Question

Two blocks $A$ and $B$ of masses $1\text{kg}$ each, are lying on a smooth horizontal surface. A spring of force constant $k=200\text{N/m}$ is fixed at one end of the block $A$. Block $B$ collides with block $A$ with an initial velocity of $v=2.0\text{m/s}$. Find the maximum compression in the spring.

• A. $0.35\text{m}$
• B. $1\text{m}$
• C. $100\text{m}$
• D. $0.1\text{m}$

Answer 1

The correct option is D $0.1\text{m}$

At maximum compression, velocity of both the blocks will be same.

According to momentum conservation principle,

$(m_A+m_B)v_o=m_{B}v$

$\Rightarrow (1+1)v_o=1\times 2$

$\therefore v_o=1\text{m/s}$

Now, according to conservation of energy principle,

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

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

$\therefore x=0.1\text{m}$