MCQ
A ball is projected at $60^o$ from horizontal at $200\, m/s$. At maximum height during its  flight it explodes into $3$ equal fragments. Out of them one part travel at $100\, m/s$  vertically up while other at $100\, m/s$ vertically down, then third part will have  speed just after explosion :-
  • A
    $100\, m/s$ horizontal
  • $300\, m/s$ horizontal
  • C
    $300\, m/s\, 60^o$ to horizontal
  • D
    $200\, m/s\, 60^o$ to horizontal

Answer

Correct option: B.
$300\, m/s$ horizontal
b
Total momentum just before explosion

$=$ Total momentum just after explosion

$\mathrm{m} \times\left(200 \cos 60^{\circ}\right) \hat{\mathrm{i}}=\frac{\mathrm{m}}{3} \times 100 \hat{\mathrm{j}}+\frac{\mathrm{m}}{3} \times 100(-\hat{\mathrm{j}})+\frac{\mathrm{m}}{3} \times \overrightarrow{\mathrm{V}}$

$\mathrm{m} \times\left(200 \cos 60^{\circ}\right) \hat{\mathrm{i}}=\frac{\mathrm{m}}{3} \times \vec{\mathrm{V}}$

$\overrightarrow{\mathrm{V}}=3\left(200 \cos 60^{\circ}\right) \hat{\mathrm{i}}$

$\overrightarrow{\mathrm{V}}=300 \mathrm{m} / \mathrm{s} \hat{\mathrm{i}}$

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