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Centre of Mass, Linear Momentum, Collision — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsCentre of Mass, Linear Momentum, Collision5 Marks
Question
The friction coefficient between the horizontal surface and each of the blocks shown in figure is 0.20. The collision between the blocks is perfectly elastic. Find the separation between the two blocks when they come o rest. Take $g = 10 m/s^2.$​​​​​​​
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Answer

Let velocity of 2kg block on reaching the 4kg block before collision $= u_1$. Given, $V_2 = 0$ (velocity of 4kg block). $\therefore$ From work energy principle,
$\Big(\frac{1}{2}\Big)\text{m}\times\text{u}_1^2-\Big(\frac{1}{2}\Big)\text{m}\times1^2=-\text{m}\times\text{ug}\times\text{s}$
$\Rightarrow\frac{\text{u}_1^2-1}{2}=-2\times5$
$\Rightarrow-16=\frac{\text{u}_1^2-1}{4}$
$\Rightarrow64\times10^{-2}=\text{u}_1^2-1$
$\Rightarrow\text{u}_1=6\text{m/s}$
Since it is a perfectly elastic collision.
Let $V_1, V_2 \rightarrow $ velocity of 2kg & 4kg block after collision.
$​​​​​​​m_1V_1 + m_2V_2 = m_1v_1 + m_2v_2$
$\Rightarrow 2 \times 0.6 + 4 \times 0 = 2v_1 + 4v_2$
$\Rightarrow v_1 + 2v_2 = 0.6 ...(1)$
Again, $V_1 - V_2 = -(u_1 - u_2) = -(0.6 - 0) = -0.6 ...(2)$
Subtracting (2) from (1)
$\Rightarrow 3v_2 = 1.2$
$\Rightarrow v_2 = 0.4 m/s.$
$\Rightarrow v_1 = -0.6 + 0.4 = -0.2$ m/s
$\therefore$ Putting work energy principle for $1^{st} $2kg block when come to rest.
$\Big(\frac{1}{2}\Big)\times2\times0^2-\Big(\frac{1}{2}\Big)\times2\times(0.2)^2$
$=-2\times0.2\times10\times\text{s}$
$\Rightarrow\Big(\frac{1}{2}\Big)\times2\times0.2\times0.2$
$=2\times0.2\times10\times\text{s}$
$\Rightarrow\text{S}_1=1\text{cm}$
Putting work energy principle for 4kg block.
$\Rightarrow\Big(\frac{1}{2}\Big)\times4\times0^2-\Big(\frac{1}{2}\Big)\times4\times(0.4)^2$
$=-4\times0.2\times10\times\text{s}$
$=2\times0.4\times0.4\times=4\times0.2\times10\times\text{s}$
$\Rightarrow\text{S}_2=4\text{cm}$
Distance between 2kg & 4kg block $= S_1 + S_2 = 1 + 4 = 5cm.$

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