SATHEE: Work Energy and Power - Result Question 42

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) $\frac{8}{9}$

(d) $\frac{5}{9}$

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Answer:

Correct Answer: 45. (b)

Solution:

(b) $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$

$\begin{aligned} Also, \frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} \\ = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2} \end{aligned}$

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, $v_{2} = \frac{4}{3} u$

$\frac{Δ K E}{K E} = \frac{1}{2} \times 2 m \times (\frac{4}{3} u)^{2}$

$= \frac{1}{2} \times 2 \times \frac{16}{9}$

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Work Energy and Power - Result Question 42

45. Body A of mass 4m moving with speed u collides with another body B of mass 2m, 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]

Answer:

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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]