Ockfalls can cause major damage to roads and infrastructure. To design mitigation bridges and barriers, engineers use the coefficient of restitution to model the behavior of the rocks. Rock A falls a distance of 20 m before striking an incline with a slope of α = 40°. Knowing that the coefficient of restitution between rock A and the incline is 0.2, determine the velocity of the rock after the impact

13.08 m/s

Explanation:

Coefficient of restitution = (speed of seperation)/(speed of approach)

= (v₁ - v₂)/(u₂ - u₁)

where v₁ = final velocity of the rock = ?

v₂ = final velocity of the inclined plane = 0 m/s

u₂ = initial velocity of the inclined plane = 0 m/s

u₁ = initial velocity of impact of the rock = ?

We can calculate the initial velocity of impact of the rock from the laws of conservation of energy.

Potential energy at the highest height for the rock = kinetic energy at impact

mgh = (1/2)mu₁²

u₁ = √(2gh)

u₁ = √(2×9.8×20) = 19.8 m/s

Using the convention of toward velocity to be negative and upward velocity to be positive.

u₁ = - 19.8 m/s

0.2 = (v₁ - 0)/(0 - (-19.8)]

v₁ = 0.2 × 19.8 = 3.96 m/s

This is the velocity after impact if it were a level surface, calculating for the inclined plane,

The components of the velocity of the rock after impact will be perpendicular to the surface of the inclined plane and parallel to the inclined plane

The component perpendicular to the plane = v₁cos 40° = 3.96 × cos 40° = 3.03 m/s

The component parallel to the plane = u₁ sin 40° = 19.8 × sin 40° = 12.73 m/s

v = √[12.73² + 3.03²] = 13.08 m/s

Hope this Helps!!!

hi

explanation: