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Q.

Two rectangular blocks A and B of masses 2kg and 3kg respectively are connected by a spring of spring constant $10.8 {Nm}^{- 1}$ and are placed on a frictionless horizontal surface. The 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

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a

0.01 m

b

0.02 m

c

0.05 m

d

0.03 m

answer is C.

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Detailed Solution

According to the law of conservation of linear momentum, we get

Question Image

$m_{A} u = (m_{A} + m_{B}) v$
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,
$\frac{1}{2} m_{A} u^{2} = \frac{1}{2} (m_{A} + m_{B}) v^{2} + \frac{1}{2} {kx}^{2}$

$\frac{1}{2} m_{A} u^{2} - \frac{1}{2} (m_{A} + m_{B}) v^{2} = \frac{1}{2} {kx}^{2}$

$\frac{1}{2} \times 2 \times {(0.15)}^{2} - \frac{1}{2} (2 + 3) {(0.06)}^{2} = \frac{1}{2} {kx}^{2}$

$0.0225 - 0.009 = \frac{1}{2} {kx}^{2} \Rightarrow 0.0135 = \frac{1}{2} {kx}^{2}$

$x = \sqrt{\frac{0.027}{k}} = \sqrt{\frac{0.027}{10.8}} = 0.05 m$

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