Work Energy and Power - Result Question 45
48. A particle of mass $5 m$ at rest suddenly breaks on its own into three fragments. Two fragments of mass $m$ each move along mutually perpendicular direction with speed $v$ each. The energy released during the process is,
[NEET Odisha 2019]
(a) $\frac{4}{3} m v^{2}$
(b) $\frac{3}{5} m v^{2}$
(c) $\frac{5}{3} m v^{2}$
(d) $\frac{3}{2} m v^{2}$
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Answer:
Correct Answer: 48. (a)
Solution:
- (a) From conservation of linear momentum.
$0=m v \hat{j}+m v \hat{i}+3 m \vec{v} _1$
$ \vec{v} _1=-\frac{v}{3}(\hat{i}+\hat{j})$
$v_1=\frac{\sqrt{2}}{3} v$
$K_f=\frac{1}{2} m v^{2}+\frac{1}{2} m v^{2}+\frac{1}{2}(3 m)(\frac{\sqrt{2}}{3})^{2} v^{2}$
$=m v^{2}+\frac{m v^{2}}{3}=\frac{4}{3} m v^{2}$
$\Delta KE=K_f-K_i=\frac{4}{3} m v^{2}$
