SEMI FINAL STAGE 2026
St. Augustine’s College: 52 points
Prempeh College: 47 points
Pope John SHS & Min. Sem.: 28 points
QUESTION
Determine the magnitude of the linear momentum of an electron of speed $1.50\times10^8$ m/s.
ANSWER: $1.58\times10^{-22}$ kg m/s
SOLUTION 1:
Whenever an electron moves at exactly $v = 0.5c$, you can use the pre-calculated constant value for its classical momentum directly.
The 0.5c Momentum Shortcut
Classical momentum ($m_e v$) at $0.5c$ is:
$m_e v = (9.11\times10^{-31}) \times (1.50\times10^8)$
$m_e v \approx 1.37\times10^{-22}$
Multiply by the exact fraction for $\gamma$ at $0.5c$, which is $\dfrac{2}{\sqrt{3}} \approx 1.15$:
$p = 1.37 \times 1.15 \times 10^{-22}$
$p = 1.58\times10^{-22}\text{ kg}\cdot\text{m/s}$
SOLUTION 2:
$p = \gamma m_e v$
$\dfrac{v}{c} = \dfrac{1.5\times10^8}{3.0\times10^8}$
$\dfrac{v}{c} = \dfrac{1}{2}$
$\gamma = \dfrac{1}{\sqrt{1 - (\dfrac{1}{2})^2}}$
$\gamma = \dfrac{1}{\sqrt{\dfrac{3}{4}}}$
$\gamma = \dfrac{2}{\sqrt{3}}$
$\gamma \approx 1.15$
Combine numbers using simple fractions ($1.5 = \dfrac{3}{2}$):
$p = 1.15 \times (9.1\times10^{-31}) \times (1.5\times10^8)$
$p = 1.15 \times \dfrac{3}{2} \times 9.1 \times 10^{-23}$
$1.15 \times 1.5 \approx 1.73$ (which is $\sqrt{3}$)
$p = 1.73 \times 9.1 \times 10^{-23}$
$p \approx 15.74 \times 10^{-23}$
$p = \dfrac{15.74}{10^{23}}$
$p \approx 1.58\times10^{-22}\text{ kg}\cdot\text{m/s}$
NOTE: With $1.73 \times 9.1 = 15.74$ the second working gives $1.57\times10^{-22}$ kg m/s; the stated $1.58\times10^{-22}$ kg m/s follows from $\gamma = \dfrac{2}{\sqrt{3}}$ and $m_e = 9.11\times10^{-31}$ kg, as in SOLUTION 1.
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PRACTICE QUESTIONS
1. QUESTION: Determine the magnitude of the linear momentum of a particle of mass $1.20\times10^{-27}$ kg moving at $2.40\times10^{8}$ m/s.
ANSWER: $4.80\times10^{-19}$ kg m/s
SOLUTION:
$\dfrac{v}{c} = \dfrac{2.40\times10^{8}}{3.00\times10^{8}}$
$\dfrac{v}{c} = \dfrac{24\times10^{7}}{3.00\times10^{8}}$
$\dfrac{v}{c} = 0.800$
$\gamma = \dfrac{1}{\sqrt{1 - (\dfrac{v}{c})^2}}$
$\gamma = \dfrac{1}{\sqrt{1 - 0.64}}$
$\gamma = \dfrac{1}{\sqrt{0.360}}$
$\gamma = \dfrac{1}{0.600}$
$\gamma = \dfrac{5}{3}$
$p = \gamma mv$
$p = \dfrac{5}{3} \times 1.20\times10^{-27} \times 2.40\times10^{8}$
$p = \dfrac{5 \times 12\times10^{-28} \times 24\times10^{7}}{3}$
$p = 5 \times 4 \times 24 \times10^{-21}$
$p = 2 \times 24 \times10^{-20}$
$p = 4.80\times10^{-19}\text{ kg m/s}$
2. QUESTION: Determine the magnitude of the linear momentum of a particle of mass $1.20\times10^{-27}$ kg moving at $1.80\times10^{8}$ m/s.
