Question

Two blocks of masses $4\text{kg}$ and $2\text{kg}$ are placed on a frictionless surface and connected by a spring. An external kick gives a velocity of $12\text{m/s}$ to the heavier block in the direction towards the lighter one. The magnitude of the velocities of two blocks in the center of mass frame, after the kick, are respectively

• A. $6\text{m/s},10\text{m/s}$
• B. $4\text{m/s},8\text{m/s}$
• C. $8\text{m/s},6\text{m/s}$
• D. $4\text{m/s},10\text{m/s}$

Answer 1

The correct option is B $4\text{m/s},8\text{m/s}$


$m_1=4\text{kg},m_2=2\text{kg}$
Velocities of $4\text{kg}$ and $2\text{kg}$ block are $12\text{m/s}$ and zero respectively.
Let the velocity of center of mass be $V_{cm}$.
$V_{cm}=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}\left[\begin{array}{c}\because v_1=12\text{m/s}\\ v_2=0\end{array}\right]$
$=\frac{4\times 12+2\times 0}{4+2}=\frac{48}{6}=8\text{m/s}$
i.e $V_{cm}=8\text{m/s}$

Velocity of $4\text{kg}$ block in $COM$ frame,
$V_1^{\prime }=v_1-V_{cm}=12-8=4\text{m/s}$
Velocity of $2\text{kg}$ block in $COM$ frame,
$V_2^{\prime }=v_2-V_{cm}=0-8\text{m/s}=-8\text{m/s}$
i.e $|V_2^{\prime }|=8\text{m/s}$

So, magnitude of velocities of block $4\text{kg}$ and $2\text{kg}$ block in COM frame are $4\text{m/s}$ and $8\text{m/s}$ respectively.