JEE MAIN - Physics (2020 - 9th January Morning Slot - No. 13)
Two particles of equal mass m have respective
initial velocities $$u\widehat i$$ and $$u\left( {{{\widehat i + \widehat j} \over 2}} \right)$$.
They collide completely inelastically. The energy lost in the process is :
initial velocities $$u\widehat i$$ and $$u\left( {{{\widehat i + \widehat j} \over 2}} \right)$$.
They collide completely inelastically. The energy lost in the process is :
$${1 \over 3}m{u^2}$$
$${1 \over 8}m{u^2}$$
$${3 \over 4}m{u^2}$$
$$\sqrt {{2 \over 3}} m{u^2}$$
Explanation
$$\overrightarrow {{P_i}} = \overrightarrow {{P_f}} $$
$$ \Rightarrow $$ mu$$\widehat i$$ + m$$\left( {{u \over 2}\widehat i + {u \over 2}\widehat j} \right)$$ = 2m$$\overrightarrow v $$
Compare both side
$$\overrightarrow v = $$ $${{3u} \over 4}\widehat i + {u \over 4}\widehat j$$
$$ \Rightarrow $$ $${\left| {\overrightarrow v } \right|^2}$$ = $${{10{u^2}} \over {16}}$$
$$\Delta $$KE = KEf – KEi
= $${1 \over 2}2m \times {{10{u^2}} \over {16}}$$ - $${1 \over 2}m{u^2}$$ - $${1 \over 2}m{\left( {{u \over {\sqrt 2 }}} \right)^2}$$
= $$ - {1 \over 8}m{u^2}$$
$$ \Rightarrow $$ mu$$\widehat i$$ + m$$\left( {{u \over 2}\widehat i + {u \over 2}\widehat j} \right)$$ = 2m$$\overrightarrow v $$
Compare both side
$$\overrightarrow v = $$ $${{3u} \over 4}\widehat i + {u \over 4}\widehat j$$
$$ \Rightarrow $$ $${\left| {\overrightarrow v } \right|^2}$$ = $${{10{u^2}} \over {16}}$$
$$\Delta $$KE = KEf – KEi
= $${1 \over 2}2m \times {{10{u^2}} \over {16}}$$ - $${1 \over 2}m{u^2}$$ - $${1 \over 2}m{\left( {{u \over {\sqrt 2 }}} \right)^2}$$
= $$ - {1 \over 8}m{u^2}$$
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