Question

Two blocks of masses of $3$ kg and $6$ kg rest on a horizontal smooth surface. The $3$ kg block is attached to a spring with a force constant $k=900N{m}^{-1}$ which is compressed $2$ m beyond the equilibrium position. The $6$ kg mass is at rest at $1$ m from mean position, $3$ kg mass strikes the $6$ kg mass and the two stick together

• A. velocity of the combined masses immediately after the collision is $10m{s}^{-1}$
• B. velocity of the combined masses immediately after the collision is $5m{s}^{-1}$
• C. amplitude of the resulting oscillation is $\sqrt{2}m$
• D. amplitude of the resulting oscillation is $\sqrt{5}/2m$

Answer 1

The correct option is A velocity of the combined masses immediately after the collision is $10m{s}^{-1}$

Using work-energy principle,
Just before collision
$\frac{1}{2}k({x_2}^2-{x_1}^2)=\frac{1}{2}m(v^2-u^2)$
$\frac{1}{2}k(2^2-1^2)=\frac{1}{2}\times 3\times \left(v^2\right)$
$\therefore v=30m/s$
Now, Using momentum conservation principle,
$m_1{u}_1+m_2{u}_2=(m_1+m_2)v$
$\therefore v=10$m/s