Question

Two blocks of masses $5\text{kg}$ and $2\text{kg}$ are placed at rest on a frictionless surface and connected by a spring. An external hit gives a velocity of $21\text{m/s}$ to the heavier block towards the lighter one. Velocities of both blocks (heavier and lighter one) in the centre of mass frame, just after the kick will be respectively:
(Consider direction of motion of heavier block as $+ve$ direction)

• A. $6\text{m/s},10\text{m/s}$
• B. $6\text{m/s},-15\text{m/s}$
• C. $15\text{m/s},15\text{m/s}$
• D. $3\text{m/s},5\text{m/s}$

Answer 1

The correct option is B $6\text{m/s},-15\text{m/s}$


$v_{\text{CM}}=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}$
$v_{\text{CM}}=\frac{(5\times 21)+0}{5+2}=15\text{m/s}$
From the frame of system's $\text{COM}$, centre of mass of system will remain at rest (no external force)
$\Rightarrow$ Velocity of the heavier block in $\text{COM}$ frame, just after the kick is
$v_1^{\prime }=v_1-v_{\text{CM}}$
$\therefore v^{\prime }=(21-15)=6\text{m/s}$
& Velocity of the lighter block in $\text{COM}$ frame ,just after the kick is
$v_2^{\prime }=v_2-v_{\text{CM}}$
$\therefore v_2^{\prime }=(0-15)=-15\text{m/s}$