Question

After a totally inelastic collision, two objects of the same masses and same initial speeds are found to move together at half of their initial speeds as shown in figure below. If $\alpha ={30}^{\circ }$, then the angle between the initial velocities of the objects is

• A. ${90}^{\circ }$
• B. ${60}^{\circ }$
• C. ${150}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is B ${60}^{\circ }$

Let the angle between initial velocities be $180-\theta$ and the situation is as shown in figure.


Momentum conservation along $x-$ axis,
$mu-mu\cos \theta =\frac{2mu\cos \alpha }{2}...\left(i\right)$
Momentum conservation along $y-$ axis,
$mu\sin \theta =\frac{2mu\sin \alpha }{2}...\left(ii\right)$

Dividing equation $\left(ii\right)$ by eqaution $\left(i\right)$
$\frac{\sin \theta }{1-\cos \theta }=\tan \alpha$

$\Rightarrow \frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{\sin }^2\frac{\theta }{2}}=\tan \alpha$
$\Rightarrow \cot \frac{\theta }{2}=\tan \alpha$
$\Rightarrow \tan (90-\frac{\theta }{2})=\tan \alpha$
$\Rightarrow \theta =2\times (90-\alpha )=2\times {60}^{\circ }={120}^{\circ }$

Therefore, angle between their velocities
$={180}^{\circ }-\theta ={180}^{\circ }-{120}^{\circ }={60}^{\circ }$