Question

A particle of mass $m_1$ (A) elastically collides with another stationary particle (B) of mass $m_2$. Then if .

• A. $\frac{m_1}{m_2}=\frac{1}{2}$ and the collision is head on, then the particles fly apart in the opposite direction with
  unequal speed.
• B. $\frac{m_1}{m_2}=\frac{1}{3}$ and the collision is head on, then the particles fly apart in the opposite direction with equal
  speed.
• C. $\frac{m_1}{m_2}=\frac{2}{1}$ and the collision is head on, then the particles fly apart in the opposite direction with
  unequal speed.
• D. $\frac{m_1}{m_2}=\frac{3}{1}$ and the collision is head on, then the particles fly apart in the opposite direction with equal
  speed.

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{m_1}{m_2}=\frac{1}{2}$ and the collision is head on, then the particles fly apart in the opposite direction with
unequal speed.
B $\frac{m_1}{m_2}=\frac{1}{3}$ and the collision is head on, then the particles fly apart in the opposite direction with equal

speed.
From above concept as the initial velocity of particle A( particle 1) is non zero and velocity of particle B( particle 2) is zero.

$v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right)u$ and $v_2=\left(\frac{2m_1}{m_1+m_2}\right)u$

Checking Option A:

$\frac{m_1}{m_2}=\frac{1}{2}\Rightarrow v_1=\frac{-u}{2}$ and $v_2=\frac{u}{2}$ particle flies apart in opposite directions.

Checking Option B:

$\frac{m_1}{m_2}=\frac{1}{3}\Rightarrow v_1=\frac{-u}{3}$ and $v_2=\frac{2u}{3}$ particle flies apart in opposite directions with equal speed.

Checking Option C:

$\frac{m_1}{m_2}=\frac{2}{1}\Rightarrow v_1=\frac{u}{3}$ and $v_2=\frac{2u}{3}$ particle travels in opposite directions.

Checking Option D:

$\frac{m_1}{m_2}=\frac{3}{1}\Rightarrow v_1=\frac{u}{2}$ and $v_2=\frac{3u}{2}$ particle travels in opposite directions.