Question

Particle $A$ makes a perfectly elastic collision with another particle $B$ at rest. After collision, they fly apart in opposite directions with equal speeds. The ratio of masses $\frac{m_A}{m_B}$ is

• A. $1:2$
• B. $1:3$
• C. $1:4$
• D. $1:\sqrt{3}$

Answer 1

The correct option is B $1:3$

Assume $u_1=$ initial speed of particle $A$
$v_1=$ speed of particle $A$ & $B$ after collision.
There is no external force, hence from linear momentum conservation :
$m_1{u}_1+0=m_1(-v_1)+m_2\left(v_1\right)$
$m_1{u}_1=(-m_1+m_2)v_1$ ... $\left(1\right)$

As we know, collision is elastic.


$\Rightarrow e=1=\frac{v_1-(-v_1)}{u_1-0}=1$
or $2v_1=u_1$ ... (2)
From equation $\left(1\right)$ and $\left(2\right)$,
$m_1\left(2v_1\right)=(-m_1+m_2)v_1$
$2m_1=-m_1+m_2$
or $\frac{m_1}{m_2}=\frac{m_A}{m_B}=\frac{1}{3}$