Question 9.48: Particle A and particle B are held together with a compresse......
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 60 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?
This problem involves both mechanical energy conservation U_{i}=K_{1}+K_{2}\,, where {{U}}_{i}=60\ J, and momentum conservation
0=m_{1}{\vec{\nu}}_{1}+m_{2}{\vec{\nu}}_{2}
where m_{2}=2m_{1}. From the second equation, we find |{\vec{\nu}}_{1}|=2|{\vec{\nu}}_{2}|, which in turn implies (since \nu_{1}=|{\vec{\nu}}_{1}| and likewise for \nu_{2}\,)
K_{1}=\frac{1}{2}m_{1}\nu_{1}^{2}=\frac{1}{2}\biggl(\frac{1}{2}m_{2}\biggr)\biggl(2\nu_{2}\biggr)^{2}=2\biggl(\frac{1}{2}m_{2}\nu_{2}^{2}\biggr)=2K_{2}.
(a) We substitute K_{1}=2K_{2} into the energy conservation relation and find
U_{i}=2K_{2}+K_{2}\Rightarrow K_{2}={\frac{1}{3}}U_{i}=20~{J}.
(b) And we obtain K_{1} = 2(20) = 40 J.