# A.5 Galilean and special relativity HL Paper 1- IBDP Physics 2025- Exam Style Questions

IBDP Physics HL 2025 -A.5 Galilean and special relativity HL  Paper 1 Exam Style Questions

## A.5 Galilean and special relativity HL  Paper 1

Reference Frames, Galilean Relativity, Special Relativity, Lorentz Transformations, Space-Time Diagrams, Simultaneity

### Question-A.5 Galilean and special relativity

A particle is kept at rest at the origin. A constant force $$\vec{F}$$ starts acting on it at $$t=0$$. Find the speed of the particle at time $$t$$.

(a) $v=\frac{F t c}{\sqrt{m_0^2 c^2+F^2 t^2}}$

(b) $v=\frac{Fc}{\sqrt{m_0^2 c^2+F^2 t^2}}$

(c) $v=\frac{F t c}{\sqrt{m_0^2 c^2+F t}}$

(d) None of these

Ans:A

The equation of motion is,
$\frac{d \vec{p}}{d t}=\vec{F}$

As the particle starts from rest and the force is always in the same direction, the motion will be along this direction only. Thus, we can write
or,
$\frac{d p}{d t}=F$
or,
$\int_0^p d p=\int_0^t F d t$
$p=F t$
or,
$\frac{m_0 v}{\sqrt{1-v^2 / c^2}}=F t$
or,
$m_0^2 v^2=F^2 t^2-\frac{F^2 t^2}{c^2} v^2$
or,
$v^2\left(m_0^2+\frac{F^2 t^2}{c^2}\right)=F^2 t^2$
or,
$v=\frac{F t c}{\sqrt{m_0^2 c^2+F^2 t^2}} .$

### Question

Choose the speed at which the mass of an electron is double of its rest mass.

(a) $c$

(b) $c\frac{\sqrt{3 }}{2}$

(c) $c\frac{1}{2}$

(d) None of these

Ans:B

The mass of an electron at speed $$v$$ is
$m=\frac{m_0}{\sqrt{1-v^2 / c^2}}$
where $$m_0$$ is its rest mass. If $$m=2 m_0$$,
or,
$2=\frac{1}{\sqrt{1-v^2 / c^2}}$
or,
\begin{aligned} & 1-\frac{v^2}{c^2}=\frac{1}{4} \\ & v=\frac{\sqrt{ } 3}{2} c \mathrm{~m} \mathrm{~s}^{-1} . \end{aligned}

Question
An experimenter measures the length of a rod. In the cases listed, all motions are with respect to the lab and parallel to the length of the rod. In which of the cases the measured length will be minimum?
(a) The rod and the experimenter move with the same speed $$v$$ in the same direction.
(b) The rod and the experimenter move with the same speed $$v$$ in opposite directions.
(c) The rod moves at speed $$v$$ but the experimenter stays at rest.
(d) The rod stays at rest but the experimenter moves with the speed $$v$$.

## Markscheme

B

The rod and the experimenter move with the same speed $$v$$ in opposite directions.

If a rod is moving with speed $$v$$ parallel to its length $$I_0$$ and the experimenter is at rest, its new length will be given as,
$l=l_o \sqrt{1-\frac{v^2}{c^2}}$

If the rod is at rest and the observer is moving with speed $$v$$ parallel to measured length of the rod, the rod’s length will be given as,
$l=l_o \sqrt{1-\frac{(-v)^2}{c^2}}=l_o \sqrt{1-\frac{v^2}{c^2}}$

If the rod and the experimenter both are moving with the same speed in the same direction, then $$I=I_0$$ while if they are moving with same speed in the opposite directions, the length of the rod will be given as,
$l=l_o \sqrt{1-\frac{(v-(-v))^2}{c^2}}=l_o \sqrt{1-\frac{4 v^2}{c^2}}$

Where, $$v<c$$
As, $$1-\frac{4 v^2}{c^2}<1-\frac{v^2}{c^2}$$
$\therefore l_o \sqrt{1-\frac{4 v^2}{c^2}}<l_o \sqrt{1-\frac{v^2}{c^2}}<l_0$

Therefore, the length will be minimum in the case when both are travelling in opposite direction.

### 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.

Ans:B

If a particle is moving at a relativistic speed $$v$$, its linear momentum (p) is given as,
\begin{aligned} & p=\frac{m_o v}{\sqrt{1-\frac{v^2}{c^2}}} \\ & \Rightarrow p=m_o v\left(1-\frac{v^2}{c^2}\right)^{\frac{-1}{2}} \end{aligned}

Expanding binomially and neglecting higher terms we have,
\begin{aligned} & p \simeq m_o v\left(1+\frac{v^2}{2 c^2}\right) \\ & \Rightarrow p \simeq m_o v+\frac{m_o v^3}{2 c^2} \end{aligned}

If the speed is doubled, such that it is travelling with speed $$2 \mathrm{v}$$, linear momentum will be given as
\begin{aligned} & p^{\prime}=\frac{m_o(2 v)}{\sqrt{1-\frac{4 v^2}{c^2}}} \\ & \Rightarrow p^{\prime}=2 m_o v\left(1-\frac{4 v^2}{c^2}\right)^{\frac{-1}{2}} \end{aligned}

Expanding binomially and neglecting higher terms we have,
\begin{aligned} & p^{\prime} \simeq 2 m_o v\left(1+\frac{4 v^2}{2 c^2}\right) \\ & \Rightarrow p^{\prime} \simeq 2 m_o v+\frac{4 m_o v^3}{c^2} \\ & \therefore p^{\prime} \simeq 2 p+\frac{3 m_o v^3}{c^2}, \frac{3 m_o v^3}{c^2}>0 \end{aligned}

Therefore, $$p^{\prime}$$ is more than double of $$p$$.

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