Question

The friction coefficient between the horizontal surface and each of the blocks shown in figure (9-E20) is 0.20. The collision between the blocks is perfectly elastic. Find the separation between the two blocks when they come to rest. Take $g-10m/s^2.$

Answer 1

Let velocity of 2 kg block on reaching the 4 kg block before collision = $u_1$

Given, $V_2=0$

(velocity of 4 kg block)

$\therefore \left(\frac{1}{2}\right)m\times u_1^2-\left(\frac{1}{2}\right)m\times (1{)}^2=-m\times mg\times S$

$\Rightarrow u_1=\sqrt{(1{)}^2-2\times 0.20\times 10\times 0.16}$

$\Rightarrow =0.6m/sec$

$\Rightarrow u_1=0.6m/s$

Since it is a perfectly elastic collision.

Let $v_1,v_2\to$ velocity of 2 kg and 4 kg block after collision,

$m_1\times u_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

$\Rightarrow 2\times 0.6+4.0=2v_1+4v_2$

$\Rightarrow 2v_1+4v_2=1.2...\left(i\right)$

Again $v_1-v_2=+(u_1-u_2)$

$=+(0.6-0)$

$=-0.6...\left(ii\right)$

Substrating (ii) from (i)

$3v_2=1.2$

$\Rightarrow v_2=0.4m/s$

$\therefore v_1=-0.6+0.4$

$=-0.2m/s$

$\therefore$ Putting work energy principle for 1st 2kg block when comes to rest

$\left(\frac{1}{2}\right)\times 2\times (0{)}^2+\left(\frac{1}{2}\right)\times 2\times (0.2{)}^2=-2\times 02\times 10\times S_1$

$\Rightarrow S_1=1cm.$

Putting work energy principle for 4 kg block

$\left(\frac{1}{2}\right)\times 4\times (0{)}^2-\left(\frac{1}{2}\right)\times 4\times (0.4{)}^2=-4\times 0.2\times 10\times S_2$

$\Rightarrow 2\times 0.4\times 0.4=4\times 0.2\times 10\times S_2$

$\Rightarrow S_2=4cm$

Distance between 2 kg and 4 kg block

$=S_1+S_2=1+4=5cm$