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Question Number 140603 by BHOOPENDRA last updated on 10/May/21

Commented by BHOOPENDRA last updated on 10/May/21

$m r . W s i r ?$
	Answered by mr W last updated on 10/May/21

	$(1)$ $m_{1} v_{1} \sin θ_{1} = m_{2} v_{2} \sin θ_{2}$ $m_{1} v_{1} \cos θ_{1} + m_{2} v_{2} \cos θ_{2} = m_{1} u$ $\frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2} = \frac{1}{2} m_{1} u^{2}$ $\sin θ_{2} = \frac{m_{1} v_{1} \sin θ_{1}}{m_{2} v_{2}} . . . (i)$ $\cos θ_{2} = \frac{m_{1} (u - v_{1} \cos θ_{1})}{m_{2} v_{2}} . . . (i i)$ $\tan θ_{2} = \frac{v_{1} \sin θ_{1}}{u - v_{1} \cos θ_{1}} . . . (i i i)$ $\frac{m_{1}^{2} v_{1}^{2} \sin^{2} θ_{1}}{m_{2}^{2} v_{2}^{2}} + \frac{m_{1}^{2} {(u - v_{1} \cos θ_{1})}^{2}}{m_{2}^{2} v_{2}^{2}} = 1$ $m_{1}^{2} v_{1}^{2} + m_{1}^{2} u^{2} - 2 m_{1}^{2} {u v}_{1} \cos θ_{1} = m_{2}^{2} v_{2}^{2}$ $m_{1}^{2} v_{1}^{2} + m_{1}^{2} u^{2} - 2 m_{1}^{2} {u v}_{1} \cos θ_{1} = m_{2} m_{1} (u^{2} - v_{1}^{2})$ $v_{1}^{2} + u^{2} - 2 {u v}_{1} \cos θ_{1} = (u^{2} - v_{1}^{2}) \frac{m_{2}}{m_{1}}$ $\cos θ_{1} = \frac{1}{2} [\frac{v_{1}}{u} (1 + \frac{m_{2}}{m_{1}}) + \frac{u}{v_{1}} (1 - \frac{m_{2}}{m_{1}})]$ $l e t k = \frac{m_{2}}{m_{1}} < 1, λ = \frac{v_{1}}{u}$ $\cos θ_{1} = \frac{1}{2} [λ (1 + k) + \frac{1}{λ} (1 - k)] . . . (i v)$ $⩾ \sqrt{1 - k^{2}}$ $i . e . θ_{1} ⩽ \cos^{- 1} \sqrt{1 - k^{2}} = \cos^{- 1} \sqrt{1 - {(\frac{m_{2}}{m_{1}})}^{2}}$ $o r θ_{1, m a x} = \cos^{- 1} \sqrt{1 - {(\frac{m_{2}}{m_{1}})}^{2}}$
	Commented by mr W last updated on 10/May/21

	$(2)$ $\tan θ_{2} = \frac{v_{1} \sin θ_{1}}{u - v_{1} \cos θ_{1}}$ $θ_{1} = 90 °$ $\tan θ_{2} = \frac{v_{1}}{u} = λ$ $\cos θ_{1} = \frac{1}{2} [λ (1 + k) + \frac{1}{λ} (1 - k)] = 0$ $\Rightarrow λ (k + 1) = \frac{1}{λ} (k - 1)$ $\Rightarrow λ^{2} = \frac{k - 1}{k + 1}$ $\Rightarrow \tan θ_{2} = λ = \sqrt{\frac{k - 1}{k + 1}}$ $\Rightarrow θ_{2} = \tan^{- 1} \sqrt{\frac{k - 1}{k + 1}} = \tan^{- 1} \sqrt{\frac{m_{2} - m_{1}}{m_{2} + m_{1}}}$
	Commented by mr W last updated on 10/May/21

	Commented by BHOOPENDRA last updated on 10/May/21

	$t h a n k s s i r$
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