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 to rest. Take $g$ = 10 $m/s^2$.

• A. 4 cm
• B. 3 cm
• C. 5 cm
• D. 6 cm

Answer 1

The correct option is C

5 cm

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,

$\left(\frac{1}{2}\right)m\times u_1^2-\left(\frac{1}{2}\right)m\times 1^2$ = $-m\times ug\times s$

$\Rightarrow 64\times {10}^{-2}$ = $u_1^2-1$ $\Rightarrow u_1$ = 0.6 m/s

Since it is a perfectly elastic collision.

Let $V_1$, $V_2$ $\to$ velocity of 2kg & 4kg block after collision.

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

$\Rightarrow$ $2\times 0.6+4\times 0$ = $2v_1+4v_2$ $\Rightarrow v_1+2v_2$ = 0.6 ..........(1)

Again, $V_1-V_2$ = $-(u_1-u_2)$ = -(0.6 - 0) = -0.6 .........(2)

Subtracting (2) from (1)

$3v_2$ = 1.2 $\Rightarrow v_2$ = 0.4 m/s.

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

$\therefore$ Putting work energy principle for 2kg block after collision

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

$\Rightarrow \left(\frac{1}{2}\right)\times 2\times 0.2\times 0.2$ = $2\times 0.2\times 10\times s$ $\Rightarrow S_1$ = 1cm.

Putting work energy principle for 4kg block.

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

$\Rightarrow (2\times 0.4\times 0.4$ = $4\times 0.2\times 10\times s$ $\Rightarrow S_2$ = 4 cm.

Distance between 2kg & 4kg block = $S_1+S_2$ = 1 + 4 = 5 cm.