Solved Numerical Problem No. 19.1 ( Unit-19: Dawn of Modern Physics); HSSC-II (Class-12); Physics (Punjab Curriculum and Text Board, Lahore) With Verified Answers

A particle called the pion lives on average only about 2.6 ｘ10⁻⁸ s when at rest in the laboratory. It then changes to another form. How long would such a particle live when shooting through space at 0.95c?[Ans. 8.3 x 10⁻⁸ s)

Given:

Proper time at rest = t₀ = 2.6 ｘ10⁻⁸ s

Observed Speed in space = v = 0.95 c

To Find:

Measured time of the pion to live in space = t = ?

Solution:

The time dilation formula in special theory of relativity is given by:

t =

\frac{t ₀}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}

$\frac {t₀}{sqrt {1 - frac {v^2}{c^2}}}$

by putting values

t =

\frac{2.6 ｘ 10 ⁻ ⁸ s}{\sqrt{1 - \frac{{(0.95 c)}^{2}}{c^{2}}}}

$\frac {2.6 ｘ10⁻⁸ s}{sqrt {1 - frac {(0.95c)^2}{c^2}}}$

t =

\frac{2.6 ｘ 10 ⁻ ⁸ s}{\sqrt{1 - \frac{0.9035 c^{2}}{c^{2}}}}

$\frac {2.6 ｘ10⁻⁸ s}{sqrt {1 - frac {0.9035 \cancel {c^2}}{\cancel {c^2}}}}$

c² will be canceled so,

t =

\frac{2.6 ｘ 10 ⁻ ⁸ s}{\sqrt{1 - 0.9025}}

$\frac {2.6 ｘ10⁻⁸ s}{sqrt {1 - 0.9025}}$

t =

\frac{2.6 ｘ 10 ⁻ ⁸ s}{\sqrt{0.0875}}

$\frac {2.6 ｘ10⁻⁸ s}{sqrt {0.0875}}$

t =

\frac{2.6 ｘ 10 ⁻ ⁸ s}{0.312}

$\frac {2.6 ｘ10⁻⁸ s}{0.312}$

t = 8.333 ｘ10⁻⁸ s -----------------Ans.

Thus, shooting a pion at speed v = 0.95 c will change to another form after an interval of 8.333 ｘ10⁻⁸ s.

The time period of a pendulum is measured to be 3s in Inertial frame of the pendulum. What is the period when measured by an observer moving with a speed of 0.95c with respect to the pendulum? (Answer: 9.65)

The length of a space ship is measured to be exactly one-third of its proper length. What is the speed of the space ship relative to the observer? (Answer: 0.9428 c)

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