A block of mass M = 10,kg rests on a horizont | vedclass

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(D) Let $M = 10\,kg$ and $m = 0.05\,kg.$ By conservation of linear momentum during the collision:$mv = (M + m)V_0$$V_0 = \frac{mv}{M + m} = \frac{0.05

Vedclass · Accepted Answer

$0.04$

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(D) Let $M = 10\,kg$ and $m = 0.05\,kg.$ By conservation of linear momentum during the collision:$mv = (M + m)V_0$$V_0 = \frac{mv}{M + m} = \frac{0.05v}{10.05} \approx \frac{0.05v}{10} = 0.005v$The block moves a distance $s = 2\,m$ before stopping due to friction. The retardation $a$ is:$a = \mu g = 0.05 	imes 10 = 0.5\,m/s^2$Using $v_f^2 - u^2 = 2as$:$0 - V_0^2 = 2(-a)s \implies V_0^2 = 2as = 2 	imes 0.5 	imes 2 = 2$$V_0 = \sqrt{2}\,m/s$Since $V_0 = \frac{0.05v}{10.05} \approx 0.005v$,we have $0.005v = \sqrt{2} \implies v = 200\sqrt{2}\,m/s$.For a freely falling object,$v_{final} = \sqrt{2gH}$. Given $v_{final} = \frac{v}{10} = \frac{200\sqrt{2}}{10} = 20\sqrt{2}\,m/s$:$20\sqrt{2} = \sqrt{2 	imes 10 	imes H}$$(20\sqrt{2})^2 = 20H$$400 	imes 2 = 20H \implies 800 = 20H \implies H = 40\,m = 0.04\,km$.

Column-$I$	Column-$II$
$(1)$ Definition of force	$(a)$ Newton's third law of motion
$(2)$ Measurement of force	$(b)$ Newton's second law of motion
	$(c)$ Newton's first law of motion

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Match the items in Column-$I$ with those in Column-$II$.

Column-$I$ Column-$II$

$(1)$ Definition of force $(a)$ Newton's third law of motion

$(2)$ Measurement of force $(b)$ Newton's second law of motion

$(c)$ Newton's first law of motion

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