Two particles of mass m and 2m moving in opposite direction collide elastically with v and 2v respectively. If the velocity of particle 1 and particle 2 is v1 and v2 respectively after collision then
Answers
Since, the Collision in elastic then the Kinetic Energy before and after the collision will remains the constant.
∴ 1/2mv² + 1/2 × 2m × (2v)² = 1/2mv₁² + 1/2 × 2m × v₂²
v²/2 + 4v² = v₁²/2 + v₂²
4.5v² = v₁²/2 + v₂² ----eq(i).
Since, the Momentum of the Body will remains the constant throughout.
∴ mv + 2m × 2v = m × v₁ + 2m × v₂
∴ v + 4v = v₁ + v₂
⇒ 5v = v₁ + v₂
25v² = v₁² + v₂² + 2v₁v₂ ---eq(ii)
Dividing eq(i) by (ii).
250/45 = (v₁² + v₂² + 2v₁v₂)/(0.5v₁² + v₂²)
50(0.5v₁² + v₂²) = 9(v₁² + v₂² + 2v₁v₂)
25v₁² + 50v₂² = 9v₁² + 9v₂² + 8v₁v₂
16v₁² + 41v₂² - 8v₁v₂ = 0
Also, Relative velocity before collision equals the Relative Velocity after the collision.
2v - v = v₁ - v₂
v = v₁ - v₂
That's all. Whatever can be find from the given condition, I have finded. Questionm is incomplete. You can find from these condition by simple puting the values, if it is asked.
Hope it helps.