Question

Particle 'A' moves with speed $10m/s$ in a frictionless circular fixed horizontal pipe of radius $5m$ and strikes with 'B' of double mass that of A. Coefficient of restitution is $\frac{1}{2}$ and particle 'A' starts its journey at $t=0$. The time at which second collision occurs is :

• A. $\frac{\pi }{2}$ sec
• B. $\frac{2\pi }{3}$ sec
• C. $\frac{5\pi }{2}$ sec
• D. $4\pi$ sec

Answer 1

Answer: (C)

$\frac{5\pi }{2}$ sec

Initial momentum of both particles before collision $=m\times 10+\left(2m\right)\times 0=m\times 10⟹\left(1\right)$

Let $v_A$ and $v_B$ are the speeds after collision of particle $A$ and $B$ respectively.

Then final momentum after collision$=m\times v_A+\left(2m\right)\times v_B⟹\left(2\right)$

By conservation of momentum :$m\times 10=m\times v_A+\left(2m\right)\times v_{B}or{v}_A+2\times v_B=10⟹\left(3\right)$

speeds before and after collision are related through coefficient of restitution $e$ as,$v_A-v_B=-e\times 10$

$v_A=v_B-10\times 0.5or{v}_A=v_B-5⟹\left(4\right)$

solving (3) and (4), we get, $v_A=0,v_B=5m/s$

Initial time taken by $A$ to hit $B$, $\frac{\pi \times 5}{10}=\frac{\pi }{2}seconds⟹\left(5\right)$

After collision, $A$ comes to rest. Time taken for $B$ to hit $A=\frac{2\pi \times 5}{5}=2\pi seconds⟹\left(6\right)$

total time =$\frac{\pi }{2}+2\pi =2.5\times \pi seconds$