Question

Two blocks of masses $m_1=2kg$ and $m_2=4kg$ are moving in the same direction with speeds $v_1=6m/s$ and $v_2=3m/s$, respectively on a frictionless surface as shown in the figure. An ideal spring with spring constant $k=30000N/m$ is attached to the back side of $m_2$. Then the maximum compression of the spring after collision will be :

• A. 0.06 m
• B. 0.04 m
• C. 0.02 m
• D. None of these

Answer 1

The correct option is C 0.02 m

By the law of conservation of linear momentum,

$m_1{v}_1+m_2{v}_2=m_{1}v+m_{2}v$

$\therefore 2\times 6+4\times 3=2v+4v$

$\therefore 12+12=6v$

$\therefore v=4m/s$

By the law of conversation of energy,

$\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2=\frac{1}{2}(m_1+m_2)v^2+\frac{1}{2}k{x}^2$

$\therefore 2\times (6{)}^2+4\times (3{)}^2=(2+4)(4{)}^2+k{x}^2$

$\therefore 72+36=96+k{x}^2$

$\therefore 108-96=k{x}^2$

$\therefore 12=30,000x^2$

$\therefore x=\sqrt{\frac{12}{30,000}}$

$\therefore x=\sqrt{\frac{1}{2,500}}$

$\therefore x=\frac{1}{50}$

$\therefore x=0.02m$

$\therefore x=2cm$

$\therefore$ option (A) is correct.