JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Perfectly Elastic Oblique Collision

Let two bodies moving as shown in figure.

By law of conservation of momentum

Along x-axis, \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}\cos \theta +{{m}_{2}}{{v}_{2}}\cos \varphi \]   ...(i)

Along y-axis, \[0={{m}_{1}}{{v}_{1}}\sin \theta -{{m}_{2}}{{v}_{2}}\sin \varphi \]          ...(ii)

By law of conservation of kinetic energy

\[\frac{1}{2}{{m}_{1}}u_{1}^{2}+\frac{1}{2}{{m}_{2}}u_{2}^{2}=\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2}\]                                        ...(iii)

In case of oblique collision it becomes difficult to solve problem unless some experimental data is provided, as in these situations more unknown variables are involved than equations formed.

Special condition : If \[{{m}_{1}}={{m}_{2}}\] and \[{{u}_{2}}=0\] substituting these values in equation (i), (ii) and (iii) we get

\[{{u}_{1}}={{v}_{1}}\cos \theta +{{v}_{2}}\cos \varphi \]                                                   ...(iv)

\[0={{v}_{1}}\sin \theta -{{v}_{2}}\sin \varphi \]           ...(v)

and \[u_{1}^{2}=v_{1}^{2}+v_{2}^{2}\]            ...(vi)

Squaring (iv) and (v) and adding we get

\[u_{1}^{2}=v_{1}^{2}+v_{2}^{2}+2{{v}_{1}}{{v}_{2}}\cos (\theta +\varphi )\]           ...(vii)

Using (vi) and (vii) we get \[\cos (\theta +\varphi )=0\]

\[\therefore \] \[\theta +\varphi =\pi /2\]

i.e. after perfectly elastic oblique collision of two bodies of equal masses (if the second body is at rest), the scattering angle \[\theta +\varphi \] would be \[{{90}^{o}}\].