Final answer:
To calculate the magnitude of the velocity of particle 1 after the collision, we use the conservation of momentum and the 1% decrease in kinetic energy. The final velocity is found to be approximately 9.95 m/s, which takes into account the energy loss.
Step-by-step explanation:
The student's question concerns a head-on collision between two particles of equal mass where one is initially moving and the other is stationary. The student is asked to calculate the magnitude of the velocity of the moving particle after the collision, given that there is a 1.0% loss in kinetic energy.
Let's assume that the two particles each have a mass m. The initial velocity of particle 1 is v1 = 10 m/s, while particle 2 is stationary, so v2 = 0 m/s. After the collision, let's denote the velocity of particle 1 as v1' and velocity of particle 2 as v2'. Because they have the same mass and the collision is head-on, by conservation of momentum:
m*v1 + m*v2 = m*v1' + m*v2'
Since v2 = 0 m/s and v1' must equal v2' due to the equal masses and one-dimensional collision, we simplify this to:
10 m/s = 2*v1'
We then solve for v1', which gives us v1' = 5 m/s. However, we are told that there is a 1% decrease in kinetic energy, which implies that the velocities are less than this result. We can calculate the initial kinetic energy (KE_initial) and the final kinetic energy (KE_final). The final kinetic energy will be 99% of the initial due to a 1% loss:
KE_initial = 1/2 * m * v1^2
KE_final = KE_initial * 0.99 = 1/2 * m * (v1')^2
We can now solve the equation for v1' and we find that:
v1' = sqrt(0.99) * v1
v1' = sqrt(0.99) * 10 m/s ≈ 9.95 m/s
Hence, the magnitude of the velocity of particle 1 after the collision is approximately 9.95 m/s.