Question

A particle of rest mass $m_0$ is travelling so that its total energy is twice its rest mass energy. It collides with another stationary particle of rest mass $m_0$ to form a new particle. What is the rest mass of the new particle?

• A. $\sqrt{6}m_0$
• B. $2m_0$
• C. $2\sqrt{3}m_0$
• D. $\sqrt{3}m_0$
• E. $3m_0$

Hint:

In perfectly inelastic relativistic collisions, the rest mass of the product is always greater than the sum of the initial rest masses because some kinetic energy is converted into mass ($E=mc^2$).

Answer 1

The Correct Option is A

Concept:
In relativistic collisions, both total energy ($E$) and momentum ($p$) are conserved. [itemsep=8pt]
• Total Energy Conservation: $E_1 + E_2 = E_{new}$
• Momentum Conservation: $\vec{p}_1 + \vec{p}_2 = \vec{p}_{new}$
• Energy-Momentum Relation: $E^2 = p^2c^2 + m_0^2c^4$

Step 1: Determine the initial energy and momentum.
For the moving particle: $E_1 = 2m_0c^2$. From the energy-momentum relation: \[ (2m_0c^2)^2 = p_1^2c^2 + m_0^2c^4 \implies 4m_0^2c^4 = p_1^2c^2 + m_0^2c^4 \implies p_1c = \sqrt{3}m_0c^2 \] For the stationary particle: $E_2 = m_0c^2$ and $p_2 = 0$.

Step 2: Apply conservation laws to the new particle.
Total Energy $E_{new} = E_1 + E_2 = 2m_0c^2 + m_0c^2 = 3m_0c^2$. Total Momentum $p_{new} = p_1 + p_2 = \sqrt{3}m_0c$.

Step 3: Find the rest mass $M$ of the new particle.
Using $E_{new}^2 = p_{new}^2c^2 + M^2c^4$: \[ (3m_0c^2)^2 = (\sqrt{3}m_0c)^2c^2 + M^2c^4 \] \[ 9m_0^2c^4 = 3m_0^2c^4 + M^2c^4 \implies 6m_0^2c^4 = M^2c^4 \implies M = \sqrt{6}m_0 \]