the particle at $t = 5 \ s$ is ............ $N-s$.

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