Two objects of masses m1 = 0.56 kg and m2 = 0.88 kg are placed on a horizontal frictionless surface and a compressed spring of force constant k = 280 N/m is placed
between them as in Fig. a. Neglect the mass of the spring. The spring is not attached to
either object and is compressed a distance of 9.8 cm. If the objects are released from rest,
find the final velocity of each object as shown in Fig. b

Explanation:

Conservation of momentum and energy.

Initial momentum is zero

0.56v₁ + 0.88v₂ = 0

v₂ = -0.56v₁ / 0.88 = -7v₁/11

Spring energy is

PS = ½(280)(0.098)² = 1.34456 J

Spring energy is converted to kinetic energy of the blocks

1.34456  = ½(0.56)v₁² + ½(0.88)v₂²

1.34456  = ½(0.56)v₁² + ½(0.88)(-7v₁/11)²

2.68912  = (0.56 + (0.88)(7/11)²)v₁²

2.68912  = 0.9163636v₁²

2.9345555 = v₁²

v₁ = 1.713054 = 1.7 m/s

v₂ = -7(1.713054)/11 = -1.0901255 = -1.1 m/s

well clustered together.

points close on a scatter diagram