Question

Particle $A$ makes a perfectly elastic collision with another particle $B$ at rest. They fly apart in opposite directions with equal speeds. The ratio of their masses $m_A/m_B$ is

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

Answer 1

Answer: (B)

According to law of conservation of linear momentum, we get
$m_1{u}_1+m_2\times 0=m_1{v}_1+m_2(-v_1)$
$m_1{u}_1=(m_1-m_2)v_1$ .....(i)
$\therefore \frac{u_1}{v_1}=\frac{m_1-m_2}{m_1}$ ....(ii)
According to law of conservation of kinetic energy, we get
$\frac{1}{2}{m}_1{u}_1^2=\frac{1}{2}(m_1+m_2)v_1^2$ ...(iii)
Divide (iii) by (i), we get
$u_1=\frac{\left(m_1+m_2\right)v_1}{m_1-m_2}$ or $\frac{u_1}{v_1}=\frac{m_1+m_2}{m_1-m_2}$ ...(iv)
From (ii) and (iv), we get $\frac{m_1-m_2}{m_1}=\frac{m_1+m_2}{m_1-m_2}$
On solving, we get $\frac{m_1}{m_2}=\frac{1}{3}$ or $\frac{m_A}{m_B}=\frac{1}{3}$