Work Energy And Power Ques 56
56. A shell of mass $200 $ $gm$ is ejected from a gun of mass $4$ $ kg$ by an explosion that generates $1.05$ $kJ$ of energy. The initial velocity of the shell is:
[2008]
(a) $100 $ $ms^{-1}$
(b) $80 $ $ms^{-1}$
(c) $40 $ $ms^{-1}$
(d) $120 $ $ms^{-1}$
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Answer:
Correct Answer: 56.(a)
Solution:
- (a) Let the initial velocity of the shell be $v$, then by the conservation of momentum $m v=M v^{\prime}$ where $v^{\prime}=$ velocity of gun.
$\therefore \quad v^{\prime}=(\frac{m}{M}) v$
Now, total K.E. $=\frac{1}{2} mv^{2}+\frac{1}{2} Mv^{\prime 2}$
$=\frac{1}{2} m v^{2}+\frac{1}{2} M(\frac{m}{M})^{2} v^{2}$
$=\frac{1}{2} mv^{2}[1+\frac{m}{M}]$
$=(\frac{1}{2} \times 0.2)(1+\frac{0.2}{4}) v^{2}=(0.1 \times 1.05) v^{2}$
But total K.E. $=1.05 kJ=1.05 \times 10^{3} $ $J$
$\therefore \quad 1.05 \times 10^{3}=0.1 \times 1.05 \times v^{2}$
$\Rightarrow \quad v^{2}=\frac{1.05 \times 10^{3}}{0.1 \times 1.05}=10^{4}$
$\therefore \quad v=10^{2}=100 $ $ms^{-1}$.
BETA
