Question

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 30 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?

Answer 1

Using the conservation of momentum,

0 = MaVa + MbVb

where

Ma = mass of particle A

Va = final velocity of particle A

Mb = mass of particle B

Vb = final velocity of particle B

Since Ma = 5(Mb), then the above formula becomes

0 = 5(Mb)(Va) + Mb(Vb)

Va = - (Vb/5)

(KE)a = kinetic energy of particle A = (1/2)(Ma)(Va)^2

(KE)b = kinetic energy of particle B = (1/2)(Mb)(Vb)^2

Taking the ratio of the two kinetic energies,

(KE)b/(KE)a = [(1/2)(Mb)(Vb)^2]/[(1/2)(Ma)(Va)^2]

and since

Ma = 5(Mb)

and

Va = -Vb/5

then the above simplifies to

(KE)b/(KE)a = [(Mb)(Vb)^2]/[(5Mb)(Vb/5)^2]

(KE)b/(KE)a = 1/(5*1/25) = 5

Therefore,

(KE)b = 5(KE)a -- call this Equation A

Using the law of conservation of energy,

58 = (KE)a + (KE)b

and since (KE)b = 5(KE)a (from Equation A), the above simplifies to

30 = (KE)a + 5(KE)a

6(KE)a = 30

(KE)a = 5 joules

Hence,

(KE)b = 5(5) = 25 joules

Kinetic energy of particle A = 5 joules

Kinetic energy of particle B = 25 joules