Question

A body 'a' experiences perfectly elastic collision with a stationary body 'b'. If after collision the bodies fly apart in opposite directions with equal velocities ,the wavelength of 'a' and 'b' is

Answer 1

The wavelength of 'a' and 'b' is 3/1

Explanation:

Before collision, the velocity of the particle a = v₁

After collision, the velocity of the particles = v

On applying conservation of momentum, we get,

Initial momentum before collision = Final momentum after collision

ma v₁ + 0 = ma v + mb v

v₁ = (ma + mb)v/ma → (equation 1)

On applying conservation of kinetic energy, we get,

1/2 ma v₁² = 1/2 ma v² + 1/2 mb v²

On substituting v₁ in above equation, we get,

1/2 ma (ma + mb)²v²/ma² = 1/2 ma v² + 1/2 mb v²

ma² + mb² - 2ma mb = ma² + ma mb

mb² = 3ma mb

∴ ma/mb = 1/3

The wavelength of particle A is given by the formula:

λa = h/(ma v)

The wavelength of particle B is given by the formula:

λb = h/(mb v)

Now,

λa/λb = h/(ma v) × (mb v)/h

λa/λb = mb/ma = 3/1

∴ λa/λb = 3/1