Question

A 0.15 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.72 m/s. It has a head-on collision with a 0.30 kg glider that is moving to the left with a speed of 0.36m/s. Suppose the collision is elastic. Find the magnitude of the final velocity (in m/s) of the 0.150kg glider. ( I got -.78 not sure where I went wrong)

Answer 1

Let us consider the right side direction as positive and the left side direction as negative.

Mass of the first glider = m₁ = 0.15 kg

Mass of the second glider = m₂ = 0.3 kg

Velocity of the first glider before the collision = V₁ = 0.72 m/s

Velocity of the second glider before the collision = V₂ = -0.36 m/s (Negative as it is moving to the left)

Velocity of the first glider after the collision = V₃

Velocity of the second glider after the collision = V₄

The collision is elastic.

Coefficient of restitution = e = 1

V₄ - V₃ = 1.08

V₄ = V₃ + 1.08

By conservation of linear momentum,

m₁V₁ + m₂V₂ = m₁V₃ + m₂V₄

(0.15)(0.72) + (0.3)(-0.36) = (0.15)V₃ + (0.3)(V₃ + 1.08)

0.108 - 0.108 = 0.15V₃ + 0.3V₃ + 0.324

0.45V₃ = -0.324

V₃ = -0.72 m/s

The negative sign indicates that the direction of velocity is to the left.

We are asked for the magnitude of the final velocity therefore it will be positive.

Magnitude of the final velocity of the 0.15 kg glider = 0.72 m/s