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 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?

Short answer

a) Kinetic energy of particle A is,$K_1=20\mathrm{J}$

b) Kinetic energy of particle B is,$K_2=40\mathrm{J}$

Step-by-step solution

Step 1: Given Data

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

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

Determining the concept

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,$0=m_1\overrightarrow{v_1}+m_2\overrightarrow{v_2}$

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

(a) Determining the kinetic energy of particle A

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

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

From second equation,

$\left|\overrightarrow{v_1}\right|=2\left|\overrightarrow{v_2}\right|$

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,

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

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

(b) Determining the kinetic energy of particle B

Now, obtain,

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

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.