Solve the following :


A body of mass $m$ makes an elastic collision with another identical body at rest. Show that if the collision is not head-on, the bodies go right angle to each other after the collision.


If $m$ is not colliding head-on, along $x$-axis, use C.O.L.M $m v \cos \theta=m\left(v_{1} \cos \alpha+v_{2} \cos \beta\right)-(1)$

Along y-axis, $-m v \sin \theta=m\left(v_{1} \sin \alpha-v_{2} \sin \beta\right)-(2)$

Use C.O.E.L

$\frac{1}{2} m v^{2}=\frac{1}{2} m\left(v_{1}^{2}+v_{2}^{2}\right)$

Squaring (1) and (2) and add them,

$v^{2}=v_{1}^{2}\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)+v_{2}^{2}\left(\cos ^{2} \beta+\sin ^{2} \beta\right)+2 v_{1} v_{2}(\cos \alpha \cos \beta-\sin \alpha \sin \beta)$

$v^{2}=v_{1}^{2}+v_{2}^{2}+2 v_{1} v_{2} \cos (\alpha+\beta)$

$\cos (\alpha+\beta)=0=\pi / 2$

$\therefore \alpha+\beta=\pi / 2$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now