Physics? I tried it and my incorrect answer is show below please help

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 3.00 times the mass of B, and the energy stored in the spring was 65 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?
My wrong answers are are 10.8333 and 54.1667

1 Answer
Ananda Dasgupta
Apr 5, 2018

#K_A = 16.25" J", qquad K_B = 48.75" J"#

Explanation:

The system composed of the two particles and the spring is acted upon by internal forces only. Thus the total momentum of the system is conserved. Since the spring is of negligible mass, this means that the final momenta #p_A# and #p_B# are equal (and opposite in direction):

#p_A = p_B#

Since kinetic energy is related to momentum by

#K = p^2/(2m)#

we have

#K_A/K_B = p_A^2/(2m_A) times (2m_B)/p_B^2 = m_B/m_A = 1/3#

The energy stored in the spring is converted to the total kinetic energy of the particles. Hence

#K_A/1 = K_B/3 = (K_A+K_B)/(1+3) = (65" J")/4 = 16.25" J"#

#K_A = 16.25" J", qquad K_B = 48.75" J"#

Related questions
Impact of this question
492 views around the world
You can reuse this answer
Creative Commons License