Work Energy and Power - Result Question 42
45. Body A of mass $4 m$ moving with speed $u$ collides with another body B of mass $2 m$, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is : [2019]
(a) $\frac{1}{9}$
(b) $\frac{8}{9}$
(c) $\frac{4}{9}$
(d) $\frac{5}{9}$
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Answer:
Correct Answer: 45. (b)
Solution:
(b) $m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2$
$ \begin{aligned} & \text{ Also, } \frac{1}{2} m_1 u_1^{2}+\frac{1}{2} m_2 u_2^{2} \\ & =\frac{1}{2} m_1 v_1^{2}+\frac{1}{2} m_2 v_2^{2} \end{aligned} $
For elastic collision
$v_2=\frac{2 m_1 u_1}{m_1+m_2}$
as $u_2=0 ; u_1=u$
$m_1=4 m$ and $m_2=2 m$
so, $v_2=\frac{4}{3} u$
$\frac{\Delta K E}{K E}=\frac{1}{2} \times 2 m \times(\frac{4}{3} u)^{2}$
$=\frac{1}{2} \times 2 \times \frac{16}{9}$
