Question

Two blocks $A$ and $B$ of masses $2kg$ and $4kg$ are connected by a light spring of force constant $4800N/m$ . The system is at rest on a smooth horizontal surface with the spring in its natural state without any compression or extension. Now a velocity of $6m/s$ is given to B in a direction away from $A$. During the subsequent motion, the maximum extension of the spring is:

• A. $5cm$
• B. $10cm$
• C. $15cm$
• D. $20cm$

Answer 1

The correct option is B $10cm$

Using conservation of momentum principle we have

$\frac{m_A{v}_A+m_B{v}_B}{m_A+m_B}=v_{CM}$.

$v_{CM}=\frac{2\times 0+4\times 6}{2+4}=4m/s$

Thus we get $v_{CM}$ as 4m/s.

Now let us sit in the centre of mass frame, i.e. moving at 4m/s away in direction of B.

B moves with a velocity of 2m/s away from CM and A moves with a velocity of 4m/s away from CM. At maximum extension, in centre of mass frame, both blocks will be stationary ( since total momentum is 0 in centre of mass frame, so if one stops, so must the other.)

Now by conserving energy,
$\frac{1}{2}{m}_A{v}_A^2+\frac{1}{2}{m}_B{v}_B^2=\frac{1}{2}k{x}^2$

$\frac{1}{2}\times 2\times 4^2+\frac{1}{2}\times 4\times 2^2=\frac{1}{2}\times 4800\times x^2$

$(16+8)J=2400x^2$

$x=0.1m=10cm$