Question

If the speed of a particle moving at a relativistic speed is doubled, its linear momentum will
(a) become double
(b) become more than double
(c) remain equal
(d) become less than double.

Answer 1

Answer:

hey guys

(remain equal)( c)

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Answer 2

Answer is Option (B)

Explanation:

• The magnitude of linear momentum is given by,

$P = mv$

• The relativistic mass of the particle is,

$m = \frac {m_o}{\sqrt 1 -\frac {v^2}{c^2}}$

• Thus, magnitude of linear momentum is,

$P = \frac {m_o v}{\sqrt 1 -\frac {v^2}{c^2}}$

$P = m_o v(1 - (\frac {v^2}{c^2}))^{\frac {-1}{2}}$

Using binomial expansion, we get,

$P = m_o + \frac {m_ov^3}{2c^2}$

When the speed of the particle is doubled then,

$P' = \frac {m_o 2v}{\sqrt 1 -\frac {4v^2}{c^2}}$

$P = 2m_ov + \frac {4m_ov^3}{2c^2}$

Thus, the new value of linear momentum is more than double.