Question

A point charge $+Q$ having mass $m$ is fixed on horizontal smooth surface. Another point charge having magnitude $+2Q$ and mass $2m$ is projected horizontal towards the charge $+Q$ from far distance with velocity $v_0$.
The impulse acting on the system of particles $(+Q\&+2Q)$ in the time interval when distance between them becomes $d$ is:

• A. $2m[\sqrt{v_0^2-\frac{2K{Q}^2}{md}}-v_0]$
• B. $2m{v}_0$
• C. $2m\left[\sqrt{v_0^2-\frac{2K{Q}^2}{md}}\right]$
• D. None of these

Answer 1

The correct option is A $2m[\sqrt{v_0^2-\frac{2K{Q}^2}{md}}-v_0]$

We know that impulse,$J=\Delta P=P_f-P_i$.......(1)


From initia state:
$P_i=2m{v}_0+m\times 0=2m{v}_0$.......(2)

On applying law of conservation of mechanical energy between initial and final state, we get


$\frac{1}{2}\times 2m\times v_0^2+\frac{1}{2}\times m\times 0^2+\frac{K\left(2Q\right)\left(Q\right)}{\infty }=\frac{1}{2}\times 2m\times v^2+\frac{1}{2}\times m\times 0^2+\frac{K\left(2Q\right)\left(Q\right)}{d}$

$\Rightarrow m{v}_0^2=m{v}^2+\frac{2K{Q}^2}{d}$

$\Rightarrow m{v}^2=m{v}_0^2-\frac{2K{Q}^2}{d}$

$\Rightarrow v=\sqrt{v_0^2-\frac{2K{Q}^2}{md}}$

$\therefore p_f=2mv+m\times 0=2m\left[\sqrt{v_0^2-\frac{2K{Q}^2}{md}}\right]$.......(3)
On putting (2) and (3) in (1), we get,
$\Delta P=2m\left[\sqrt{v_0^2-\frac{2K{Q}^2}{md}}\right]-2m{v}_0$
$\Rightarrow \Delta P=2m[\sqrt{v_0^2-\frac{2K{Q}^2}{md}}-v_0]$