Question

Particle $\text{A}$ moves with speed $10\text{m/s}$ in a frictionless circular fixed horizontal pipe of radius $5\text{m}$ and strikes with $\text{B}$ of double mass that of $\text{A}$. Coefficient of restitution is $\frac{1}{2}$ and particle $\text{A}$ starts its journey at $t=0$. The time at which the second collision occurs is

• A. $\frac{\pi }{2}\text{s}$
• B. $\frac{2\pi }{3}\text{s}$
• C. $\frac{5\pi }{2}\text{s}$
• D. $1\text{s}$

Answer 1

The correct option is C $\frac{5\pi }{2}\text{s}$

For first collision

velocity of particle $\text{A},v_1=10\text{m/s}$. So,

$t_1=\frac{\pi r}{v}=\frac{\pi \left(5\right)}{10}=\frac{\pi }{2}\text{s}$

Let the mass of particle $\text{A}$ be $m$ and $2m$ of particle $\text{B}$.

So, Momentum $P_i$ before first collision will be,

$P_i=m_1{v}_1+m_2{v}_2=(m\times 10)+(2m\times 0)=10m$

And, momentum $P_f$ after collision will be

$P_f=m_1{v}_1+m_2{v}_2=m{v}_1+2m{v}_2$

Apply conservation of momentum,

$P_i=P_f$

$10m=m{v}_1+2m{v}_2$

$\Rightarrow v_1+2v_2=10......\left(1\right)$

Using formula of coefficient of restitution,

Velocity of separation = $e\times$velocity of approach

$v_2-v_1=\frac{1}{2}\times \left(10\right)$

$\Rightarrow v_2-v_1=5......\left(2\right)$

from Eqs.(1) and (2),

$v_1=0;v_2=5\text{m/s}$

Thus velocity of body A becomes zero after first collision and body B travels whole circumference for
2nd collision.
For second collision,

$\therefore t_2=\frac{2\pi \left(5\right)}{5}=2\pi \text{s}$

$\therefore$ total time, $t=t_1+t_2$

$\Rightarrow t=\frac{\pi }{2}+2\pi =2.5\pi \text{s}$

So, option $\left(c\right)$ is correct.