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5.work,energy,power and collision — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysics5.work,energy,power and collision1 Mark
MCQ
A projectile moving vertically upwards with a velocity of $200\, ms^{-1}$ breaks into two equal parts at a height of $490\, m$. One part starts moving vertically upwards with a velocity of $400\, ms^{-1}$. How much time it will take, after the break up with the other part to hit the ground? .............. $\mathrm{s}$
  • A
    $2\sqrt {10}$
  • B
    $5$
  • ✓
    $10$
  • D
    $\sqrt {10}$
✓

Answer

Correct option: C.
$10$
c
$YX$

$v = 400 m/s$

$Mass \,before\, explosion = m$

$490\, m\, and\, velocity \,v = 200\, m/s\, (vertically)$

Momentum before explosion

 $=$Momentum after explosion

$\begin{array}{l}
m \times 200\hat j\, = \frac{m}{2} \times 400\hat j + \frac{m}{2}\,v\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{m}{2}\left( {400\,\hat j + v} \right)
\end{array}$

$\begin{array}{l}
 \Rightarrow \,400\,\hat j - 400\hat j = v\\
\therefore \,\,v = 0
\end{array}$

i.e., the velocity of the other part of the mass, $v = 0$

Let time taken to reach the earth by this part be $t$

Applying formula, $h = ut +$ $\frac{1}{2}g{t^2}$

$\begin{array}{l}
490 = 0 + \frac{1}{2} \times 9.8 \times {t^2}\\
 \Rightarrow \,{t^2} = \frac{{980}}{{9.8}} = 100\\
\therefore \,\,t = \sqrt {100}  = 10\sec 
\end{array}$

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