Perfectly Elastic Oblique Collision
Category : JEE Main & Advanced
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 \[{{90}^{o}}\].
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