Question

A block of mass m1 = 2.3 kg initially moving to the right with a speed of 4.8 m/s on a frictionless, horizontal track collides with a spring attached to a second block of mass m2 = 3.5 kg initially moving to the left with a speed of 2.7 m/s. The spring constant is 580N/m. What if m1 is initially moving at 3.6 m/s while m2 is initially at rest? (a) Find the maximum spring compression in this case. x = 0.176 m (b) What will be the individual velocities of the two masses (v1 and v2) after the spring extended fully again? (That is, when the two masses separate from each other after the collision is complete.) v1=? v2=?

Answer 1

Masses of the blocks are m₁=2.3 kg , m₂ =3.5 kg

speed of block 1 is v₁ =3.6 m/s and v₂=0 m/s (block 2 is at rest initially)

a)

Conserving momentum of the two blocks before and just after collision ( when the combined mass moves with same velocity V,

m₁v₁ +m₂v₂ =(m₁+m₂) V

2.3 x 3.6 + 3.5 x 0 = (2.3+3.5) V

V=0.27414 m/s

Conserving energy of the two blocks before and just after collision,

(1/2) m₁v₁² +(1/2)m₂v₂² =(1/2) kx² + (1/2) (m₂+m₂) V²

x² = [2.3 x 3.6² + 0 - (2.3+3.5)x(0.27414)² ] / [580]

x = 0.176 m

Maximum spring compression is x = 0.176 m

b) Since there is no loss of energy due to friction and heat, the collision is elastic,

Final speeds of blocks for elastic collision,

V₁ =  [v₁(m₁-m₂)+2m₂v₂] / [m₁+m₂]

V₁ = [3.6 (2.3-3.5) +2x3.5x0] / [ 2.3+3.5] = - 0.745 m/s

V₂ = [ v₂ (m₂-m₁) +2m₁v₁] / [ [m₁+m₂ ]

V₂ = [ 0 (3.5-2.3) +2x2.3x3.6] / [2.3+3.5] = 2.855 m/s

V₁ = -0.745 m/s and V₂ = 2.855 m/s