Question
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 at right angle to each other after the collision.
✓
Answer
Since it is not an head on collision, the two bodies move in different dimensions. Let $V_1, V_2$ → velocities of the bodies vector collision. Since, the collision is elastic. Applying law of conservation of momentum on X-direction. 
$\text{mu}_1+\text{mx}_0=\text{mv}_1\cos\alpha+\text{mv}_2\cos\beta$
$\Rightarrow\text{v}_1\cos\text{a}+\text{v}_2\cos\text{b}=\text{u}_1\ \dots(1)$
Putting law of conservation of momentum in y direction.$0=\text{mv}_1\sin\alpha-\text{mv}_2\sin\beta$
$\Rightarrow\text{v}_1\sin\alpha=\text{v}_2\sin\beta\ \dots(2)$
Again $\frac{1}2{}\text{mu}_1^2+0=\frac{1}{2}\text{mv}_1^2+\frac{1}{2}\text{mv}_2^2$$\Rightarrow\text{u}_1^2=\text{v}_1^2+\text{v}_2^2\ \dots(3)$
Squaring equation (1)$\text{u}_1^2=\text{v}_1^2\cos^2\alpha+\text{v}_2^2\cos^2\beta+2\text{v}_1\text{v}_2\cos\alpha\cos\beta$
Equating (1) & (3)$\text{v}_1^2+\text{v}_2^2=\text{v}_1^2\cos^2\alpha+\text{v}_2^2\cos^2\beta+2\text{v}_1\text{v}_2\cos\alpha\cos\beta$
$\Rightarrow\text{v}_1^2\sin^2\alpha+\text{v}_2^2\sin^2\beta=2\text{v}_1\text{v}_2\cos\alpha\cos\beta$
$\Rightarrow\text{2v}_1^2\sin^2\alpha=2\times\text{v}_1\times\frac{\text{v}_1\sin\alpha}{\sin\beta}\times\cos\alpha\cos\beta$
$\Rightarrow\sin\alpha\sin\beta=\cos\alpha\cos\beta$
$\Rightarrow\cos\alpha\cos\beta-\sin\alpha\sin\beta=0$
$\Rightarrow\cos(\alpha+\beta)=0=\cos90^\circ$
$\Rightarrow(\alpha+\beta)=90 ^\circ$
$\text{mu}_1+\text{mx}_0=\text{mv}_1\cos\alpha+\text{mv}_2\cos\beta$
$\Rightarrow\text{v}_1\cos\text{a}+\text{v}_2\cos\text{b}=\text{u}_1\ \dots(1)$
Putting law of conservation of momentum in y direction.$0=\text{mv}_1\sin\alpha-\text{mv}_2\sin\beta$
$\Rightarrow\text{v}_1\sin\alpha=\text{v}_2\sin\beta\ \dots(2)$
Again $\frac{1}2{}\text{mu}_1^2+0=\frac{1}{2}\text{mv}_1^2+\frac{1}{2}\text{mv}_2^2$$\Rightarrow\text{u}_1^2=\text{v}_1^2+\text{v}_2^2\ \dots(3)$
Squaring equation (1)$\text{u}_1^2=\text{v}_1^2\cos^2\alpha+\text{v}_2^2\cos^2\beta+2\text{v}_1\text{v}_2\cos\alpha\cos\beta$
Equating (1) & (3)$\text{v}_1^2+\text{v}_2^2=\text{v}_1^2\cos^2\alpha+\text{v}_2^2\cos^2\beta+2\text{v}_1\text{v}_2\cos\alpha\cos\beta$
$\Rightarrow\text{v}_1^2\sin^2\alpha+\text{v}_2^2\sin^2\beta=2\text{v}_1\text{v}_2\cos\alpha\cos\beta$
$\Rightarrow\text{2v}_1^2\sin^2\alpha=2\times\text{v}_1\times\frac{\text{v}_1\sin\alpha}{\sin\beta}\times\cos\alpha\cos\beta$
$\Rightarrow\sin\alpha\sin\beta=\cos\alpha\cos\beta$
$\Rightarrow\cos\alpha\cos\beta-\sin\alpha\sin\beta=0$
$\Rightarrow\cos(\alpha+\beta)=0=\cos90^\circ$
$\Rightarrow(\alpha+\beta)=90 ^\circ$
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