Question

Two particles of masses $m_1$ & $m_2$ and velocities $u_1$ and $\left(\alpha u_1\right)(\alpha \ne 0)$ make an elastic head on collision. If the initial kinetic energies of the two particles are equal and $m_1$ comes to rest after collision, then

• A. $\frac{m_2}{m_1}=9+2\sqrt{2}$
• B. $\frac{m_2}{m_1}=3-2\sqrt{2}$
• C. $\frac{m_1}{m_2}=3-2\sqrt{2}$
• D. $\frac{m_1}{m_2}=9+\sqrt{2}$

Answer 1

The correct option is C $\frac{m_1}{m_2}=3-2\sqrt{2}$

Given,
Two particles of mass $m_1$ & $m_2$ with velocities $u_1$ & $\alpha u_1$ respectively.
If the initial kinetic energies of the two particles are equal, and after the collision, $m_1$ comes to rest.
$\frac{1}{2}{m}_1{u}_1^2=\frac{1}{2}{m}_2(\alpha u_1{)}^2$
$\Rightarrow \frac{m_1}{m_2}={\alpha }^2$ ... (1)

By the conservation of linear momentum,
$m_1{u}_1+m_2\left(\alpha u_1\right)=0+m_2{v}_2$
Here, $v_2=$ velocity of $m_2$ particle after collision
Using eq. (1)
$m_2{\alpha }^2{u}_1+m_2\alpha u_1=m_2{v}_2$
$\Rightarrow v_2=u_1\alpha (\alpha +1)$ ... (2)

Collision is elastic, hence the kinetic energy of the particles before and after the collision will be equal.
$\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2(\alpha u_1{)}^2=\frac{1}{2}{m}_2{v}_2^2$
On putting $v_2$ from equation $\left(2\right)$
$m_1{u}_1^2+m_2{\alpha }^2{u}_1^2=m_2{u}_1^2{\alpha }^2(\alpha +1{)}^2$
$\Rightarrow \frac{m_1}{m_2}={\alpha }^4+2{\alpha }^3$
From equation $\left(1\right)$,
${\alpha }^4+2{\alpha }^3={\alpha }^2$
${\alpha }^2+2\alpha -1=0$
$\Rightarrow \alpha =-1+\sqrt{2}$ (taking +ve root)
$\Rightarrow \frac{m_1}{m_2}={\alpha }^2=(-1+\sqrt{2}{)}^2$
Hence, $\frac{m_1}{m_2}=3-2\sqrt{2}$