Question Number 140603 by BHOOPENDRA last updated on 10/May/21
Commented by BHOOPENDRA last updated on 10/May/21
$${mr}.{W}\:{sir}? \\ $$
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 (1/2)m_1 v_1 ^2 +(1/2)m_2 v_2 ^2 =(1/2)m_1 u^2 sin θ_2 =((m_1 v_1 sin θ_1 )/(m_2 v_2 )) ...(i) cos θ_2 =((m_1 (u−v_1 cos θ_1 ))/(m_2 v_2 )) ...(ii) tan θ_2 =((v_1 sin θ_1 )/(u−v_1 cos θ_1 )) ...(iii) ((m_1 ^2 v_1 ^2 sin^2 θ_1 )/(m_2 ^2 v_2 ^2 ))+((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 −2m_1 ^2 uv_1 cos θ_1 =m_2 ^2 v_2 ^2 m_1 ^2 v_1 ^2 +m_1 ^2 u^2 −2m_1 ^2 uv_1 cos θ_1 =m_2 m_1 (u^2 −v_1 ^2 ) v_1 ^2 +u^2 −2uv_1 cos θ_1 =(u^2 −v_1 ^2 )(m_2 /m_1 ) cos θ_1 =(1/2)[(v_1 /u)(1+(m_2 /m_1 ))+(u/v_1 )(1−(m_2 /m_1 ))] let k=(m_2 /m_1 )<1, λ=(v_1 /u) cos θ_1 =(1/2)[λ(1+k)+(1/λ)(1−k)] ...(iv) ≥(√(1−k^2 )) i.e. θ_1 ≤cos^(−1) (√(1−k^2 ))=cos^(−1) (√(1−((m_2 /m_1 ))^2 )) or θ_(1,max) =cos^(−1) (√(1−((m_2 /m_1 ))^2 ))](https://www.tinkutara.com/question/Q140611.png)
$$\left(\mathrm{1}\right) \\ $$$${m}_{\mathrm{1}} {v}_{\mathrm{1}} \mathrm{sin}\:\theta_{\mathrm{1}} ={m}_{\mathrm{2}} {v}_{\mathrm{2}} \mathrm{sin}\:\theta_{\mathrm{2}} \\ $$$${m}_{\mathrm{1}} {v}_{\mathrm{1}} \mathrm{cos}\:\theta_{\mathrm{1}} +{m}_{\mathrm{2}} {v}_{\mathrm{2}} \mathrm{cos}\:\theta_{\mathrm{2}} ={m}_{\mathrm{1}} {u} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{m}_{\mathrm{1}} {v}_{\mathrm{1}} ^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{m}_{\mathrm{2}} {v}_{\mathrm{2}} ^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{m}_{\mathrm{1}} {u}^{\mathrm{2}} \\ $$$$\mathrm{sin}\:\theta_{\mathrm{2}} =\frac{{m}_{\mathrm{1}} {v}_{\mathrm{1}} \mathrm{sin}\:\theta_{\mathrm{1}} }{{m}_{\mathrm{2}} {v}_{\mathrm{2}} }\:\:\:…\left({i}\right) \\ $$$$\mathrm{cos}\:\theta_{\mathrm{2}} =\frac{{m}_{\mathrm{1}} \left({u}−{v}_{\mathrm{1}} \mathrm{cos}\:\theta_{\mathrm{1}} \right)}{{m}_{\mathrm{2}} {v}_{\mathrm{2}} }\:\:\:…\left({ii}\right) \\ $$$$\mathrm{tan}\:\theta_{\mathrm{2}} =\frac{{v}_{\mathrm{1}} \mathrm{sin}\:\theta_{\mathrm{1}} }{{u}−{v}_{\mathrm{1}} \mathrm{cos}\:\theta_{\mathrm{1}} }\:\:\:…\left({iii}\right) \\ $$$$\frac{{m}_{\mathrm{1}} ^{\mathrm{2}} {v}_{\mathrm{1}} ^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta_{\mathrm{1}} }{{m}_{\mathrm{2}} ^{\mathrm{2}} {v}_{\mathrm{2}} ^{\mathrm{2}} }+\frac{{m}_{\mathrm{1}} ^{\mathrm{2}} \left({u}−{v}_{\mathrm{1}} \mathrm{cos}\:\theta_{\mathrm{1}} \right)^{\mathrm{2}} }{{m}_{\mathrm{2}} ^{\mathrm{2}} {v}_{\mathrm{2}} ^{\mathrm{2}} }=\mathrm{1} \\ $$$${m}_{\mathrm{1}} ^{\mathrm{2}} {v}_{\mathrm{1}} ^{\mathrm{2}} +{m}_{\mathrm{1}} ^{\mathrm{2}} {u}^{\mathrm{2}} −\mathrm{2}{m}_{\mathrm{1}} ^{\mathrm{2}} {uv}_{\mathrm{1}} \mathrm{cos}\:\theta_{\mathrm{1}} ={m}_{\mathrm{2}} ^{\mathrm{2}} {v}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$${m}_{\mathrm{1}} ^{\mathrm{2}} {v}_{\mathrm{1}} ^{\mathrm{2}} +{m}_{\mathrm{1}} ^{\mathrm{2}} {u}^{\mathrm{2}} −\mathrm{2}{m}_{\mathrm{1}} ^{\mathrm{2}} {uv}_{\mathrm{1}} \mathrm{cos}\:\theta_{\mathrm{1}} ={m}_{\mathrm{2}} {m}_{\mathrm{1}} \left({u}^{\mathrm{2}} −{v}_{\mathrm{1}} ^{\mathrm{2}} \right) \\ $$$${v}_{\mathrm{1}} ^{\mathrm{2}} +{u}^{\mathrm{2}} −\mathrm{2}{uv}_{\mathrm{1}} \mathrm{cos}\:\theta_{\mathrm{1}} =\left({u}^{\mathrm{2}} −{v}_{\mathrm{1}} ^{\mathrm{2}} \right)\frac{{m}_{\mathrm{2}} }{{m}_{\mathrm{1}} } \\ $$$$\mathrm{cos}\:\theta_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{v}_{\mathrm{1}} }{{u}}\left(\mathrm{1}+\frac{{m}_{\mathrm{2}} }{{m}_{\mathrm{1}} }\right)+\frac{{u}}{{v}_{\mathrm{1}} }\left(\mathrm{1}−\frac{{m}_{\mathrm{2}} }{{m}_{\mathrm{1}} }\right)\right] \\ $$$${let}\:{k}=\frac{{m}_{\mathrm{2}} }{{m}_{\mathrm{1}} }<\mathrm{1},\:\lambda=\frac{{v}_{\mathrm{1}} }{{u}} \\ $$$$\mathrm{cos}\:\theta_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\lambda\left(\mathrm{1}+{k}\right)+\frac{\mathrm{1}}{\lambda}\left(\mathrm{1}−{k}\right)\right]\:\:\:…\left({iv}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\geqslant\sqrt{\mathrm{1}−{k}^{\mathrm{2}} } \\ $$$${i}.{e}.\:\theta_{\mathrm{1}} \leqslant\mathrm{cos}^{−\mathrm{1}} \sqrt{\mathrm{1}−{k}^{\mathrm{2}} }=\mathrm{cos}^{−\mathrm{1}} \sqrt{\mathrm{1}−\left(\frac{{m}_{\mathrm{2}} }{{m}_{\mathrm{1}} }\right)^{\mathrm{2}} } \\ $$$${or}\:\theta_{\mathrm{1},{max}} =\mathrm{cos}^{−\mathrm{1}} \sqrt{\mathrm{1}−\left(\frac{{m}_{\mathrm{2}} }{{m}_{\mathrm{1}} }\right)^{\mathrm{2}} } \\ $$
Commented by mr W last updated on 10/May/21
![(2) tan θ_2 =((v_1 sin θ_1 )/(u−v_1 cos θ_1 )) θ_1 =90° tan θ_2 =(v_1 /u)=λ cos θ_1 =(1/2)[λ(1+k)+(1/λ)(1−k)]=0 ⇒λ(k+1)=(1/λ)(k−1) ⇒λ^2 =((k−1)/(k+1)) ⇒tan θ_2 =λ=(√((k−1)/(k+1))) ⇒θ_2 =tan^(−1) (√((k−1)/(k+1)))=tan^(−1) (√((m_2 −m_1 )/(m_2 +m_1 )))](https://www.tinkutara.com/question/Q140612.png)
$$\left(\mathrm{2}\right) \\ $$$$\mathrm{tan}\:\theta_{\mathrm{2}} =\frac{{v}_{\mathrm{1}} \mathrm{sin}\:\theta_{\mathrm{1}} }{{u}−{v}_{\mathrm{1}} \mathrm{cos}\:\theta_{\mathrm{1}} } \\ $$$$\theta_{\mathrm{1}} =\mathrm{90}° \\ $$$$\mathrm{tan}\:\theta_{\mathrm{2}} =\frac{{v}_{\mathrm{1}} }{{u}}=\lambda \\ $$$$\mathrm{cos}\:\theta_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\lambda\left(\mathrm{1}+{k}\right)+\frac{\mathrm{1}}{\lambda}\left(\mathrm{1}−{k}\right)\right]=\mathrm{0} \\ $$$$\Rightarrow\lambda\left({k}+\mathrm{1}\right)=\frac{\mathrm{1}}{\lambda}\left({k}−\mathrm{1}\right) \\ $$$$\Rightarrow\lambda^{\mathrm{2}} =\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}} \\ $$$$\Rightarrow\mathrm{tan}\:\theta_{\mathrm{2}} =\lambda=\sqrt{\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}} \\ $$$$\Rightarrow\theta_{\mathrm{2}} =\mathrm{tan}^{−\mathrm{1}} \sqrt{\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}}=\mathrm{tan}^{−\mathrm{1}} \sqrt{\frac{{m}_{\mathrm{2}} −{m}_{\mathrm{1}} }{{m}_{\mathrm{2}} +{m}_{\mathrm{1}} }} \\ $$
Commented by mr W last updated on 10/May/21
Commented by BHOOPENDRA last updated on 10/May/21
$${thanks}\:{sir} \\ $$