Before You Begin

Question #42020

Difficulty: 8/10 | 🟢 Free

The Question

A particle of rest mass $M$ moving with initial momentum $p_0 = \frac{4}{3}Mc$ decays into two identical product particles, each of rest mass $m$ . The Q-value of this decay process is given as $Q = mc^2$ . What is the maximum possible transverse momentum (perpendicular to the initial direction of motion) of one of the product particles in the laboratory frame?

Options

A.
$\frac{\sqrt{5}}{2} \frac{Q}{c}$
B.
$\frac{5\sqrt{5}}{6} \frac{Q}{c}$
C.
$\frac{3\sqrt{5}}{10} \frac{Q}{c}$
D.
$\sqrt{3} \frac{Q}{c}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Dimensional Independence vs. Scalar/Vector Nature|Q-value Calculation|Mass Defect vs. Binding Energy

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of relativistic mass-energy equivalence and vector momentum conservation.

The trap exploits the misconception that the scalar Q-value can be directly used to find final laboratory momenta via non-relativistic kinetic energy formulas ( $p = \sqrt{2mK}$ ) or by incorrectly applying the Lorentz factor $\gamma$ to transverse momentum components.

Students often fail to recognize that while total energy and longitudinal momentum are coupled via Lorentz transformations, transverse momentum is strictly invariant.

By forcing the analysis into the Center of Momentum (CM) frame, the problem highlights the 'energy-momentum mismatch' trap where students neglect the vector nature of the decay products' momenta.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Units and Measurements - Class 11 Part 1
Concept	Dimensional Independence vs. Scalar/Vector Nature
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free