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 4.00 times the mass of B, and the energy stored in the spring was 85 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

let mass of B=m kg.

then mass of A=4*m

as initial total momentum is 0,

total final momentum will also be 0(due to conservation of momentum)

so if velocity of mass A is v m/s, then velocity of B=-(momentum of A)/mass of B=(-4*m*v/m)=-4*v m/s

using conservation of energy principle:

initial potential energy of the spring=kinetic energy of A + kinetic energy of B

==>85=0.5*(4*m)*v^2+0.5*m*(4*v)^2=10*m*v^2

==>m*v^2=85/10=8.5 ...(1)

part a:

kinetic energy of A=0.5*4*m*v^2=2*m*v^2=2*8.5=17 J

part b:

kinetic energy of particle B=0.5*m*(4*v)^2=8*m*v^2=8*8.5=68 J