Question

Two blocks $\text{A}$ and $\text{B}$ of equal mass $m=1.0$$\text{kg}$ are lying on a smooth horizontal surface as shown in figure. A spring of force constant $k=200\text{N/m}$ is fixed at one end of block $\text{A}$. Block $\text{B}$ collides with block $\text{A}$ with velocity $v_0=2.0$$\text{m/s}$. Find the maximum compression of the spring.

• A. $10\text{cm}$
• B. $5\text{cm}$
• C. $15\text{cm}$
• D. $20\text{cm}$

Answer 1

The correct option is A $10\text{cm}$

Given,
mass of both blocks, $m_A=m_B=1\text{kg}$.
initial velocity of block $\text{A},u_A=0$
initial velocity of block $\text{B},u_B=v_0$

At the maximum compression $\left(x_m\right)$ velocity of both the blocks is same, say it is $v$.


Applying conservation of linear momentum, we have
$m_{A}v+m_{B}v=m_A\times 0+m_B{v}_o$

$\Rightarrow (1.0+1.0)v=\left(1.0\right)v_0$

$v=\frac{v_0}{2}=\frac{2.0}{2}=1.0\text{m/s}$

Using conservation of mechanical energy, we have
$\frac{1}{2}{m}_B{v}_0^2=\frac{1}{2}(m_A+m_B)v^2+\frac{1}{2}k{x}_m^2$

Substituing the values, we get
$\frac{1}{2}\times \left(1\right)\times (2.0{)}^2=\frac{1}{2}\times (1.0+1.0)\times (1.0{)}^2+\frac{1}{2}\times \left(200\right)\times x_m^2$

$\Rightarrow 2=1.0+100x_m^2$

$\Rightarrow x_m=0.1\text{m}=10.0\text{cm}$

Hence, option (a) is the correct answer.

why this question: This is a multi-conceptual question. It requires application of linear momentum conservation as well as energy conservation principle.