Work Energy and Power - Result Question 57
«««< HEAD:content/english/neet-pyq-chapterwise/physics/work-energy-and-power/work-energy-and-power-result-question-57.md
60. 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:
======= ####60. 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:
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/work-energy-and-power/work-energy-and-power—result-question-57.md (a) $100 ms^{-1}$
(b) $80 ms^{-1}$
(c) $40 ms^{-1}$
(d) $120 ms^{-1}$
[2008]
Show Answer
Answer:
Correct Answer: 60. (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. $=1.05 kJ=1.05 \times 10^{3} J$
$\therefore 1.05 \times 10^{3}=0.1 \times 1.05 \times v^{2}$
$\Rightarrow v^{2}=\frac{1.05 \times 10^{3}}{0.1 \times 1.05}=10^{4}$
$\therefore v=10^{2}=100 ms^{-1}$.
