When a boy starts running at a speed of 12 km h^ -1 , by what fraction does his mass increase?

Correct answer: C. 6.17 × 10^ -17 Speed v = 12 km h^ -1 = 12 × (5)/(18) m s^ -1 = (10)/(3) m s^ -1 The relativistic mass is given by m = m_0 1 - (v^2)/(c^2) The fractional increase in mass is (Δ m)/(m_0) = (m - m_0)/(m_0) = (1 - (v^2)/(c^2))^ -1/2 - 1 Since v c , using the binomial expansion: (Δ m)/(m_0) (1 + (1)/(2)(v^2)/(c^2)) - 1 = (1)/(2)(v^2)/(c^2) Substituting the values: (Δ m)/(m_0) = (1)/(2) × (((10)/(3))^2)/((3 × 10^8)^2) (Δ m)/(m_0) = (1)/(2) × 100 9 × 9 × 10^ 16 = 100 162 × 10^ 16 (Δ m)/(m_0) = (1)/(1.62) × 10^ -16 6.17 × 10^ -17

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: The Special Theory of Relativity

Concepts Of Physics MCQ Edition [Volume 2] · Physics · The Special Theory of Relativity

When a boy starts running at a speed of 12 km h^ -1 , by what...

When a boy starts running at a speed of

12 \text{ km h}^{-1}

, by what fraction does his mass increase?

Options

A $1.23 \times 10^{-16}$
B $8.00 \times 10^{-16}$
C $6.17 \times 10^{-17}$
D $1.60 \times 10^{-15}$

Correct answer

C. $6.17 \times 10^{-17}$

Step-by-step solution

Speed

v = 12 \text{ km h}^{-1} = 12 \times \dfrac{5}{18} \text{ m s}^{-1} = \dfrac{10}{3} \text{ m s}^{-1}

The relativistic mass is given by

m = \dfrac{m_0}{\sqrt{1 - \dfrac{v^2}{c^2}}}

The fractional increase in mass is

\dfrac{\Delta m}{m_0} = \dfrac{m - m_0}{m_0} = \left(1 - \dfrac{v^2}{c^2}\right)^{-1/2} - 1

Since

v \ll c

, using the binomial expansion:

\dfrac{\Delta m}{m_0} \approx \left(1 + \dfrac{1}{2}\dfrac{v^2}{c^2}\right) - 1 = \dfrac{1}{2}\dfrac{v^2}{c^2}

Substituting the values:

\dfrac{\Delta m}{m_0} = \dfrac{1}{2} \times \dfrac{\left(\dfrac{10}{3}\right)^2}{(3 \times 10^8)^2}

\dfrac{\Delta m}{m_0} = \dfrac{1}{2} \times \dfrac{100}{9 \times 9 \times 10^{16}} = \dfrac{100}{162 \times 10^{16}}

\dfrac{\Delta m}{m_0} = \dfrac{1}{1.62} \times 10^{-16} \approx 6.17 \times 10^{-17}

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