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 2.00 times the mass of B, and the energy stored in the spring was 50 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

a)

let mass of B = M_B = m

then mass of A = M_A = 2m              ( since mass of A is 2.00 times the mass of B )

V_A = speed of mass A

V_B = speed of mass B

using conservation of momentum ::

initial total momentum before release = final total momentum after release

initial total momentum before release = 0               ( since both masses were at rest)

final Total momentum = M_A V_A + M_B ( -V_B)           (since particles moves opposite to each other)

so   0 =   M_A V_A + M_B ( -V_B)   

inserting the values

0 = (2m) V_A - m V_B

V_B = 2 V_A

Using conservation of energy

spring energy = KE of A   + KE of B

50 = 1/2 M_A V_A² + 1/2 M_B V_B²

100 = M_A V_A² + M_B V_B²

inserting the values

100 = M_A V_A² + (M_A/2 ) (2V_A)²                   ( since M_B = M_A/2   , V_B = 2V_A)

100/3 = M_A V_A²

dividing both side by 2

50/3 = 1/2 M_A V_A²

KE of A = 50/3 joules

b)

kinetic energy of B = 1/2 M_B V_B²

kinetic energy of B = 1/2 (M_A/2) (2V_A)² = 2 ( 1/2 M_A V_A² )

kinetic energy of B = = 2 ( 50/3)   = 100 /3