Question

Consider elastic collision of a particle of mass $m$ moving with a velocity $u$ with another particle of the same mass at rest. After the collision the projectile and the stuck particle move in directions making angles ${\theta }_1$ and ${\theta }_2$ respectively with the initial direction of motion. The sum of the angles ${\theta }_1+{\theta }_2$ is

• A. ${45}^{\circ }$
• B. ${90}^{\circ }$
• C. ${135}^{\circ }$
• D. ${180}^{\circ }$

Answer 1

The correct option is B ${90}^{\circ }$

Given $m_1=m_2$
We will apply the principle of conservation of momentum in the mutually perpendicular $di{r}^n$.
Along x-axis, $m_1{u}_1=m_1{v}_1\cos \theta +m_2{v}_2\cos \varphi$
or $u_1=v_1\cos \theta +v_2\cos \varphi ....\left(i\right)$
Along y-axis, $0=m_1{v}_1\sin \theta -m_2{v}_2\sin \varphi$
or $0=v_1\sin \theta -v_2\sin \varphi ....\left(ii\right)$
Again for elastic collision, kinetic energy is conserved
$\Rightarrow \frac{1}{2}m{u}_1^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2$
or $u_1^2=v_1^2+v_2^2.....\left(iii\right)$
Squaring and adding (i) and (ii), we get
$u_1^2=v_1^2({\cos }^2\theta +{\sin }^2\theta )+v_2^2({\cos }^2\varphi +{\sin }^2\varphi )+2v_1{v}_2\cos \theta \cos \varphi -2v_1{v}_2\sin \theta \sin \varphi$
or $u_1^2=v_1^2+v_2^2+2v_1{v}_2\cos (\theta +\varphi ....(iv)$
Using (iii) and (iv), we get
$\cos (\theta +\varphi )=0=\cos \frac{\pi }{2}\Rightarrow \theta +\varphi =\frac{\pi }{2}$
Note: This is a standard case of oblique collision.