At the same time another body of mass 2 kg is dropped along the same line from a height of 25m. If they collide elastically, find the velocities of body A and B just after collision.

Given:-

$ Mass, m_A=2kg=m_B$

The collision will occur when $h_A=h_B $ (Say at point p)

At point P,

Displacement of A, $S_A=x$ (say)

$\therefore $ Displacement of A, $S_A=-(25-x)=(x-25)$

Using $v^2=u^2+2as,$

For body A,

$v_A^2=20^2-2 × 9.81 x \\ \therefore x=\dfrac {400-v_A^2}{19.62} ---------(i)$

For body B, $v_B^2 = -2 × 9.81 × (x-25) = -2 × 9.81 [\dfrac {400-v_A^2}{19.62-25}] (from (i)) \\ \therefore v_B^2=v_A^2+90.5 ---------(ii) $

Using $v=u+at$

For body A,

$ v_A=20-9.81t \\ \therefore t= \dfrac {20-v_A}{9.81} ----------(iii) $

For body B,

$ v_B=-9.81t \\ \therefore v_B=-9.81(\dfrac {20-v_A}{9.81}) \\ v_B=v_A-20 -------(iv)$

Substituting (iv) in (iii), we get,

$ (v_A-20)^2=v_A^2+90.5 \\ \therefore v_A^2-40v_A+400=v_A^2+90.5 \\ \therefore v_A=7.738 m/s \\ \therefore v_B=12.263 m/s [from (iv)] $

Let $v^1_A$ and $v^1_B$ be the velocities of A and B after collision, respectively.

$\because $ The collision is elastic,

$\therefore v^1_A = v_A (\dfrac {m_A-m_B}{m_A-m_B }) + v_B (\dfrac {2m_B}{m_A-m_B }) \\ =7.738(\dfrac {2-2}{2+2}) + (-12.263)( \dfrac {2×2}{2+2}) \\ \therefore v^1_A =-12.263 m/s \Rightarrow Ans \\ \therefore v^1_B = v_B (\dfrac {m_B-m_A}{m_A-m_B }) + v_A (\dfrac {2m_A}{m_B-m_A}) \\ =(-12.263)(\dfrac {2-2}{2+2})+7.738(\dfrac {2×2}{2+2}) \\ \therefore v^1_A = 7.738 m/s \Rightarrow Ans$

Note:-if you can’t remember the above formulae, you can get the answer by following the steps :-

I]Use the Law of Conservation of Momentum

$m_A v_A+m_B v_B=m_A v^1_A+m_B v^1_B $

(Use negative signs properly)

II]For elastic collisions, kinetic energy is conserved,

$\dfrac 12 m_A v_A^2+ \dfrac 12 m_B v_B^2= \dfrac 12 m_A (v^1_A)^2+ \dfrac 12 m_B (v^1_B)^2$

The above steps will give two equations and two unknowns.