In a collinear collision, a particle with an initial speed v0 strikes a stationary particle of the same mass. If the final total kinetic energy is 50% of the initial kinetic energy, what is the final speed of the first particle?

Let m be the mass of each particle. The initial kinetic energy is given by:

KE_initial = (1/2)mv0^2

Since the second particle is stationary, its initial kinetic energy is 0. After the collision, the total kinetic energy is 50% of the initial kinetic energy:

KE_final = 0.5 * KE_initial = 0.5 * (1/2)mv0^2 = (1/4)mv0^2

Let v1 and v2 be the final velocities of the first and second particles, respectively. By conservation of momentum:

mv0 = mv1 + mv2

v0 = v1 + v2

By conservation of kinetic energy (in this case, 50% is retained):

(1/4)mv0^2 = (1/2)mv1^2 + (1/2)mv2^2

v0^2 = 2(v1^2 + v2^2)

We have a system of two equations with two unknowns:

1. v0 = v1 + v2
2. v0^2 = 2(v1^2 + v2^2)

From equation (1), v2 = v0 - v1. Substitute this into equation (2):

v0^2 = 2(v1^2 + (v0 - v1)^2)

v0^2 = 2(v1^2 + v0^2 - 2v0v1 + v1^2)

v0^2 = 4v1^2 - 4v0v1 + 2v0^2

0 = 4v1^2 - 4v0v1 + v0^2

This is a quadratic equation in v1. We can solve it using the quadratic formula:

v1 = [4v0 ± √(16v0^2 - 16v0^2)] / 8

v1 = [4v0 ± 0] / 8

v1 = v0/2

Therefore, the final speed of the first particle is v0/2.