A 40 g ball dropped from a certain height bo | vedclass

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(B) The mass of the ball is $m = 40 \ g = 0.04 \ kg$. The initial velocity just before impact is $v_i = -10 \ m/s$ (taking downward as negative). Sinc

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$50$

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(B) The mass of the ball is $m = 40 \ g = 0.04 \ kg$. The initial velocity just before impact is $v_i = -10 \ m/s$ (taking downward as negative). Since the collision is elastic (no loss of mechanical energy),the final velocity just after impact is $v_f = +10 \ m/s$. The change in momentum is $\Delta p = m(v_f - v_i) = 0.04 	imes (10 - (-10)) = 0.04 	imes 20 = 0.8 \ kg \cdot m/s$. According to the impulse-momentum theorem,the impulse $J = F_{avg} 	imes \Delta t = \Delta p$. Given $F_{avg} = 16 \ N$,we have $16 	imes \Delta t = 0.8$. $\Delta t = \frac{0.8}{16} = 0.05 \ s$. Converting to milliseconds,$\Delta t = 0.05 	imes 1000 = 50 \ ms$.

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