IBDP Physics- A.5 Galilean and special relativity- IB Style Questions For HL Paper 1A -FA 2025
▶️ Answer/Explanation
Use relativistic velocity addition: \( u = \frac{v + u’}{1 + \frac{vu’}{c^2}} \)
Let X’s frame be the rest frame. In X’s frame, the spacecraft has velocity +0.50c.
In Y’s frame, the spacecraft has velocity -0.50c.
We want the relative speed of Y with respect to X.
Using the velocity addition formula: \( 0.50c = \frac{v + (-0.50c)}{1 + \frac{v(-0.50c)}{c^2}} \)
Solving: \( 0.50 = \frac{v – 0.50}{1 – 0.50v/c} \)
\( 0.50(1 – 0.50v/c) = v – 0.50 \)
\( 0.50 – 0.25v/c = v – 0.50 \)
\( 1.00 = 1.25v/c \)
\( v = 0.80c \)
✅ Answer: (B)
▶️ Answer/Explanation
The relativistic equation of motion is \( \dfrac{d\vec{p}}{dt}=\vec{F} \).
Since the particle starts from rest and the force is constant in direction, motion is along the force direction, so \( \dfrac{dp}{dt}=F \).
Integrate from \(0\) to \(t\):
\( \int_0^{p} dp = \int_0^{t} F\,dt \Rightarrow p = Ft \).
Relativistic momentum is \( p=\gamma m_0 v=\dfrac{m_0 v}{\sqrt{1-\dfrac{v^2}{c^2}}} \).
So \( \dfrac{m_0 v}{\sqrt{1-\dfrac{v^2}{c^2}}}=Ft \).
Square both sides:
\( \dfrac{m_0^2 v^2}{1-\dfrac{v^2}{c^2}} = F^2 t^2 \).
Rearrange:
\( m_0^2 v^2 = F^2 t^2 \left(1-\dfrac{v^2}{c^2}\right) \).
\( m_0^2 v^2 = F^2 t^2 – \dfrac{F^2 t^2}{c^2}v^2 \).
Bring \(v^2\) terms together:
\( v^2\left(m_0^2 + \dfrac{F^2 t^2}{c^2}\right)=F^2 t^2 \).
Therefore,
\( v^2=\dfrac{F^2 t^2}{m_0^2+\dfrac{F^2 t^2}{c^2}} \) and \( v=\dfrac{Ftc}{\sqrt{m_0^2c^2+F^2t^2}} \).
✅ Answer: (A)
▶️ Answer/Explanation
The relativistic mass of a particle moving at speed \(v\) is given by \( m=\dfrac{m_0}{\sqrt{1-\dfrac{v^2}{c^2}}} \), where \(m_0\) is the rest mass.
If the mass of the electron is double its rest mass, then \( m = 2m_0 \).
Hence, \( 2 = \dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}} \).
Squaring both sides gives \( 1-\dfrac{v^2}{c^2} = \dfrac{1}{4} \).
Therefore, \( \dfrac{v^2}{c^2} = \dfrac{3}{4} \) and \( v = \dfrac{\sqrt{3}}{2}c \).
✅ Answer: (B)