Date | November 2012 | Marks available | 5 | Reference code | 12N.3.HL.TZ0.16 |
Level | Higher level | Paper | Paper 3 | Time zone | Time zone 0 |
Command term | Calculate and Determine | Question number | 16 | Adapted from | N/A |
Question
This question is about rest mass and relativistic energy.
(i) Define the rest mass of a particle.
(ii) The rest mass of a particle is said to be an invariant quantity. State, with reference to special relativity, what is meant by the term invariant.
In a thought experiment, two particles X and Y, each of rest mass 380 MeVc–2, are approaching each other head on.
The speed of X and of Y is 0.60 c relative to a laboratory.
(i) Calculate the momentum of X in the frame of reference in which Y is at rest.
(ii) As a result of the collision a single particle Z is formed. Determine the rest mass of Z. The gamma factor for a speed of 0.60 c is 1.25.
Markscheme
(i) the mass of an object in its rest frame / the mass as measured by an observer at rest with respect to the body;
(ii) a quantity that is the same for all observers/reference frames;
(i) speed of X relative to Y is \(\left( {\frac{{0.60{\rm{c}} - \left( { - 0.60{\rm{c}}} \right)}}{{1 + {{0.60}^2}}}} \right) = 0.882{\rm{c}}\);
gamma factor at this speed is \(\gamma = \frac{1}{{\sqrt {1 - {{0.882}^2}} }} = 2.12\);
momentum is then p=γmv= 2.12×380×0.882c=710MeVc−1;
Award [3] for a bald correct answer between 700MeVc−1 and 713MeVc−1 due to rounding.
(ii) \(M{c^2} = 2 \times \gamma m{c^2} = 2 \times \frac{5}{4} \times 380\);
⇒M=950MeVc−2;
Award [2] for a bald correct answer.