Question

A particle of mass m₁ is moving with a velocity v₁ and another particle of mass m₂ is moving with a velocity v₂. Both of them have the same momentum but their different kinetic energies are E₁ and E₂ respectively. f m₁ > m₂ then :

• A. \(\frac{E_1}{E_2}=\frac{m_1}{m_2}\)
• B. E₁ > E₂
• C. E₁ = E₂
• D. E₁< E

Answer 1

The Correct Option is D

 To solve this problem, let's apply the concepts of momentum and kinetic energy.

We are given:

• Both particles have the same momentum. Therefore, for both particles:
• \(p_1 = p_2 \implies m_1 v_1 = m_2 v_2\)
• The kinetic energies are \(E_1\) for the first particle and \(E_2\) for the second particle.

We need to compare their kinetic energies using the relationship between momentum and energy.

Kinetic Energy Calculation:

• The kinetic energy for a particle is given by the formula: \(E = \frac{1}{2}mv^2\)
• Substituting in terms of momentum: \(v = \frac{p}{m}\)

For the first particle:

• \(E_1 = \frac{1}{2}m_1 \left(\frac{p}{m_1}\right)^2 = \frac{p^2}{2m_1}\)

For the second particle:

• \(E_2 = \frac{1}{2}m_2 \left(\frac{p}{m_2}\right)^2 = \frac{p^2}{2m_2}\)

Comparison of Kinetic Energies:

• Since the momentum is the same for both particles, \(p^2\) is common.
• Thus, the kinetic energy comparison becomes:
• \(\frac{E_1}{E_2} = \frac{p^2/2m_1}{p^2/2m_2} = \frac{m_2}{m_1}\)
• Given that \(m_1 > m_2\), this implies \(\frac{m_2}{m_1} < 1\)
• Therefore, \(E_1 < E_2\)

This shows that the kinetic energy of the particle with a larger mass is less than that of the particle with a smaller mass, given they have the same momentum.

Thus, the correct answer is: \(E_1 < E_2\) (which matches option E₁< E).