Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes them apart and they then fly off in opposite directions, free of the spring. The mass of A is 2.00 times the mass of B, and the energy stored in the spring was 35 J. Assume that the spring has negligible mass and that all its stored energy is transferred to the particles. Once that transfer is complete, what are the kinetic energies of (a) particle A and (b) particle B

Explanation:

Given:

Final speed of mass A = Va

Final speed of mass B = Vb

Mass of A = Ma

Mass of B = Mb

Ma = 2 × Mb

By conservation of linear momentum,

0 = Ma × Va + Mb × Vb

0 =  2 × Mb × Va + Mb × Vb

Vb = - 2 × Va              

Energy of the spring, U = 1/2 × k × x^2

1/2 k x² = 1/2 × Ma × Va² + 1/2 × Mb × Vb²

35 = 1/2 × Ma × Va² + 1/2 × Mb × Vb²

Ma × Va² + Mb × Vb² = 70

2 × Mb (-Vb/2)² + Mb × Vb² = 70

1/2 × Mb × Vb² + Mb × Vb² = 70

3/2 × Mb × Vb² = 70

Mb × Vb² = 140/3

= 46.7 J

Ma = 2 × Mb and Vb = - 2 × Va

Ma/2 × (4 × Va²) = 140/3

Ma × Va² = 70/3

Kinetic energy of mass A, KEa = 1/2 × Ma × Va² = 23.3 J

Kinetic energy of mass B = 1/2 × Mb × Vb² = 46.7 J

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm