Question

In an elastic head-on collision between two particles,

• A. velocity of separation is equal to the velocity of approach
• B. velocity of the target is always more than the velocity of the projectile
• C. the maximum velocity of the target is double to that of the projectile
• D. maximum transfer of kinetic energy occurs when masses of both projectile and target are equal

Answer 1

The correct option is A velocity of separation is equal to the velocity of approach

We know that, in an elastic collision, both momentum and kinetic energy are conserved
$\therefore m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2\dots \left(i\right)$
$\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2\dots \left(ii\right)$

From (ii)
$m_1{u}_1^2+m_2{u}_2^2=m_1{v}_1^2+m_2{v}_2^2$
$\Rightarrow m_1(u_1^2-v_1^2)=m_2(u_2^2-v_2^2)\dots \left(iii\right)$

From (i)
$m_1(u_1-v_1)=m_2(v_2-u_2)\dots \left(iv\right)$

From (iii) and (iv):
$u_1+v_1=u_2+v_2$
$\Rightarrow u_1-u_2=v_2-v_1$