Question

Two blocks on a frictionless table are held at rest with a compressed spring between them. When the blocks are released, the spring pushes them apart and they move in opposite directions. The spring has a spring constant of 215N / m and was compressed by 0.20 m. If one of the blocks moves at 1.50 m/s and the other moves away at 3.00 m/s , what is the mass of the heavier block in kg? (Hint: mechanical energy in this system is conserved)

Answer 1

Let m₁ and m₂ are masses of blocks . Let m₁ be heavier than m₂

By Conservation of energy , (1/2) k x² = (1/2) m₁ v₁² + (1/2) m₂ v₂² ..........(1)

Where k = 215 N/m is spring constant , x = 0.2 m is compressed distance of spring , v₁ is speed of block of mass m₁ , v₂ is speed of block of mass m₂

By Conservation of momentum , m₁ v₁ - m₂ v₂ = 0

( LHS is total momentum immediately after blocks are released and RHS is initial momentum    which is zero ) .

hence we get , m₁ = ( v₂ / v₁ ) m₂ = ( 3 / 1.5 ) m₂ = 2 m₂ ..............(2)

Using eqn.(2) and by substituting values of k and x , we rewrite eqn.(1) as

8.6 = m₂ { ( 2 $\times$ 1.5 $\times$ 1.5 ) + ( 3 $\times$ 3 ) }   or    m₂ = 0.637 kg

Heavier mass m₁ = 2 m₂ = 2 $\times$ 0.637 = 1.274 kg