ANSWER: $2.70\times10^{-19}$ kg m/s
SOLUTION:
$\dfrac{v}{c} = \dfrac{1.80\times10^{8}}{3.00\times10^{8}}$
$\dfrac{v}{c} = \dfrac{18\times10^{7}}{3.00\times10^{8}}$
$\dfrac{v}{c} = 0.600$
$\gamma = \dfrac{1}{\sqrt{1 - (\dfrac{v}{c})^2}}$
$\gamma = \dfrac{1}{\sqrt{1 - 0.36}}$
$\gamma = \dfrac{1}{\sqrt{0.640}}$
$\gamma = \dfrac{1}{0.800}$
$\gamma = 1.25$
$p = \gamma mv$
$p = 1.25 \times 1.20\times10^{-27} \times 1.80\times10^{8}$
$p = \dfrac{5 \times 12\times10^{-28} \times 18\times10^{7}}{4}$
$p = 5 \times 3 \times 18 \times10^{-21}$
$p = 9 \times 3 \times10^{-20}$
$p = 2.70\times10^{-19}\text{ kg m/s}$
3. QUESTION: Determine the magnitude of the linear momentum of a particle of mass $1.25\times10^{-27}$ kg moving at $2.40\times10^{8}$ m/s.
ANSWER: $5.00\times10^{-19}$ kg m/s
SOLUTION:
$\dfrac{v}{c} = \dfrac{2.40\times10^{8}}{3.00\times10^{8}}$
$\dfrac{v}{c} = \dfrac{24\times10^{7}}{3.00\times10^{8}}$
$\dfrac{v}{c} = 0.800$
$\gamma = \dfrac{1}{\sqrt{1 - (\dfrac{v}{c})^2}}$
$\gamma = \dfrac{1}{\sqrt{1 - 0.64}}$
$\gamma = \dfrac{1}{\sqrt{0.360}}$
$\gamma = \dfrac{1}{0.600}$
$\gamma = \dfrac{5}{3}$
$p = \gamma mv$
$p = \dfrac{5}{3} \times 1.25\times10^{-27} \times 2.40\times10^{8}$
$p = \dfrac{5 \times 125\times10^{-29} \times 24\times10^{7}}{3}$
$p = 5 \times 125 \times 8 \times10^{-22}$
$p = 5.00\times10^{-19}\text{ kg m/s}$
4. QUESTION: Determine the magnitude of the linear momentum of a particle of mass $1.50\times10^{-27}$ kg moving at $2.40\times10^{8}$ m/s.
ANSWER: $6.00\times10^{-19}$ kg m/s
SOLUTION:
$\dfrac{v}{c} = \dfrac{2.40\times10^{8}}{3.00\times10^{8}}$
$\dfrac{v}{c} = \dfrac{24\times10^{7}}{3.00\times10^{8}}$
$\dfrac{v}{c} = 0.800$
$\gamma = \dfrac{1}{\sqrt{1 - (\dfrac{v}{c})^2}}$
$\gamma = \dfrac{1}{\sqrt{1 - 0.64}}$
$\gamma = \dfrac{1}{\sqrt{0.360}}$
$\gamma = \dfrac{1}{0.600}$
$\gamma = \dfrac{5}{3}$
$p = \gamma mv$
$p = \dfrac{5}{3} \times 1.50\times10^{-27} \times 2.40\times10^{8}$
$p = \dfrac{5 \times 15\times10^{-28} \times 24\times10^{7}}{3}$
$p = 5 \times 5 \times 24 \times10^{-21}$
$p = 6.00\times10^{-19}\text{ kg m/s}$
5. QUESTION: Determine the magnitude of the linear momentum of a particle of mass $1.60\times10^{-27}$ kg moving at $2.40\times10^{8}$ m/s.
ANSWER: $6.40\times10^{-19}$ kg m/s
SOLUTION:
$\dfrac{v}{c} = \dfrac{2.40\times10^{8}}{3.00\times10^{8}}$
$\dfrac{v}{c} = \dfrac{24\times10^{7}}{3.00\times10^{8}}$
$\dfrac{v}{c} = 0.800$
$\gamma = \dfrac{1}{\sqrt{1 - (\dfrac{v}{c})^2}}$
$\gamma = \dfrac{1}{\sqrt{1 - 0.64}}$
$\gamma = \dfrac{1}{\sqrt{0.360}}$
$\gamma = \dfrac{1}{0.600}$
$\gamma = \dfrac{5}{3}$
$p = \gamma mv$
$p = \dfrac{5}{3} \times 1.60\times10^{-27} \times 2.40\times10^{8}$
$p = \dfrac{5 \times 16\times10^{-28} \times 24\times10^{7}}{3}$
$p = 5 \times 16 \times 8 \times10^{-21}$
$p = 8 \times 8 \times10^{-20}$
$p = 6.40\times10^{-19}\text{ kg m/s}$
6. QUESTION: Determine the magnitude of the linear momentum of a particle of mass $1.60\times10^{-27}$ kg moving at $1.80\times10^{8}$ m/s.
