Question

After perfectly inelastic collision between two identical particles moving with same speed in different directions the speed of the particles becomes half the initial speed. Find the angle between the two before collision.

• A. $120^{\circ }$
• B. $140^{\circ }$
• C. $90^{\circ }$
• D. $100^{\circ }$

Answer 1

Answer: (A)

Let $\theta$ be the desired angle. Linear momentum of the system will remain conserved. Hence,
$P^2=P_1^2+P_2^2+2P_1{P}_{2}cos\theta$

or ${\left(\left\{2m\left(\frac{v}{2}\right)\right\}\right)}^2={\left(mv\right)}^2+{\left(mv\right)}^2+2\left(mv\right)\left(mv\right)cos\theta$
or $1=1+1+2cos\theta$ or $cos\theta =-\frac{1}{2}$
$\therefore \theta =120^{\circ }$