Question

Three objects $A$, $B$ and $C$ are kept in a straight line on a frictionless horizontal surface. These have masses $m$, $2m$ and $m$ respectively. The object $A$ moves towards $B$ with a speed of $9\text{m/s}$ and makes an elastic collision with it. Thereafter, $B$ makes completely inelastic collision with $C$. All motions occur on the same straight line. Find the final speed (in m/s) of object $C$.

Answer 1

Velocity of COM for block A and B will be,
$V_{co{m}_{AB}}=\frac{m\left(V_A\right)+2m\left(V_B\right)}{m+2m}=\frac{m\times 9}{3m}=3m/s$
Hence block A will be appearing moving right at speed of (9-3=6 m/s) and block B moving left at speed of 3 m/s w.r.t COM of A and B before collision.
Since the collision is elastic, block A and B will be moving in opposite direction with same initial velocity.
Hence velocity of block B$=V_B+V_{co{m}_{AB}}=3+3=6m/s$
The collision between B and C is inelastic and since no external force acts on the system, net momentum will be conserved. Let after collision both B and C start moving at speed $V$
Therefore, Initial momentum = final momentum
$m_B\left(u_B\right)+m_C\left(u_C\right)=m_B\left(V\right)+m_C\left(V\right)$
$2m\times 6+0=2mV+mV$
$\Rightarrow V=4m/s$