Question

Two blocks of masses m₁ and m₂ are connected by a spring of spring constant k (See figure). The block of mass m₂ is given a sharp impulse so that it acquires a velocity v₀ towards right. Find (a) the velocity of the centre of mass, (b) the maximum elongation that the spring will suffer.

Answer 1

Given,
Velocity of mass, m₂ = v₀
Velocity of mass, m₁ = 0 

(a) Velocity of centre of mass is given by,
\[v_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}\]

\[\Rightarrow v_{cm} = \frac{m_1 \times 0 + m_2 \times v_0}{m_1 + m_2}\]

\[ \Rightarrow v_{cm} = \frac{m_2 v_0}{m_1 + m_2}\]

(b) Let the maximum elongation in spring be x.

The spring attains maximum elongation when velocities of both the blocks become equal to the velocity of centre of mass.
i.e. v₁ = v₂ = v_cm

On applying the law of conservation of energy, we can write:
Change in kinetic energy = Potential energy stored in spring

\[\Rightarrow \frac{1}{2} m_2 v_0^2 - \frac{1}{2}( m_1 + m_2 ) \left( \frac{m_2 v_0}{m_1 + m_2} \right)^2 = \frac{1}{2}k x^2 \]

\[ \Rightarrow m_2 v_0^2 \left( 1 - \frac{m_2}{m_1 + m_2} \right) = k x^2\] 

`Rightarrow = V_o[(m_1m_2)/((m_1+m_2)K)]^(1/2)`