Question

A stationary particle explodes into two particles of masses m₁ and m₂ which move in opposite directions with velocities v₁ and v₂. The ratio of their kinetic energies \(\frac{E_1}{E_2}\) is:

• A. \(\frac{m_2}{m_1}\)
• B. \(\frac{m_1}{m_2}\)
• C. 1
• D. \(\frac{m_1v_2}{m_2v_1}\)

Answer 1

The Correct Option is A

To solve the given problem, we need to understand the relationship between the masses of the two particles and their velocities after the explosion. The key here is the conservation of momentum and the expression for kinetic energy.

Initially, the system is stationary, which means the initial momentum is zero. After the explosion, due to the law of conservation of momentum:

m_1v_1 = m_2v_2

This equation tells us that the momentum of the two particles is equal and opposite.

Now, we need to find the ratio of their kinetic energies, \( \frac{E_1}{E_2} \). The formula for kinetic energy is given by:

E = \frac{1}{2}mv^2

Thus, for the two particles, the kinetic energies are:

For particle 1:

E_1 = \frac{1}{2}m_1v_1^2

For particle 2:

E_2 = \frac{1}{2}m_2v_2^2

Hence, the ratio of their kinetic energies is:

\(\frac{E_1}{E_2} = \frac{\frac{1}{2}m_1v_1^2}{\frac{1}{2}m_2v_2^2} = \frac{m_1v_1^2}{m_2v_2^2}\)

Substituting the relation from the conservation of momentum equation \(v_1 = \frac{m_2}{m_1}v_2\) (since \(m_1v_1 = m_2v_2\)), we get:

v_1^2 = \left(\frac{m_2}{m_1}\right)^2v_2^2

Now substitute \(v_1^2\) into the energy ratio:

\(\frac{E_1}{E_2} = \frac{m_1\left(\frac{m_2}{m_1}\right)^2v_2^2}{m_2v_2^2} = \frac{m_2^2}{m_1m_2} = \frac{m_2}{m_1}\)

Hence, the ratio of their kinetic energies is \(\frac{m_2}{m_1}\), which is option 1.