Question

Two blocks A and B of equal mass m=1 kg are lying on a smooth horizontal surface as shown below A spring of force constant k = 200 N/m is fixed at one end of block 'A'. Block 'B' collides with 'A' with velocity V₀ = 2 m/s. The maximum compression of the spring is

• A. 10 cm
• B. 12 cm
• C. 14 cm
• D. 16 cm

Answer 1

The correct option is A

10 cm

At maximum compression ($x_m$) velocity of both the blocks is same , say it is 'V'. Applying conservation of linear momentum , we have

($m_A$ + $m_B$) V = $m_b$ $V_0$

(1 + 1) V = (1) $V_0$ $\Rightarrow$ V = 1 m/s

Using conservation of mechanical energy , we have

$\frac{1}{2}$ ($m_B$ ${V_0}^2$ = $\frac{1}{2}$ (($m_B$ + ($m_A$) $V^2$ + $\frac{1}{2}$ K ${x_m}^2$

$\Rightarrow$ $\frac{1}{2}$ (1)$2^2$ = $\frac{1}{2}$ (1 + 1) $1^2$ + $\frac{1}{2}$ (200) ${x_m}^2$

$\Rightarrow$ $x_m$ = 0.1 m = 10 cm