# Solve the following :

Question:

The friction coefficient between the horizontal surface and each of the blocks is shown in figure is $0.20$. The collision between the blocks is perfectly elastic. Find the separation between the two blocks when they come to rest. Take $g=10 \mathrm{~m} / \mathrm{s}^{2}$.

Solution:

Using work energy principle $\mathrm{V}=$ velocity of $2 \mathrm{~kg}$ near collision

$\mathrm{V}=$ velocity of $2 \mathrm{~kg}$ near collision

$\frac{1}{2} m\left(V^{2}-1^{2}\right)=-m \times g \mu \times 10.16$ $V^{2}-1=-(1.6 \times 0.2) \times 2=-0.64$ $V^{2}=1-0.64=0.36 \Rightarrow V=0.6 \mathrm{~m} / \mathrm{s}$ Use C.O.L.M, $\left\{V_{1}, V_{2}=\right.$ final velocities $\}$ $2 \times 0.6+0=m_{1} V_{1}+m_{2} V_{2}$ $2 \times 0.6+0=m_{1} V_{1}+m_{2} V_{2}$

$V_{1}+2 V_{2}=0.6$

As collision is elastic, $e=1$

$V_{2}-V_{1}=0.6-0=0.6$

Solving (1) and (2)

$3 V_{2}=1.2$

Use C.O.E.L for $1^{\text {st }}$ block,

$\frac{1}{2}(2)\left((0)^{2}-(0.2)^{2}\right)=-2 \times 0.2 \times 10 \times x_{1}$

$x_{1}=4 \mathrm{~cm}$

Distance between them $=4+1=5 \mathrm{~cm}$