Question

The friction coefficient between the horizontal surface and each of the blocks shown in the 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 $g=10m/s$

Answer 1

Given,

Velocity of lock is $V_1=1m{s}^{-1}$

Velocity of 4$kg$ block, $V_2=0$

Let velocity of 2 kg block on reaching the

From work energy principle,

$\left(1/2\right)m\times u_1^2-\left(1/2\right)m\times 1^2$

$\Rightarrow \frac{u_1^2-1}{2}=-2\times 5\Rightarrow \frac{u_1^2-1}{2}=-2\times 5$

$\Rightarrow 64\times {10}^{-2}=u_1^2-1\Rightarrow 64\times {10}^{-2}=u_1^2-1$

$\Rightarrow {\text{u}}_1=0.6\text{m}/\text{s}.........\text{(1)}$

Since, it is a perfectly elastic collision.

Let $V_1,V_2\to$ velocity of $2kgand4kg$ block after collision.

From conservation of momentum

$m_1{V}_1+m_2{V}_2=m_1{V}_1+m_2{V}_2$

$\Rightarrow 2\times 0.6+4\times 0=2v_1+4v_2\Rightarrow v_1+2v_2=0.6$

$Again,V_1-V_2=-(u_1-u_2)=-(0.6-0)=-0.6$

Subtracting ( 2) from (1)

$3v_2=1.2$

$\therefore v_1=-0.6+0.4=-0.2m/s$

Putting work energy principle for 1st 2 kg block when come to rest.

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

$\Rightarrow \left(1/2\right)\times 2\times 0.2\times 0.2=2\times 0.2\times 10\times s$

$\Rightarrow {\text{S}}_1=1\text{cm}.$ ……. (3)

Putting work energy principle for 4$kg$ block.

$\left(1/2\right)\times 4\times 0^2-\left(1/2\right)\times 4\times (0.4{)}^2=-4\times 0.2\times 10\times s$

Putting work energy principle for 4$kg$ block.

$\left(1/2\right)\times 4\times 0^2-\left(1/2\right)\times 4\times (0.4{)}^2=-4\times 0.2\times 10\times s$

$\Rightarrow 2\times 0.4\times 0.4=4\times 0.2\times 10\times s$

$\Rightarrow S_2=4cm$…… (4)

From (3) and (4)

Distance between$\text{2kg}and4\text{kg}$ block $=S_1+S_2=1+4=5cm.$