Question

Two rectangular blocks $A$ and $B$ of masses $2\,kg$ and $3\,kg$ respectively are connected by a spring of spring constant $10.8\, N$ $m^{-1}$ are placed on a frictionless horizontal surface. Block $A$ was given an initial velocity of $0.15 \,ms^{-1}$ 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 $0.15 \,ms ^{-1}$ 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, $0.15\, ms ^{-1}$. 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}=0.06\, ms ^{-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}}=0.05\, m$