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Three particles of masses $10g, 20g$ and $40g$ are moving with velocities $10\widehat i,10\widehat j$ and $10\widehat k$ $m/s$ respectively. If due to some mutual interaction, the first particle comes to rest and the velocity of second particle becomes $\left( {3\widehat i + 4\widehat j\,\,} \right)\, m/s$, then the velocity of third particle is
$\widehat i + \widehat j + 5\widehat k\,\,$
$\widehat j + 10\widehat k\,\,$
$\widehat i + \widehat j + 10\widehat k\,\,$
$\widehat i + 3\widehat j + 10\widehat k\,\,$
Solution
From conservation of momentum,
$\mathrm{m}_{1} \overrightarrow{\mathrm{u}}_{1}+\mathrm{m}_{2} \overrightarrow{\mathrm{u}}_{2}+\mathrm{m}_{3} \overrightarrow{\mathrm{u}}_{3}=\mathrm{m}_{1} \overrightarrow{\mathrm{v}}_{1}+\mathrm{m}_{2} \overrightarrow{\mathrm{v}}_{2}+\mathrm{m}_{3} \overrightarrow{\mathrm{v}}_{3}$
$10(10 \hat{\mathrm{i}})+20(10 \hat{\mathrm{j}})+40(10 \hat{\mathrm{k}})$
$=10(0)+20(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}})+40 \overrightarrow{\mathrm{v}}_{3}$
$\Rightarrow 40 \overrightarrow{\mathrm{V}}_{3}=40 \mathrm{i}+120 \mathrm{j}+400 \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{v}}_{3}=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}$
