MCQ
A bullet fired into a fixed target loses half of its velocity after penetrating $3\,cm$. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion?........$cm$
- A$1.5$
- ✓$1$
- C$3$
- D$2$
✓
Answer
Correct option: B.
$1$
b
(b) Let initial velocity of the bullet $= u$
(b) Let initial velocity of the bullet $= u$
After penetrating $3\, cm$ its velocity becomes $\frac{u}{2}$
From ${v^2} = {u^2} - 2as$
${\left( {\frac{u}{2}} \right)^2} = {u^2} - 2a\,(3)$
$⇒$ $6a = \frac{{3{u^2}}}{4}$ $⇒$ $a = \frac{{{u^2}}}{8}$
Let further it will penetrate through distance $x$ and stops at point $C$.
For distance $BC$, $v = 0,\,u = u/2,\,s = x,\,a = {u^2}/8$
From ${v^2} = {u^2} - 2as$ $⇒$ $0 = {\left( {\frac{u}{2}} \right)^2} - 2\left( {\frac{{{u^2}}}{8}} \right)\,.\,x$ $⇒$ $x = 1\,cm$.
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