Using the second law of motion,derive the rel | vedclass

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(B) Derivation: According to Newton's second law,$F \propto \frac{dp}{dt}$. Since $p = mv$,$F = k \frac{d(mv)}{dt} = kma$. Taking $k=1$,we get $F = ma

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$10^{5} \, N, 10^{-4} \, s$

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(B) Derivation: According to Newton's second law,$F \propto \frac{dp}{dt}$. Since $p = mv$,$F = k \frac{d(mv)}{dt} = kma$. Taking $k=1$,we get $F = ma$.$(i)$ Given: $m = 10 \, g = 0.01 \, kg$,$u = 10^3 \, m \, s^{-1}$,$v = 0$,$s = 5 \, cm = 0.05 \, m$.Using $v^2 - u^2 = 2as$:$0^2 - (10^3)^2 = 2 \cdot a \cdot 0.05$$-10^6 = 0.1 \cdot a$$a = -10^7 \, m \, s^{-2}$.Resistive force $F = |ma| = 0.01 	imes 10^7 = 10^5 \, N$.$(ii)$ Using $v = u + at$:$0 = 10^3 + (-10^7) \cdot t$$10^7 \cdot t = 10^3$$t = 10^{-4} \, s$.

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