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 5.00 times the mass of B, and the energy stored in the spring was 46 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 each particle?

Particle A: K.E = 38.3 J

Particle B: K.E = 7.67 J

Explanation:

Given that 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 5.00 times the mass of B, and the energy stored in the spring was 46 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 each particle?

The total kinetic energy of Particle A and particle B should be equal to 46 J

Since the mass of A is 5.00 times the mass of B, then, the K.E energy of each particle can be calculated by using ratio.

5 + 1 = 6

K.E for particle A

5 /6 × 46 = 230 / 6 = 38.3 J

K.E for particle B

1/6 × 46 = 46 / 6 = 7.67 J

because growing process had stoped after 4-5 year

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//