Question

Two rectangular blocks $A$ and $B$ of masses $2kg$ and $3kg$ respectively are connected by a spring of spring constant $10.8N$ $m^{-1}$ are placed on a frictionless horizontal surface. Block $A$ was given an initial velocity of $0.15m{s}^{-1}$ in the direction shown in the figure. The maximum compression of the spring during the motion is

• A. 0.03 m
• B. 0.01 m
• C. 0.02 m
• D. 0.05 m

Answer 1

Answer: (D)

As the block $A$ moves with velocity $0.15m{s}^{-1}$ it compresses the spring which pushes $B$ towards right. $A$ goes on compressing the spring till the velocity acquired by $B$ becomes equal to the velocity of $A$, ie, $0.15m{s}^{-1}$. Let this velocity be $v$. Now, spring is in a state of maximum compression. Let $x$ be the maximum compression at this stage.
According to the law of conservation of linear momentum, we get
$m_{A}u=\left(m_A+m_B\right)v$
or$v=\frac{m_{A}u}{m_A+m_B}$
$=\frac{2\times 0.15}{2+3}=0.06m{s}^{-1}$
According to the law of conservation of energy.
$\frac{1}{2}{m}_A{u}^2=\frac{1}{2}\left(m_A+m_B\right)v^2+\frac{1}{2}k{x}^2$
$\frac{1}{2}{m}_A{u}^2-\frac{1}{2}\left(m_A+m_B\right)v^2=\frac{1}{2}k{x}^2$
$\frac{1}{2}\times 2\times (0.15{)}^2-\frac{1}{2}(2+3)(0.06{)}^2=\frac{1}{2}k{x}^2$
$0.0225-0.009=\frac{1}{2}k{x}^2$ or $0.0135=\frac{1}{2}k{x}^2$
or $x=\sqrt{\frac{0.027}{k}}=\sqrt{\frac{0.027}{10.8}}=0.05m$