Question

Two blocks $A$ and $B$ 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 $A$. Block $B$ collides with block $A$ with velocity $v_0=2.0\text{m/s}$. Find the maximum compression of the spring.

• A. $2\text{cm}$
• B. $5\text{cm}$
• C. $8\text{cm}$
• D. $10\text{cm}$

Answer 1

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

As at maximum compression $\left(x_m\right)$ velocity of both the blocks be same. Lets it is $v$.


Applying conservation of linear momentum, we have

$m_B{v}_0=(m_A+m_B)v$

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

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

Using conservation of mechanical energy, we have

$\frac{1}{2}{m}_B{v}_0^2=\frac{1}{2}(m_A+m_B)v^2+\frac{1}{2}k{x}_m^2$

Substituting the values, we get

$\frac{1}{2}\times \left(1\right)\times (2.0{)}^2=\frac{1}{2}\times (1.0+1.0)\times (1.0{)}^2+\frac{1}{2}\times \left(200\right)\times x_m^2$

$\Rightarrow 2.0=1.0+100x_m^2$

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

Hence, option (d) is the correct answer.