Two particles, of mass m and mass nm, undergo a head-on, fully elastic collision. (Here n is just a unitless number.) Before colliding the particles approach one another with equal speed v. Snapshots of the particles before, during, and after the collision are shown above. After the collision the first particle moves away with speed 0.750v in the exact opposite direction, and the velocity of the second particle is unknown. What is the value of n?

To determine the value of n, we can use the principles of conservation of momentum and kinetic energy. Let's break down the problem step by step:

1. Before the collision:
Both particles are moving towards each other with equal speed v. Since the masses are m and nm, the momenta of the particles are mv and nmv respectively before the collision.

2. During the collision:
During a head-on, fully elastic collision, the relative speed of approach is twice the original speed v. Therefore, the relative speed during the collision is 2v.
According to conservation of momentum, the total momentum after the collision should be equal to the total momentum before the collision. Since the first particle moves away with speed 0.750v in the exact opposite direction, its momentum is -0.750mv.
Therefore, the momentum of the second particle, which is moving in the positive direction, is (nmv - 0.750mv).

3. After the collision:
The first particle moves away with speed 0.750v, so its momentum is -0.750mv.
The unknown velocity of the second particle can be represented as v2.
Its momentum after the collision is mv2.

Now we can equate the total momentum before and after the collision:

(mv) + (nmv - 0.750mv) = (-0.750mv) + (mv2)

Simplifying the equation:

mv + nmv - 0.750mv = -0.750mv + mv2
(1 + n - 0.750)mv = (-0.750m + m)v2

Dividing both sides by mv:

1 + n - 0.750 = -0.750 + v2

Simplifying further:

1 + n - 0.750 = v2 - 0.750
n = v2 - 1

Therefore, the value of n is equal to v2 - 1.