91Ó°ÊÓ

The energy stored in the spring was,$U_i=60\mathrm{J}$

The mass of A is,$m_2=2m_1$

By using the conservation of momentum and the conservation of mechanical energy, find the kinetic energies of particle A and particle B. According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.According tothe conservation of mechanical energy, if an isolated system is subject only to conservative forces, then the mechanical energy is constant.

Formulae are as follow:

1. The mechanical energy conservation,$U_i=K_1+K_2$
2. The momentum conservation,$0=m_1\stackrel{→}{v_1}+m_2\stackrel{→}{v_2}$

where, m₁, m₂ are masses,$\stackrel{→}{v_1,}\stackrel{→}{v_2}$are velocity vectors, K₁, K₂ are kinetic energies and U_i is mechanical energy.

Note that this problem involves both mechanical energy conservation and the momentum conservation. That is, the mechanical energy conservation,

$$\begin{gathered} U_i=K_1+K_2,\mathrm{where}{U}_i=60\mathrm{J} \\ \mathrm{And},\mathrm{momentum}\mathrm{conservation}, \\ 0=m_1\stackrel{→}{v_1}+m_2\stackrel{→}{v_2} \\ Where,m_2=2m_1 \end{gathered}$$

From second equation,

$\left|\stackrel{→}{v_1}\right|=2\left|\stackrel{→}{v_2}\right|$

This implies that,

$$\begin{gathered} K_1=\frac{1}{2}{m}_1{v}_1^2 \\ {K}_1=\frac{1}{2}\left(\frac{1}{2}{m}_2\right){\left(2v_2\right)}^2 \\ {K}_1=2\left(\frac{1}{2}{m}_2{v}_2^2\right) \\ {K}_1=2K_2 \end{gathered}$$

Now, substitute $K_1=2K_2$into the energy conservation relation,

$$\begin{gathered} U_i=2K_2+K_1 \\ {U}_i=3K_2 \\ {K}_2=\frac{1}{3}{U}_i \\ {K}_2=\frac{1}{3}×60 \\ {K}_2=20\mathrm{J} \end{gathered}$$

Hence, the kinetic energy of particle A is 20 J .

Now, obtain,

$$\begin{gathered} K_1=2K_2 \\ {K}_1=2×20 \\ {K}_1=40\mathrm{J} \end{gathered}$$

Hence,the kinetic energy of particle B is 40 J.

Therefore, by using the conservation of momentum and conservation mechanical energy, the kinetic energies of both the particles can be found.