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Relativistic Momentum Problem?
Consider a one-dimensional collision at relativistic speeds between two particles with masses m1 and m2. Particle 1 is initially moving with a speed of 0.717c and collides with particle 2, which is initially at rest. After the collision, particle 1 recoils with speed 0.495c, while particle 2 starts moving with a speed of 0.279c. What is the ratio m2/m1?
and... that gives a non-real answer.
1 Answer
- electron1Lv 78 years agoFavorite Answer
Relativistic momentum = (m * v) ÷ √(1 – v^2/c^2)
For Particle 1, v/2/c^2 = (0.717c)^2 / c^2 = 0.717^2
Final relativistic momentum of Particle 1 = (m1 * 0.717) ÷ √(1 – 0.717^2)
Since Particle 1 recoils, it is moving in the opposite direction. The velocity of Particle 1 after the collision is -0.495c.
For Particle 1, v/2/c^2 = (-0.495c^2/c^2 = 0.495^2
Final momentum of Particle 1 = (m1 * -0.495) ÷ √(1 – 0.495^2)
Final momentum of Particle 2 = (m2 * 0.279) ÷ √(1 – 279^2)
Total final momentum = Initial momentum
(m1 * -0.495) ÷ √(1 – 0.495^2) + (m2 * 0.279) ÷ √(1 – 279^2) = (m1 * 0.717) ÷ √(1 – 0.717^2)
Add (m1 * 0.495) ÷ √(1 – 0.495^2) to both sides.
(m2 * 0.279) ÷ √(1 – 279^2) = (m1 * 0.717) ÷ √(1 – 0.717^2) + (m1 * 0.495) ÷ √(1 – 0.495^2)
m2 * [(0.279 ÷ √(1 – 279^2)] = m1 *[ (0.717 ÷ √(1 – 0.717^2) + (0.495 ÷ √(1 – 0.495^2)]
m2/m1 = [ (0.717 ÷ √(1 – 0.717^2) + (0.495 ÷ √(1 – 0.495^2)] ÷ [(0.279 ÷ √(1 – 0.279^2)]
