Question

Two blocks, of masses $M$ and $2 M$, are connected to a light spring of spring constant $K$ that has one end fixed, as shown in figure. The horizontal surface and the pulley are frictionless. The blocks are released from rest when the spring is non-deformed. The string is light. Select the correct statement(s).

• A. Maximum extension in the spring is $\frac{4 M g}{K}$
• B. Maximum kinetic energy of the system is $\frac{2 M^{2} g^{2}}{K}$
• C. Maximum energy stored in the spring is four times that of maximum kinetic energy of the system
• D. When kinetic energy of the system is maximum energy stored in the spring is $\frac{4 M^{2} g^{2}}{K}$

Answer 1

Answer: (A), (B), (C)

Maximum extension will be at the moment when both masses stop momentarily after going down. Applying W-E theorem from starting to that instant,
$k_{f}-k_{i}=W_{g r .}+W_{ sp }+W_{\text {ten }}$
$0-0=2 M g x+\left(-\frac{1}{2} K x^{2}\right)+0$
$x=\frac{4 M g}{K}$
System will have maximum KE when net force on the system becomes zero.
$\therefore 2 M g=T \text { and } T=k x$
$\Rightarrow x=\frac{2 M g}{K}$
Hence, $KE$ will be maximum when $2 M$ mass has gone down by $2 M g / K$. Applying W/E theorem,
$ k_{f}-0 =2 M g \frac{2 M g}{K}-\frac{1}{2} K \cdot \frac{4 M^{2} g^{2}}{K^{2}}$
$k_{f} =\frac{2 M^{2} g^{2}}{K^{2}}$
Maximum energy of spring
$\frac{1}{2} K \cdot\left(\frac{4 M g}{K}\right)^{2}=\frac{8 M^{2} g^{2}}{K}$
Therefore maximum spring energy $=4 \times$ maximum $KE$ When $KE$ is maximum $x=\frac{2 M g}{K}$.
Spring energy $=\frac{1}{2} \cdot K \cdot \frac{4 M^{2} g^{2}}{K^{2}}=\frac{2 M^{2} g^{2}}{K^{2}}$ i.e. (4) is wrong.