Sunday
March 26, 2017

Post a New Question

Posted by on .

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 132 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 A and B in Joules?

  • Physics Mechanics - ,

    first conservation of momentum

    MassA*Va+Massb*Vb=0
    3Va+Vb=0
    Vb=-3Va

    then energy:
    132=1/2 MassA*Va^2+1/2 MassB*Vb^2
    264=massB (3Va^2+Vb^2) or
    264=MassB (3Vb^2/9+Vb^2)) or
    264=4/3 MassB*Vb^2
    so you know then that KE massb is

    KEmassB=132*3/4 so the remaining 1/4 must be KEmassA

    check my work

  • Physics Mechanics - ,

    Thanks a lot. Your answer are both right. Thank you so very much.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question