ANSWER: $3.60\times10^{-19}$ kg m/s
SOLUTION:
$\dfrac{v}{c} = \dfrac{1.80\times10^{8}}{3.00\times10^{8}}$
$\dfrac{v}{c} = \dfrac{18\times10^{7}}{3.00\times10^{8}}$
$\dfrac{v}{c} = 0.600$
$\gamma = \dfrac{1}{\sqrt{1 - (\dfrac{v}{c})^2}}$
$\gamma = \dfrac{1}{\sqrt{1 - 0.36}}$
$\gamma = \dfrac{1}{\sqrt{0.640}}$
$\gamma = \dfrac{1}{0.800}$
$\gamma = 1.25$
$p = \gamma mv$
$p = 1.25 \times 1.60\times10^{-27} \times 1.80\times10^{8}$
$p = \dfrac{5 \times 16\times10^{-28} \times 18\times10^{7}}{4}$
$p = 5 \times 4 \times 18 \times10^{-21}$
$p = 2 \times 18 \times10^{-20}$
$p = 3.60\times10^{-19}\text{ kg m/s}$
7. QUESTION: Determine the magnitude of the linear momentum of a particle of mass $2.00\times10^{-27}$ kg moving at $2.40\times10^{8}$ m/s.
ANSWER: $8.00\times10^{-19}$ kg m/s
SOLUTION:
$\dfrac{v}{c} = \dfrac{2.40\times10^{8}}{3.00\times10^{8}}$
$\dfrac{v}{c} = \dfrac{24\times10^{7}}{3.00\times10^{8}}$
$\dfrac{v}{c} = 0.800$
$\gamma = \dfrac{1}{\sqrt{1 - (\dfrac{v}{c})^2}}$
$\gamma = \dfrac{1}{\sqrt{1 - 0.64}}$
$\gamma = \dfrac{1}{\sqrt{0.360}}$
$\gamma = \dfrac{1}{0.600}$
$\gamma = \dfrac{5}{3}$
$p = \gamma mv$
$p = \dfrac{5}{3} \times 2.00\times10^{-27} \times 2.40\times10^{8}$
$p = \dfrac{5 \times 2\times10^{-27} \times 24\times10^{7}}{3}$
$p = 5 \times 2 \times 8 \times10^{-20}$
$p = 8.00\times10^{-19}\text{ kg m/s}$
8. QUESTION: Determine the magnitude of the linear momentum of a particle of mass $2.00\times10^{-27}$ kg moving at $1.80\times10^{8}$ m/s.
ANSWER: $4.50\times10^{-19}$ kg m/s
SOLUTION:
$\dfrac{v}{c} = \dfrac{1.80\times10^{8}}{3.00\times10^{8}}$
$\dfrac{v}{c} = \dfrac{18\times10^{7}}{3.00\times10^{8}}$
$\dfrac{v}{c} = 0.600$
$\gamma = \dfrac{1}{\sqrt{1 - (\dfrac{v}{c})^2}}$
$\gamma = \dfrac{1}{\sqrt{1 - 0.36}}$
$\gamma = \dfrac{1}{\sqrt{0.640}}$
$\gamma = \dfrac{1}{0.800}$
$\gamma = 1.25$
$p = \gamma mv$
$p = 1.25 \times 2.00\times10^{-27} \times 1.80\times10^{8}$
$p = \dfrac{5 \times 2\times10^{-27} \times 18\times10^{7}}{4}$
$p = \dfrac{5 \times 18}{2} \times10^{-20}$
$p = 5 \times 9 \times10^{-20}$
$p = 4.50\times10^{-19}\text{ kg m/s}$
9. QUESTION: Determine the magnitude of the linear momentum of a particle of mass $2.50\times10^{-27}$ kg moving at $2.40\times10^{8}$ m/s.
ANSWER: $1.00\times10^{-18}$ kg m/s
SOLUTION:
$\dfrac{v}{c} = \dfrac{2.40\times10^{8}}{3.00\times10^{8}}$
$\dfrac{v}{c} = \dfrac{24\times10^{7}}{3.00\times10^{8}}$
$\dfrac{v}{c} = 0.800$
$\gamma = \dfrac{1}{\sqrt{1 - (\dfrac{v}{c})^2}}$
$\gamma = \dfrac{1}{\sqrt{1 - 0.64}}$
$\gamma = \dfrac{1}{\sqrt{0.360}}$
$\gamma = \dfrac{1}{0.600}$
$\gamma = \dfrac{5}{3}$
$p = \gamma mv$
$p = \dfrac{5}{3} \times 2.50\times10^{-27} \times 2.40\times10^{8}$
$p = \dfrac{5 \times 25\times10^{-28} \times 24\times10^{7}}{3}$
$p = 5 \times 25 \times 8 \times10^{-21}$
$p = 1.00\times10^{-18}\text{ kg m/s}$