A particle of mass m1 experiences a perfectly elastic collision with a particle of mass m2 at rest. What fraction of the incident energy does the incident particle lose if a) it recoils at right angle to the original direction and b) the collision is head-on?

To determine the fraction of the incident energy that the incident particle loses in each scenario, we need to consider the conservation of momentum and kinetic energy.

a) In the first scenario, where the incident particle recoils at a right angle to the original direction, we can use the conservation of momentum to find the velocity of the incident particle after the collision.

Let's denote the final velocities of the particles as v1 and v2 (for the incident and stationary particles, respectively), in their respective directions. Since the collision is perfectly elastic, the total momentum before and after the collision must be conserved:

m1 * u1 + m2 * u2 = m1 * v1 + m2 * v2,

where u1 and u2 are the initial velocities of the particles.

Since the stationary particle is at rest, u2 = 0, and the equation simplifies to:

m1 * u1 = m1 * v1.

The incident particle loses energy if its final velocity (v1) is smaller than its initial velocity (u1). Since the incident particle recoils, its final velocity in the original direction is zero (v1 = 0). Thus, all of the incident energy is lost in this scenario.

b) In the second scenario, where the collision is head-on, we can again use the conservation of momentum to find the velocity of the incident particle after the collision.

Applying the conservation of momentum:

m1 * u1 + m2 * u2 = m1 * v1 + m2 * v2.

Since the particles are moving in opposite directions with opposite velocities (u2 = -u1), the equation becomes:

m1 * u1 - m2 * u1 = m1 * v1 - m2 * v2,
(u1 * (m1 - m2)) = (m1 * v1 - m2 * v2).

To determine the fraction of energy lost, we need to compare the initial kinetic energy (1/2 * m1 * u1^2) to the final kinetic energy (1/2 * m1 * v1^2) after the collision.

Squaring both sides of the equation and dividing by 2, we obtain:

(1/2 * (m1 - m2) * u1^2) = (1/2 * m1 * v1^2 - 1/2 * m2 * v2^2).

Since the stationary particle is at rest, v2 = 0, and the equation becomes:

(1/2 * (m1 - m2) * u1^2) = (1/2 * m1 * v1^2).

The fraction of the incident energy lost can be found by comparing the initial and final kinetic energies:

Fraction of energy lost = 1 - (final kinetic energy / initial kinetic energy)
= 1 - [(1/2 * m1 * v1^2) / (1/2 * m1 * u1^2)]
= 1 - (v1^2 / u1^2).

Please note that to find the fraction of energy lost in this scenario, we need the specific values for the masses and initial velocities of the particles.