JEE MAIN - Physics (2016 - 9th April Morning Slot - No. 2)
Which of the following option correctly describes the variation of the speed v and acceleration ‘a’ of a point mass falling vertically in a viscous medium that applies a force F = − kv, where ‘k’ is a constant, on the body ? (Graphs are schematic and not drawn to scale)
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Explanation
Equation of motion for the mass,
ma = mg $$-$$ kv
$$ \Rightarrow $$ $${{dv} \over {dt}} = {{mg - kv} \over m}$$
$$ \Rightarrow $$ $$\int\limits_0^v {{{dv} \over {mg - kv}}} = {1 \over m}\int\limits_0^t {dt} $$
$$ \Rightarrow $$ $$ - {1 \over k}\left[ {\ln \left( {mg - kv} \right)} \right]_0^v = {t \over m}$$
$$ \Rightarrow $$ $$\ln \left( {{{mg - kv} \over {mg}}} \right) = - {{kt} \over m}$$
$$ \Rightarrow $$ $$1 - {{kv} \over {mg}} = {e^{ - {{kt} \over m}}}$$
$$ \Rightarrow $$ $${{kv} \over {mg}} = 1 - {e^{ - {{kt} \over m}}}$$
$$ \Rightarrow $$ $$v = {{mg} \over k}\left( {1 - {e^{ - {{kt} \over m}}}} \right)$$
ma $$=$$ mg $$-$$ k $$ \times $$ $${{mg} \over k}$$ (1 $$-$$ e$$^{ - {{kt} \over m}}$$)
$$=$$ mg $$-$$ mg + mge$$^{ - {{kt} \over m}}$$
a $$=$$ g e$$^{ - {{kt} \over m}}$$
ma = mg $$-$$ kv
$$ \Rightarrow $$ $${{dv} \over {dt}} = {{mg - kv} \over m}$$
$$ \Rightarrow $$ $$\int\limits_0^v {{{dv} \over {mg - kv}}} = {1 \over m}\int\limits_0^t {dt} $$
$$ \Rightarrow $$ $$ - {1 \over k}\left[ {\ln \left( {mg - kv} \right)} \right]_0^v = {t \over m}$$
$$ \Rightarrow $$ $$\ln \left( {{{mg - kv} \over {mg}}} \right) = - {{kt} \over m}$$
$$ \Rightarrow $$ $$1 - {{kv} \over {mg}} = {e^{ - {{kt} \over m}}}$$
$$ \Rightarrow $$ $${{kv} \over {mg}} = 1 - {e^{ - {{kt} \over m}}}$$
$$ \Rightarrow $$ $$v = {{mg} \over k}\left( {1 - {e^{ - {{kt} \over m}}}} \right)$$
ma $$=$$ mg $$-$$ k $$ \times $$ $${{mg} \over k}$$ (1 $$-$$ e$$^{ - {{kt} \over m}}$$)
$$=$$ mg $$-$$ mg + mge$$^{ - {{kt} \over m}}$$
a $$=$$ g e$$^{ - {{kt} \over m}}$$
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